Your FREE AP Physics 1 Practice Test 2026 – 230+ Q&A

AP Physics 1 Practice Questions

1. A car accelerates uniformly from $12\ \text{m/s}$ to $30\ \text{m/s}$ in $6.0\ \text{s}$. How far does it travel during this time?
   • $126\ \text{m}$
   • $180\ \text{m}$
   • $72\ \text{m}$
   • $252\ \text{m}$

   Correct answer: $126\ \text{m}$

   The car travels $126\ \text{m}$. For uniform acceleration the distance equals the average of the initial and final velocities multiplied by the time, so the average velocity of $21\ \text{m/s}$ times $6.0\ \text{s}$ gives $21\ \text{m/s} \times 6.0\ \text{s} = 126\ \text{m}$. Using only the final velocity of $30\ \text{m/s}$ times $6.0\ \text{s}$ would wrongly give $180\ \text{m}$, and using only the initial velocity would underestimate the distance.

2. A ball is thrown straight up and returns to the thrower's hand. Ignoring air resistance, how does the speed of the ball when it returns to the launch height compare with its launch speed?
   • The return speed is greater than the launch speed
   • The return speed is half the launch speed
   • The return speed is zero
   • The return speed equals the launch speed

   Correct answer: The return speed equals the launch speed

   The return speed equals the launch speed. Gravity removes speed on the way up at the same rate it adds speed on the way down, so by symmetry the ball arrives back at the launch height moving just as fast as when it left, only in the opposite direction. The speed is not zero at that height; zero speed occurs only momentarily at the very top of the path.

3. An object moving in a straight line has a velocity versus time graph that is a straight line rising from $0\ \text{m/s}$ to $12\ \text{m/s}$ over $4.0\ \text{s}$. What is the magnitude of the object's acceleration?
   • $48\ \text{m/s}^2$
   • $0.33\ \text{m/s}^2$
   • $12\ \text{m/s}^2$
   • $3.0\ \text{m/s}^2$

   Correct answer: $3.0\ \text{m/s}^2$

   The acceleration is $3.0\ \text{m/s}^2$. On a velocity versus time graph the acceleration is the slope of the line, found as the rise of $12\ \text{m/s}$ divided by the run of $4.0\ \text{s}$, which gives $\frac{12\ \text{m/s}}{4.0\ \text{s}} = 3.0\ \text{m/s}^2$. Multiplying the two values would incorrectly yield $48\ \text{m/s}^2$, and dividing time by velocity would give the reciprocal of the correct answer.

4. Which of the following quantities is correctly described as a vector?
   • The distance a hiker walks along a winding trail
   • The time elapsed during a race
   • The displacement of a car from its garage
   • The speed shown on a speedometer

   Correct answer: The displacement of a car from its garage

   Displacement is the vector quantity in this list. Displacement specifies both how far and in which direction an object's position has changed, giving it the magnitude and direction required of a vector. Distance, elapsed time, and speedometer speed each carry only a magnitude with no direction, so they are scalars.

5. A walkway moves at $1.5\ \text{m/s}$ relative to the ground. A person walks on the walkway at $1.0\ \text{m/s}$ relative to the walkway but in the direction opposite to the walkway's motion. What is the person's velocity relative to the ground?
   • $0.5\ \text{m/s}$ in the walkway's direction
   • $2.5\ \text{m/s}$ in the walkway's direction
   • $1.0\ \text{m/s}$ in the walkway's direction
   • $0.5\ \text{m/s}$ opposite the walkway's direction

   Correct answer: $0.5\ \text{m/s}$ in the walkway's direction

   The person moves at $0.5\ \text{m/s}$ in the walkway's direction relative to the ground. Because the person walks opposite to the walkway, the person's $1.0\ \text{m/s}$ subtracts from the walkway's $1.5\ \text{m/s}$, leaving a net $1.5\ \text{m/s} - 1.0\ \text{m/s} = 0.5\ \text{m/s}$ in the walkway's direction. Adding the two speeds to get $2.5\ \text{m/s}$ would be correct only if the person walked the same way the walkway moves.

6. A particle moving along a straight line has a velocity versus time graph consisting of a line that slopes downward and crosses below the time axis. While the velocity value is negative, how is the particle moving?
   • It is moving in the negative direction
   • It is at rest
   • It is moving in the positive direction
   • It is speeding up in the positive direction

   Correct answer: It is moving in the negative direction

   While its velocity is negative, the particle is moving in the negative direction. The sign of the velocity value indicates the direction of motion, so a negative value means the object travels the opposite way from the chosen positive direction. A negative velocity does not mean the object is at rest or slowing; those depend on the velocity reaching zero or on the relationship between velocity and acceleration signs.

7. A stone is launched horizontally from a cliff at $15\ \text{m/s}$. Ignoring air resistance, what is its horizontal velocity $2.0\ \text{s}$ after launch?
   • $35\ \text{m/s}$
   • $15\ \text{m/s}$
   • $30\ \text{m/s}$
   • $0\ \text{m/s}$

   Correct answer: $15\ \text{m/s}$

   The horizontal velocity remains $15\ \text{m/s}$. With no horizontal force acting on the stone, its horizontal velocity stays unchanged for the entire flight regardless of how much time passes. The value $35\ \text{m/s}$ mistakenly adds the vertical speed gained from gravity, but horizontal and vertical motions are independent and should not be combined this way for the horizontal component.

8. A sprinter runs the first $100\ \text{m}$ of a straight track in $12\ \text{s}$ and then walks the next $20\ \text{m}$ back toward the start in $8.0\ \text{s}$. What is the magnitude of the sprinter's average velocity over the entire $20\ \text{s}$?
   • $6.0\ \text{m/s}$
   • $5.0\ \text{m/s}$
   • $1.0\ \text{m/s}$
   • $4.0\ \text{m/s}$

   Correct answer: $4.0\ \text{m/s}$

   The average velocity magnitude is $4.0\ \text{m/s}$. Average velocity uses net displacement, and moving $100\ \text{m}$ forward then $20\ \text{m}$ back leaves a net displacement of $100\ \text{m} - 20\ \text{m} = 80\ \text{m}$, which divided by the total $20\ \text{s}$ gives $\frac{80\ \text{m}}{20\ \text{s}} = 4.0\ \text{m/s}$. Dividing the total path length of $120\ \text{m}$ by $20\ \text{s}$ would instead give the average speed of $6.0\ \text{m/s}$, a different quantity.

9. A rock is dropped from rest and falls freely. Taking $g = 10\ \text{m/s}^2$, how far does it fall during the first $2.0\ \text{s}$?
   • $40\ \text{m}$
   • $10\ \text{m}$
   • $20\ \text{m}$
   • $5.0\ \text{m}$

   Correct answer: $20\ \text{m}$

   The rock falls $20\ \text{m}$. Starting from rest, the distance fallen equals one-half times the acceleration times the time squared, so $\frac{1}{2} \times 10\ \text{m/s}^2 \times (2.0\ \text{s})^2 = 20\ \text{m}$. Forgetting the factor of one-half would give $40\ \text{m}$, and forgetting to square the time would give only $10\ \text{m}$.

10. Two projectiles are launched from the same height at the same instant with no air resistance: one is dropped straight down from rest and the other is thrown horizontally. Which reaches the ground first?
    • The dropped one, because it has no horizontal motion
    • The thrown one, because it covers more ground
    • They reach the ground at the same time
    • The thrown one, because horizontal speed adds to the fall

    Correct answer: They reach the ground at the same time

    The two projectiles reach the ground at the same time. The time to fall depends only on the vertical motion, and both objects start with zero vertical velocity and fall the same height under the same downward acceleration. The horizontal velocity of the thrown object has no effect on its vertical fall, illustrating the independence of horizontal and vertical motion.

11. An airplane has an airspeed of $200\ \text{m/s}$ pointed due east, and a wind blows toward the east at $30\ \text{m/s}$ relative to the ground. What is the airplane's speed relative to the ground?
    • $170\ \text{m/s}$
    • $230\ \text{m/s}$
    • $200\ \text{m/s}$
    • $30\ \text{m/s}$

    Correct answer: $230\ \text{m/s}$

    The airplane's ground speed is $230\ \text{m/s}$. Because the airplane's velocity through the air and the wind's velocity both point east, the two velocities add to give $200\ \text{m/s} + 30\ \text{m/s} = 230\ \text{m/s}$ relative to the ground. Subtracting to get $170\ \text{m/s}$ would apply only if the wind blew toward the west, opposing the plane's motion.

12. An object moves with constant velocity along a straight line. Which statement correctly describes its acceleration and the appearance of its position versus time graph?
    • Acceleration is nonzero and the position graph is curved
    • Acceleration is zero and the position graph is a straight sloped line
    • Acceleration is zero and the position graph is horizontal
    • Acceleration is constant and the position graph is a parabola

    Correct answer: Acceleration is zero and the position graph is a straight sloped line

    Constant velocity means zero acceleration and a straight sloped position versus time graph. Because the velocity does not change, there is no acceleration, and equal changes in position occur in equal time intervals, producing a line with constant slope. A horizontal position graph would represent an object at rest, while a curved or parabolic graph would indicate changing velocity and nonzero acceleration.

13. A net force of $24\ \text{N}$ acts on a $6.0\ \text{kg}$ cart on a frictionless surface. What is the magnitude of the cart's acceleration?
    • $144\ \text{m/s}^2$
    • $0.25\ \text{m/s}^2$
    • $4.0\ \text{m/s}^2$
    • $18\ \text{m/s}^2$

    Correct answer: $4.0\ \text{m/s}^2$

    The acceleration is $4.0\ \text{m/s}^2$. Newton's second law states that acceleration equals the net force divided by the mass, so $a = \frac{24\ \text{N}}{6.0\ \text{kg}} = 4.0\ \text{m/s}^2$. Multiplying the force and mass would wrongly give $144$, and dividing mass by force would give the reciprocal value of $0.25$.

14. A book rests on a table. According to Newton's third law, what is the reaction force to the gravitational pull of Earth on the book?
    • The normal force of the table pushing up on the book
    • The weight of the table pressing on the floor
    • The friction force between the book and the table
    • The gravitational pull of the book on Earth

    Correct answer: The gravitational pull of the book on Earth

    The reaction to Earth's gravitational pull on the book is the book's gravitational pull on Earth. Newton's third law pairs act on different objects and are the same type of force, so the partner of Earth pulling the book down is the book pulling Earth up with equal magnitude. The normal force from the table is a separate contact force that happens to balance the weight but is not its third-law partner.

15. A $2.0\ \text{kg}$ block is pulled across a horizontal floor at constant velocity by a horizontal rope. The coefficient of kinetic friction is $0.30$ and $g = 10\ \text{m/s}^2$. What is the tension in the rope?
    • $20\ \text{N}$
    • $6.0\ \text{N}$
    • $0.60\ \text{N}$
    • $60\ \text{N}$

    Correct answer: $6.0\ \text{N}$

    The tension is $6.0\ \text{N}$. At constant velocity the net force is zero, so the rope tension must equal the kinetic friction force, which is the coefficient of $0.30$ times the normal force; on a horizontal floor the normal force equals the weight of $2.0\ \text{kg} \times 10\ \text{m/s}^2 = 20\ \text{N}$, so the friction force is $0.30 \times 20\ \text{N} = 6.0\ \text{N}$. The value $20\ \text{N}$ is just the weight, not the friction force.

16. A ball is tied to a string and whirled in a horizontal circle at constant speed. What is the direction of the net force acting on the ball?
    • Tangent to the circle in the direction of motion
    • Outward, away from the center of the circle
    • Toward the center of the circle
    • Zero, since the speed is constant

    Correct answer: Toward the center of the circle

    The net force points toward the center of the circle. An object moving in a circle is constantly changing direction, which is a centripetal acceleration directed inward, so by Newton's second law the net force must also point inward toward the center. There is no real outward force on the ball; the often-imagined outward push is not an actual force acting on the object.

17. A spring with spring constant $250\ \text{N/m}$ is stretched $0.040\ \text{m}$ from its natural length. According to Hooke's law, what is the magnitude of the restoring force exerted by the spring?
    • $10\ \text{N}$
    • $6250\ \text{N}$
    • $0.16\ \text{N}$
    • $290\ \text{N}$

    Correct answer: $10\ \text{N}$

    The restoring force is $10\ \text{N}$. Hooke's law states that the spring force magnitude equals the spring constant times the displacement from the natural length, so $250\ \text{N/m} \times 0.040\ \text{m} = 10\ \text{N}$. Dividing the spring constant by the displacement would wrongly give $6250$, and the spring constant and displacement are multiplied, not added.

18. An elevator of total mass $800\ \text{kg}$ accelerates upward at $2.0\ \text{m/s}^2$. Taking $g = 10\ \text{m/s}^2$, what is the tension in the supporting cable?
    • $8000\ \text{N}$
    • $6400\ \text{N}$
    • $9600\ \text{N}$
    • $1600\ \text{N}$

    Correct answer: $9600\ \text{N}$

    The cable tension is $9600\ \text{N}$. Newton's second law applied to the elevator gives tension minus weight equal to mass times acceleration, so the tension is the weight of $800 \times 10$ plus the mass times the upward acceleration of $800 \times 2.0$, which is $8000 + 1600 = 9600\ \text{N}$. The value $8000\ \text{N}$ would be correct only if the elevator were not accelerating.

19. Two carts of equal mass, one moving fast and one moving slowly, collide head-on. During the collision, how does the force the fast cart exerts on the slow cart compare with the force the slow cart exerts on the fast cart?
    • The fast cart exerts a larger force on the slow cart
    • The slow cart exerts a larger force on the fast cart
    • Neither cart exerts a force until they stop moving
    • The forces are equal in magnitude and opposite in direction

    Correct answer: The forces are equal in magnitude and opposite in direction

    The two forces are equal in magnitude and opposite in direction. Newton's third law guarantees that the force each object exerts on the other in a collision is always equal and opposite, regardless of the objects' speeds or masses. The fast cart may experience a larger change in motion if it has different mass, but the mutual contact forces themselves are always paired and equal.

20. A $1500\ \text{kg}$ car rounds a flat circular curve of radius $50\ \text{m}$ at a constant speed of $10\ \text{m/s}$. What is the magnitude of the net centripetal force required to keep the car on the curve?
    • $300\ \text{N}$
    • $3000\ \text{N}$
    • $15000\ \text{N}$
    • $750\ \text{N}$

    Correct answer: $3000\ \text{N}$

    The required centripetal force is $3000\ \text{N}$. The centripetal force equals the mass times the speed squared divided by the radius, so $F = \frac{1500 \times 10^2}{50} = \frac{1500 \times 100}{50} = 3000\ \text{N}$. Forgetting to square the speed would give $300\ \text{N}$, far too small to keep the car turning.

21. The center of mass of a uniform meter stick is at its $50\ \text{cm}$ mark. If a small lump of clay is stuck onto the $90\ \text{cm}$ end, where does the center of mass of the stick-and-clay system move?
    • It stays at the $50\ \text{cm}$ mark
    • It moves toward the $90\ \text{cm}$ end
    • It moves toward the $10\ \text{cm}$ end
    • It moves off the stick entirely

    Correct answer: It moves toward the $90\ \text{cm}$ end

    The center of mass moves toward the $90\ \text{cm}$ end. The center of mass is the mass-weighted average position of a system, so adding extra mass at one end pulls the average position in that direction. The shift is toward the added clay at the $90\ \text{cm}$ mark, though it remains on the stick because the clay is only a small fraction of the total mass.

22. An astronaut measures that two planets exert equal gravitational forces on a standard test mass placed at their surfaces. Planet B has twice the radius of Planet A. By Newton's law of gravitation, how does the mass of Planet B compare with the mass of Planet A?
    • Planet B has twice the mass of Planet A
    • Planet B has the same mass as Planet A
    • Planet B has four times the mass of Planet A
    • Planet B has half the mass of Planet A

    Correct answer: Planet B has four times the mass of Planet A

    Planet B must have four times the mass of Planet A. Newton's law of gravitation makes the surface force proportional to the planet's mass divided by the square of its radius ($F \propto \frac{M}{r^2}$), so doubling the radius reduces the force by a factor of four unless the mass also increases by a factor of four to compensate. Only a fourfold mass increase keeps the surface gravitational force equal to Planet A's.

23. A $5.0\ \text{kg}$ object hangs at rest from a single vertical rope. Taking $g = 10\ \text{m/s}^2$, what is the tension in the rope?
    • $50\ \text{N}$
    • $5.0\ \text{N}$
    • $0.50\ \text{N}$
    • $100\ \text{N}$

    Correct answer: $50\ \text{N}$

    The tension is $50\ \text{N}$. Because the object hangs at rest the net force is zero, so the upward tension must exactly balance the downward weight, which is the mass of $5.0\ \text{kg}$ times $g$ of $10\ \text{m/s}^2$, giving $5.0\ \text{kg} \times 10\ \text{m/s}^2 = 50\ \text{N}$. The mass alone of $5.0$ is not a force, and doubling to $100\ \text{N}$ would imply an upward acceleration rather than equilibrium.

24. A $10\ \text{kg}$ crate sits on a horizontal floor. The coefficient of static friction is $0.40$ and $g = 10\ \text{m/s}^2$. What is the maximum horizontal force that can be applied before the crate begins to slide?
    • $100\ \text{N}$
    • $40\ \text{N}$
    • $4.0\ \text{N}$
    • $250\ \text{N}$

    Correct answer: $40\ \text{N}$

    The maximum force before sliding is $40\ \text{N}$. The largest static friction force equals the coefficient of static friction times the normal force, and on a horizontal floor the normal force equals the weight of $10 \times 10 = 100\ \text{N}$, so the maximum static friction is $0.40 \times 100\ \text{N} = 40\ \text{N}$. Any applied force above this threshold overcomes static friction and the crate starts to move.

25. A $3.0\ \text{kg}$ block on a frictionless surface is pushed by two horizontal forces along the same line: $18\ \text{N}$ to the right and $6.0\ \text{N}$ to the left. What is the magnitude and direction of the block's acceleration?
    • $8.0\ \text{m/s}^2$ to the right
    • $4.0\ \text{m/s}^2$ to the right
    • $4.0\ \text{m/s}^2$ to the left
    • $24\ \text{m/s}^2$ to the right

    Correct answer: $4.0\ \text{m/s}^2$ to the right

    The acceleration is $4.0\ \text{m/s}^2$ to the right. The net force is found by combining the opposing forces, giving $18\ \text{N} - 6.0\ \text{N} = 12\ \text{N}$ to the right, and dividing this net force by the $3.0\ \text{kg}$ mass gives $\frac{12\ \text{N}}{3.0\ \text{kg}} = 4.0\ \text{m/s}^2$ in the direction of the larger force. Adding the forces to get $24\ \text{N}$ would ignore that they point in opposite directions.

26. A horse pulls a cart forward and the cart accelerates. A student argues the cart should not move because the cart pulls back on the horse with an equal and opposite force. Why does the cart actually accelerate?
    • The horse pulls harder on the cart than the cart pulls on the horse
    • Newton's third law does not apply when objects are moving
    • The cart's backward pull on the horse is much smaller than claimed
    • The action-reaction forces act on different objects, so they do not cancel on the cart

    Correct answer: The action-reaction forces act on different objects, so they do not cancel on the cart

    The cart accelerates because the action-reaction forces act on different objects and therefore cannot cancel each other on a single object. The horse's pull acts on the cart while the cart's reaction acts on the horse, so when analyzing the cart only the forces on the cart matter, and the forward pull combined with friction from the ground produces a nonzero net force. Third-law pairs are always equal but never cancel because they are exerted on two separate objects.

27. A small object of mass $m$ is at the top of a vertical circular loop, moving fast enough to maintain contact with the track. Which forces act on the object at the very top of the loop?
    • Only the normal force, pointing upward
    • Gravity and the normal force, both pointing downward
    • Gravity downward and the normal force upward
    • An outward centrifugal force balancing gravity

    Correct answer: Gravity and the normal force, both pointing downward

    At the top of the loop both gravity and the normal force point downward. Gravity always pulls the object toward Earth, and at the top of the loop the track is above the object, so the track pushes it inward and downward toward the center of the circle. Both forces therefore add together to supply the centripetal force, and no real outward force acts on the object.

28. A $4.0\ \text{kg}$ object is acted on by a single force that varies in time. At one instant the acceleration is measured to be $7.5\ \text{m/s}^2$. What is the magnitude of the net force on the object at that instant?
    • $0.53\ \text{N}$
    • $11.5\ \text{N}$
    • $30\ \text{N}$
    • $1.875\ \text{N}$

    Correct answer: $30\ \text{N}$

    The net force is $30\ \text{N}$. Newton's second law states that the net force equals mass times acceleration, so $F = 4.0\ \text{kg} \times 7.5\ \text{m/s}^2 = 30\ \text{N}$. Dividing the mass by the acceleration or the acceleration by the mass would produce the much smaller incorrect values, since force is the product of the two.

29. Two identical springs, each with spring constant $k$, are connected side by side in parallel and a single object is hung from both. Compared with hanging the object from one spring alone, how does the effective stiffness of the parallel pair behave?
    • The pair is twice as stiff, with effective constant $2k$
    • The pair is half as stiff, with effective constant $k/2$
    • The pair has the same stiffness $k$ as a single spring
    • The pair has zero effective stiffness

    Correct answer: The pair is twice as stiff, with effective constant $2k$

    Two identical springs in parallel are twice as stiff, with an effective constant of $2k$. When springs share a load side by side, each stretches by the same amount and each contributes its own restoring force, so the total force for a given stretch is doubled, which by Hooke's law means the effective spring constant doubles. The parallel arrangement therefore resists stretching more strongly than a single spring.

30. Two objects are separated by a distance $r$ and attract each other gravitationally with a force $F$. If the distance between them is tripled while their masses stay the same, what is the new gravitational force?
    • $F$ divided by $9$
    • $F$ divided by $3$
    • $F$ multiplied by $3$
    • $F$ multiplied by $9$

    Correct answer: $F$ divided by $9$

    The new gravitational force is $\frac{F}{9}$. Newton's law of gravitation makes the force inversely proportional to the square of the separation distance, so tripling the distance reduces the force by a factor of $3^2 = 9$. Simply dividing by three would be the result if the force fell off linearly, but the inverse-square relationship makes the drop much steeper.

31. A skydiver of constant mass falls and eventually reaches terminal velocity, descending at a steady speed. At terminal velocity, what is true about the forces acting on the skydiver?
    • The upward air resistance is greater than the downward weight
    • The downward weight is greater, so the skydiver keeps speeding up
    • Only weight acts, since air resistance disappears at high speed
    • The upward air resistance equals the downward weight, giving zero net force

    Correct answer: The upward air resistance equals the downward weight, giving zero net force

    At terminal velocity the upward air resistance equals the downward weight, so the net force is zero. A steady constant speed means there is no acceleration, and by Newton's second law zero acceleration requires zero net force, which happens when the drag force has grown to exactly balance gravity. If the forces were unbalanced the skydiver would still be accelerating rather than falling at a constant speed.

32. Which expression correctly gives the translational kinetic energy of an object of mass $m$ moving with speed $v$?
    • $\frac{1}{2} m v^2$
    • $m v$
    • $m g v$
    • $2 m v^2$

    Correct answer: $\frac{1}{2} m v^2$

    The translational kinetic energy is $\frac{1}{2} m v^2$. Kinetic energy measures the energy an object has because of its motion, and it grows with the square of the speed while increasing in direct proportion to the mass. The quantity $m v$ is momentum rather than energy, so it does not describe kinetic energy.

33. A $1500\ \text{kg}$ car traveling at $20\ \text{m/s}$ slows to $10\ \text{m/s}$. By how much does its translational kinetic energy decrease?
    • $75000\ \text{J}$
    • $150000\ \text{J}$
    • $225000\ \text{J}$
    • $15000\ \text{J}$

    Correct answer: $225000\ \text{J}$

    The kinetic energy decreases by $225000\ \text{J}$. The initial kinetic energy is $\frac{1}{2} \times 1500 \times 20^2 = 300000\ \text{J}$, and the final is $\frac{1}{2} \times 1500 \times 10^2 = 75000\ \text{J}$, so the difference is $300000\ \text{J} - 75000\ \text{J} = 225000\ \text{J}$. Halving the speed does not halve the kinetic energy because energy depends on the square of the speed, so cutting the speed in half drops the kinetic energy to one quarter of its original value.

34. The net work done on an object over some interval is found to be $0\ \text{J}$. According to the work-energy theorem, what must be true of the object's speed over that interval?
    • Its speed must be zero throughout the interval
    • Its speed must increase steadily
    • Its speed must decrease to half its initial value
    • Its speed at the end equals its speed at the start

    Correct answer: Its speed at the end equals its speed at the start

    The object's final speed equals its initial speed. The work-energy theorem states that the net work equals the change in kinetic energy, so zero net work means zero change in kinetic energy and therefore an unchanged speed. The object need not be at rest; it can be moving the whole time as long as its speed does not change overall.

35. A constant horizontal force of $40\ \text{N}$ is applied to a $5.0\ \text{kg}$ box, pushing it $6.0\ \text{m}$ across a frictionless floor. If the box started from rest, what is its kinetic energy after the $6.0\ \text{m}$?
    • $200\ \text{J}$
    • $240\ \text{J}$
    • $120\ \text{J}$
    • $48\ \text{J}$

    Correct answer: $240\ \text{J}$

    The box has $240\ \text{J}$ of kinetic energy. With a frictionless floor the net work equals the applied force times the distance, so $40\ \text{N} \times 6.0\ \text{m} = 240\ \text{J}$ of work, and by the work-energy theorem this entire amount becomes kinetic energy because the box started from rest. The mass would only be needed to find the final speed, not the kinetic energy itself.

36. A $2.0\ \text{kg}$ object is lifted straight up at constant speed to a height of $3.0\ \text{m}$. Taking $g = 10\ \text{m/s}^2$, how much gravitational potential energy does it gain relative to its starting point?
    • $6.0\ \text{J}$
    • $60\ \text{J}$
    • $20\ \text{J}$
    • $30\ \text{J}$

    Correct answer: $60\ \text{J}$

    The object gains $60\ \text{J}$ of gravitational potential energy. Gravitational potential energy near Earth's surface is mass times $g$ times height, so $2.0\ \text{kg} \times 10\ \text{m/s}^2 \times 3.0\ \text{m} = 60\ \text{J}$. Multiplying only mass by height while forgetting $g$ would give $6.0\ \text{J}$, which leaves out the gravitational field strength that the energy depends on.

37. Two identical crates are raised to the top of a building. One is lifted straight up by a crane and the other is pushed up a long frictionless ramp to the same final height. Ignoring friction, how do their gains in gravitational potential energy compare?
    • The crane-lifted crate gains more because the path is shorter
    • The ramp-pushed crate gains more because the path is longer
    • Neither gains potential energy because they end up at rest
    • Their gains in gravitational potential energy are equal

    Correct answer: Their gains in gravitational potential energy are equal

    The two crates gain equal gravitational potential energy. Gravitational potential energy depends only on the change in vertical height and the mass, not on the path taken to get there, so reaching the same final height yields the same energy gain. The ramp requires a smaller force over a longer distance, but the product that determines potential energy comes out the same as the direct lift.

38. A spring with spring constant $800\ \text{N/m}$ is compressed $0.10\ \text{m}$ from its natural length. How much elastic potential energy is stored in the spring?
    • $80\ \text{J}$
    • $4.0\ \text{J}$
    • $8.0\ \text{J}$
    • $40\ \text{J}$

    Correct answer: $4.0\ \text{J}$

    The spring stores $4.0\ \text{J}$ of elastic potential energy. Elastic potential energy equals one-half times the spring constant times the square of the displacement, so $\frac{1}{2} \times 800 \times (0.10)^2 = \frac{1}{2} \times 800 \times 0.01 = 4.0\ \text{J}$. Forgetting to square the small displacement would wrongly give $40\ \text{J}$, greatly overstating the stored energy.

39. A spring is stretched from its natural length by a distance $x$, storing a certain amount of elastic potential energy. If the same spring is instead stretched by $3x$, how does the stored elastic potential energy change?
    • It triples
    • It stays the same
    • It becomes nine times as large
    • It doubles

    Correct answer: It becomes nine times as large

    The stored elastic potential energy becomes nine times as large. Elastic potential energy is proportional to the square of the displacement from the natural length, so tripling the stretch multiplies the energy by $3^2 = 9$. A simple tripling would occur only if the energy were directly proportional to the stretch, but the squared dependence makes the increase much steeper.

40. A ball is dropped from rest from a height of $1.8\ \text{m}$. Ignoring air resistance and taking $g = 10\ \text{m/s}^2$, what is its speed just before it hits the ground?
    • $18\ \text{m/s}$
    • $6.0\ \text{m/s}$
    • $36\ \text{m/s}$
    • $3.6\ \text{m/s}$

    Correct answer: $6.0\ \text{m/s}$

    The ball is moving at $6.0\ \text{m/s}$ just before impact. By conservation of energy the gravitational potential energy $mgh$ converts entirely into kinetic energy $\frac{1}{2}mv^2$, and the mass cancels, leaving $v = \sqrt{2gh}$. That is $\sqrt{2 \times 10 \times 1.8} = \sqrt{36} = 6.0\ \text{m/s}$.

41. A roller coaster car starts from rest at the top of a frictionless hill of height $45\ \text{m}$. Taking $g = 10\ \text{m/s}^2$, what is its speed at the bottom of the hill?
    • $30\ \text{m/s}$
    • $45\ \text{m/s}$
    • $21\ \text{m/s}$
    • $90\ \text{m/s}$

    Correct answer: $30\ \text{m/s}$

    The car reaches $30\ \text{m/s}$ at the bottom. On a frictionless track all the gravitational potential energy at the top converts into kinetic energy at the bottom, so $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 45} = \sqrt{900} = 30\ \text{m/s}$. The shape of the hill does not matter because only the change in height determines the final speed.

42. A $0.50\ \text{kg}$ cart moving at $4.0\ \text{m/s}$ on a frictionless track runs into and compresses a spring of spring constant $200\ \text{N/m}$, momentarily coming to rest. What is the maximum compression of the spring?
    • $0.20\ \text{m}$
    • $0.40\ \text{m}$
    • $0.10\ \text{m}$
    • $0.04\ \text{m}$

    Correct answer: $0.20\ \text{m}$

    The spring compresses a maximum of $0.20\ \text{m}$. At maximum compression the cart is momentarily at rest, so all of its kinetic energy $\frac{1}{2} \times 0.50 \times 4.0^2 = 4.0\ \text{J}$ has converted into elastic potential energy $\frac{1}{2} \times 200 \times x^2$. Setting $4.0 = 100 x^2$ gives $x^2 = 0.04$ and $x = 0.20\ \text{m}$.

43. A motor lifts a $50\ \text{kg}$ load straight up at a steady speed of $2.0\ \text{m/s}$. Taking $g = 10\ \text{m/s}^2$, what is the power output of the motor?
    • $100\ \text{W}$
    • $250\ \text{W}$
    • $500\ \text{W}$
    • $1000\ \text{W}$

    Correct answer: $1000\ \text{W}$

    The motor delivers $1000\ \text{W}$ of power. Power is the rate of doing work, and for a constant-velocity lift it equals $P = Fv$; the force needed to raise the load at steady speed equals its weight $50 \times 10 = 500\ \text{N}$, and multiplying by the $2.0\ \text{m/s}$ speed gives $1000\ \text{W}$. Using only the mass and speed without including $g$ would leave out the gravitational force the motor must overcome.

44. Which statement best describes what the physical quantity power measures?
    • The total energy an object possesses at an instant
    • The force needed to move an object a fixed distance
    • The rate at which energy is transferred or work is done
    • The product of force and the time it acts

    Correct answer: The rate at which energy is transferred or work is done

    Power is the rate at which energy is transferred or work is done. It tells how quickly energy is delivered or used, defined as work divided by the time taken, with units of watts equal to joules per second $\left(\text{W} = \text{J/s}\right)$. Total energy at an instant and force over a distance are different quantities, so they do not capture the time-rate idea that defines power.

45. A horizontal force of $30\ \text{N}$ is used to push a crate $5.0\ \text{m}$ across a floor, and friction does negative work on the crate. If the crate moves at constant speed, how much net work is done on the crate over the $5.0\ \text{m}$?
    • $150\ \text{J}$
    • $0\ \text{J}$
    • $75\ \text{J}$
    • $30\ \text{J}$

    Correct answer: $0\ \text{J}$

    The net work on the crate is $0\ \text{J}$. Because the crate moves at constant speed its kinetic energy does not change, and by the work-energy theorem the net work must therefore be zero. The $150\ \text{J}$ of positive work done by the push ($30\ \text{N} \times 5.0\ \text{m}$) is exactly canceled by an equal amount of negative work done by friction, leaving no net work.

46. A pendulum bob swings back and forth without friction. As the bob moves from its highest point down to the lowest point of the swing, what energy transformation occurs?
    • Kinetic energy converts into gravitational potential energy
    • Elastic potential energy converts into kinetic energy
    • Total mechanical energy increases steadily during the descent
    • Gravitational potential energy converts into kinetic energy

    Correct answer: Gravitational potential energy converts into kinetic energy

    Gravitational potential energy converts into kinetic energy as the bob descends. At the highest point the bob is momentarily at rest with maximum gravitational potential energy, and as it falls toward the lowest point that stored energy turns into motion, giving the bob its greatest speed at the bottom. With no friction the total mechanical energy stays constant rather than increasing during the swing.

47. A $1200\ \text{kg}$ car and a $600\ \text{kg}$ car are both raised to a height of $4.0\ \text{m}$. Taking $g = 10\ \text{m/s}^2$, how does the gravitational potential energy gained by the heavier car compare with that of the lighter car?
    • The heavier car gains the same potential energy
    • The heavier car gains half the potential energy
    • The heavier car gains twice the potential energy
    • The heavier car gains four times the potential energy

    Correct answer: The heavier car gains twice the potential energy

    The heavier car gains twice the gravitational potential energy. Gravitational potential energy $U = mgh$ is directly proportional to mass when the height and gravitational field are the same, so doubling the mass from $600\ \text{kg}$ to $1200\ \text{kg}$ doubles the energy gained. The heavier car gains $48000\ \text{J}$ while the lighter car gains $24000\ \text{J}$, confirming the factor of two.

48. Two motors each lift identical $20\ \text{kg}$ boxes to the top of a shelf $3.0\ \text{m}$ high. Motor X does the job in $5.0\ \text{s}$ and Motor Y does it in $10\ \text{s}$. Taking $g = 10\ \text{m/s}^2$, how do the work done and the power compare for the two motors?
    • Motor X does more work but both have equal power
    • Both do equal work, and Motor X has greater power
    • Both do equal work and have equal power
    • Motor Y does more work and has greater power

    Correct answer: Both do equal work, and Motor X has greater power

    Both motors do the same work, but Motor X has greater power. The work to lift each box equals $mgh = 20 \times 10 \times 3.0 = 600\ \text{J}$, the same for both since the boxes and heights match. Power is work divided by time, so the faster Motor X delivers $600\ \text{J}$ in $5.0\ \text{s}$ for $120\ \text{W}$ while Motor Y delivers the same energy in $10\ \text{s}$ for only $60\ \text{W}$.

49. A block slides along a horizontal surface and gradually comes to a stop because of friction. From an energy standpoint, what happens to the block's kinetic energy?
    • It is converted into thermal energy through friction
    • It is stored as gravitational potential energy
    • It is destroyed and removed from the universe
    • It is converted into elastic potential energy in the block

    Correct answer: It is converted into thermal energy through friction

    The kinetic energy is converted into thermal energy through friction. Energy is conserved overall, so the block's energy of motion does not vanish but is transferred into heat as the rubbing surfaces warm up. The block does not rise, so no gravitational potential energy is gained, and energy is never truly destroyed, only changed in form.

50. Which expression correctly gives the linear momentum of an object of mass $m$ moving with velocity $v$?
    • The product of mass $m$ and velocity $v$
    • $\frac{1}{2} m v^2$
    • Mass $m$ divided by velocity $v$
    • Mass $m$ times the acceleration of the object

    Correct answer: The product of mass $m$ and velocity $v$

    Linear momentum is the product of mass and velocity, $p = mv$. Momentum describes how much motion an object carries and combines both how massive it is and how fast it moves, so it grows in direct proportion to each. The expression $\frac{1}{2} m v^2$ is the kinetic energy, which is a different quantity and is not what momentum measures.

51. A $0.45\ \text{kg}$ soccer ball moves to the right at $8.0\ \text{m/s}$. What is the magnitude of its linear momentum?
    • $14.4\ \text{kg}\cdot\text{m/s}$
    • $0.056\ \text{kg}\cdot\text{m/s}$
    • $3.6\ \text{kg}\cdot\text{m/s}$
    • $18\ \text{kg}\cdot\text{m/s}$

    Correct answer: $3.6\ \text{kg}\cdot\text{m/s}$

    The momentum magnitude is $3.6\ \text{kg}\cdot\text{m/s}$. Linear momentum equals mass times velocity, so $p = 0.45 \times 8.0 = 3.6\ \text{kg}\cdot\text{m/s}$. Using $\frac{1}{2} m v^2$ would instead give the kinetic energy of $14.4\ \text{J}$, which is a different physical quantity.

52. The impulse delivered to an object by a net force is best described as which of the following?
    • The net force divided by the time over which it acts
    • The net force multiplied by the time over which it acts
    • The net force multiplied by the distance over which it acts
    • The change in kinetic energy of the object

    Correct answer: The net force multiplied by the time over which it acts

    Impulse is the net force multiplied by the time over which it acts, $J = F \Delta t$. It measures the cumulative effect of a force applied over a time interval, and by the impulse-momentum theorem it equals the change in the object's momentum. Force multiplied by distance instead defines work, which relates to energy rather than to momentum.

53. A $1200\ \text{kg}$ car traveling at $15\ \text{m/s}$ comes to a complete stop. What is the magnitude of the change in the car's momentum?
    • $80\ \text{kg}\cdot\text{m/s}$
    • $1200\ \text{kg}\cdot\text{m/s}$
    • $135000\ \text{kg}\cdot\text{m/s}$
    • $18000\ \text{kg}\cdot\text{m/s}$

    Correct answer: $18000\ \text{kg}\cdot\text{m/s}$

    The change in momentum has a magnitude of $18000\ \text{kg}\cdot\text{m/s}$. The car's initial momentum is $1200 \times 15 = 18000\ \text{kg}\cdot\text{m/s}$, and stopping brings the final momentum to zero, so the magnitude of the change equals that initial value. The value $135000$ would come from computing kinetic energy instead of momentum.

54. A $0.15\ \text{kg}$ baseball moving toward a bat at $30\ \text{m/s}$ is hit straight back at $40\ \text{m/s}$ in the opposite direction. Taking the original direction of motion as positive, what is the magnitude of the impulse delivered to the ball?
    • $1.5\ \text{kg}\cdot\text{m/s}$
    • $10.5\ \text{kg}\cdot\text{m/s}$
    • $6.0\ \text{kg}\cdot\text{m/s}$
    • $4.5\ \text{kg}\cdot\text{m/s}$

    Correct answer: $10.5\ \text{kg}\cdot\text{m/s}$

    The impulse magnitude is $10.5\ \text{kg}\cdot\text{m/s}$. By the impulse-momentum theorem the impulse equals the change in momentum; the velocity goes from $+30\ \text{m/s}$ to $-40\ \text{m/s}$, a change of $70\ \text{m/s}$. Multiplying the $0.15\ \text{kg}$ mass by this $70\ \text{m/s}$ change gives $0.15 \times 70 = 10.5\ \text{kg}\cdot\text{m/s}$. Treating the reversal as only a $10\ \text{m/s}$ change would wrongly give $1.5\ \text{kg}\cdot\text{m/s}$.

55. Why does bending the knees when landing from a jump reduce the force felt by the legs?
    • It increases the impulse so the momentum change is larger
    • It increases the time over which the momentum changes, lowering the force
    • It reduces the total change in momentum that must occur
    • It converts the momentum into potential energy stored in the muscles

    Correct answer: It increases the time over which the momentum changes, lowering the force

    Bending the knees lowers the force by lengthening the time over which the momentum changes. The same change in momentum must occur to bring the body to rest, and since impulse equals $F \Delta t$, spreading that fixed momentum change over a longer time reduces the average force required. The total momentum change is set by the landing speed and mass, so it is the stretched time, not a smaller momentum change, that softens the impact.

56. Two ice skaters of equal mass stand at rest facing each other on frictionless ice. They push off, and one skater moves to the right at $1.5\ \text{m/s}$. What is the velocity of the other skater?
    • $1.5\ \text{m/s}$ to the right
    • $3.0\ \text{m/s}$ to the left
    • $1.5\ \text{m/s}$ to the left
    • Zero, because the pushes cancel

    Correct answer: $1.5\ \text{m/s}$ to the left

    The other skater moves at $1.5\ \text{m/s}$ to the left. The system started at rest with zero total momentum, and with no external horizontal force the total momentum stays zero, so the two equal-mass skaters must move with equal and opposite momenta. Because their masses are the same, the second skater moves at the same speed of $1.5\ \text{m/s}$ but in the opposite direction.

57. A $2.0\ \text{kg}$ cart moving at $3.0\ \text{m/s}$ collides with and sticks to a stationary $4.0\ \text{kg}$ cart on a frictionless track. What is the speed of the combined carts immediately after the collision?
    • $1.5\ \text{m/s}$
    • $3.0\ \text{m/s}$
    • $1.0\ \text{m/s}$
    • $0.5\ \text{m/s}$

    Correct answer: $1.0\ \text{m/s}$

    The combined carts move at $1.0\ \text{m/s}$. Momentum is conserved, so the initial momentum $2.0 \times 3.0 = 6.0\ \text{kg}\cdot\text{m/s}$ must equal the final momentum carried by the total mass of $6.0\ \text{kg}$; dividing gives $\frac{6.0}{6.0} = 1.0\ \text{m/s}$. The speed drops below the original $3.0\ \text{m/s}$ because the same momentum is now shared by a larger combined mass.

58. A $5.0\ \text{kg}$ rifle fires a $0.020\ \text{kg}$ bullet at $400\ \text{m/s}$. Ignoring external forces during firing, what is the magnitude of the recoil speed of the rifle?
    • $1.6\ \text{m/s}$
    • $8.0\ \text{m/s}$
    • $0.10\ \text{m/s}$
    • $400\ \text{m/s}$

    Correct answer: $1.6\ \text{m/s}$

    The rifle recoils at $1.6\ \text{m/s}$. The rifle and bullet start with zero total momentum, so the bullet's forward momentum $0.020 \times 400 = 8.0\ \text{kg}\cdot\text{m/s}$ must be balanced by an equal and opposite momentum of the rifle; dividing by the rifle's mass gives $\frac{8.0}{5.0} = 1.6\ \text{m/s}$. The much greater mass of the rifle is why its recoil speed is far smaller than the bullet's speed.

59. Which statement correctly distinguishes a perfectly elastic collision from a perfectly inelastic collision?
    • Momentum is conserved only in elastic collisions, not inelastic ones
    • Kinetic energy is conserved in an elastic collision but not in a perfectly inelastic collision
    • The objects always stick together in an elastic collision
    • Total energy is lost in an elastic collision but conserved in an inelastic collision

    Correct answer: Kinetic energy is conserved in an elastic collision but not in a perfectly inelastic collision

    The key difference is that kinetic energy is conserved in a perfectly elastic collision but not in a perfectly inelastic one. In both kinds of collisions total momentum is conserved, but an elastic collision preserves the total kinetic energy while an inelastic collision converts some of it into other forms such as heat or deformation. The objects stick together in a perfectly inelastic collision, not in an elastic one.

60. Two identical carts approach each other on a frictionless track, each moving at the same speed of $2.0\ \text{m/s}$ but in opposite directions. They collide and stick together. What is the speed of the combined carts immediately after the collision?
    • $4.0\ \text{m/s}$ in the direction of the first cart
    • $2.0\ \text{m/s}$ in the direction of the first cart
    • $1.0\ \text{m/s}$ in the direction of the second cart
    • Zero, the combined carts are momentarily at rest

    Correct answer: Zero, the combined carts are momentarily at rest

    The combined carts are momentarily at rest, so their speed is zero. Because the two identical carts carry equal and opposite momenta, the total momentum of the system before the collision is zero, and conservation of momentum requires the total momentum afterward to remain zero. With the masses joined and zero total momentum, the combined object cannot be moving immediately after the collision.

61. In a perfectly inelastic collision between two carts on a frictionless track, the total momentum of the system is found to be conserved, yet the total kinetic energy after the collision is less than before. How is this possible without violating conservation of momentum?
    • Momentum conservation requires kinetic energy to also be conserved, so the data must be wrong
    • Some kinetic energy is transformed into other forms such as heat and deformation, while momentum stays constant
    • Momentum is only approximately conserved when kinetic energy decreases
    • The lost kinetic energy is carried away as additional momentum

    Correct answer: Some kinetic energy is transformed into other forms such as heat and deformation, while momentum stays constant

    The kinetic energy decreases because some of it is transformed into other forms such as heat and deformation, even though momentum stays constant. Momentum conservation follows from the absence of external forces and does not require kinetic energy to be preserved, so the two principles are independent. Energy overall is still conserved, but in an inelastic collision part of the kinetic energy becomes non-mechanical energy rather than remaining as motion.

62. A wheel starts from rest and reaches an angular velocity of $24\ \text{rad/s}$ after rotating with constant angular acceleration for $6.0\ \text{s}$. What is the magnitude of its angular acceleration?
    • $4.0\ \text{rad/s}^2$
    • $0.25\ \text{rad/s}^2$
    • $18\ \text{rad/s}^2$
    • $144\ \text{rad/s}^2$

    Correct answer: $4.0\ \text{rad/s}^2$

    The angular acceleration is $4.0\ \text{rad/s}^2$. With constant angular acceleration, $\alpha = \frac{\Delta \omega}{\Delta t} = \frac{24 - 0}{6.0} = 4.0\ \text{rad/s}^2$. This is the rotational analog of linear $a = \frac{\Delta v}{\Delta t}$.

63. Torque is best described as which of the following?
    • The rotational analog of mass that resists changes in spin
    • The rotational analog of force, equal to the product of a force and its lever arm
    • The total angular displacement of a rotating object
    • The rate at which angular velocity changes with time

    Correct answer: The rotational analog of force, equal to the product of a force and its lever arm

    Torque is the rotational analog of force, equal to the product of a force and its lever arm. It measures a force's tendency to produce rotation about an axis and is calculated as $\tau = r F \sin\theta$, where the lever arm is the perpendicular distance from the axis to the line of action of the force. The rotational analog of mass is moment of inertia, not torque.

64. A force of $12\ \text{N}$ is applied perpendicular to the end of a wrench at a distance of $0.25\ \text{m}$ from the bolt it is turning. What is the magnitude of the torque exerted on the bolt?
    • $48\ \text{N}\cdot\text{m}$
    • $12\ \text{N}\cdot\text{m}$
    • $3.0\ \text{N}\cdot\text{m}$
    • $0.30\ \text{N}\cdot\text{m}$

    Correct answer: $3.0\ \text{N}\cdot\text{m}$

    The torque is $3.0\ \text{N}\cdot\text{m}$. Because the force is applied perpendicular to the wrench, the lever arm equals the full distance from the axis, so $\tau = rF = 0.25 \times 12 = 3.0\ \text{N}\cdot\text{m}$. Applying the force perpendicular to the lever arm maximizes the torque for a given force and distance.

65. A uniform beam $4.0\ \text{m}$ long is balanced on a pivot at its center. A $30\ \text{N}$ weight hangs $1.0\ \text{m}$ to the left of the pivot. To keep the beam in rotational equilibrium, where should a $15\ \text{N}$ weight be hung on the right side of the pivot?
    • $0.50\ \text{m}$ from the pivot
    • $1.0\ \text{m}$ from the pivot
    • $4.0\ \text{m}$ from the pivot
    • $2.0\ \text{m}$ from the pivot

    Correct answer: $2.0\ \text{m}$ from the pivot

    The $15\ \text{N}$ weight should hang $2.0\ \text{m}$ from the pivot. Rotational equilibrium requires equal and opposite torques, so the counterclockwise torque ($30 \times 1.0 = 30\ \text{N}\cdot\text{m}$) must equal the clockwise torque ($15 \times r$). Solving $15 r = 30$ gives $r = 2.0\ \text{m}$. The smaller weight must sit farther out to produce the same torque.

66. Moment of inertia is the rotational quantity that plays the same role in rotational motion that which quantity plays in linear motion?
    • Mass
    • Velocity
    • Force
    • Acceleration

    Correct answer: Mass

    Moment of inertia is the rotational analog of mass. Just as mass measures an object's resistance to changes in its linear motion, moment of inertia measures an object's resistance to changes in its rotational motion. It depends not only on the total mass but also on how that mass is distributed relative to the axis of rotation.

67. Two solid disks have the same total mass, but disk A has twice the radius of disk B. For rotation about an axis through the center of each disk, how does the moment of inertia of disk A compare with that of disk B?
    • It is twice as large
    • It is four times as large
    • It is half as large
    • It is the same

    Correct answer: It is four times as large

    Disk A's moment of inertia is four times as large. For a solid disk, the moment of inertia is $I = \frac{1}{2}MR^2$, so with equal mass it scales with the square of the radius. Doubling the radius multiplies $R^2$ by four, so the moment of inertia increases by a factor of four. This shows why mass distributed farther from the axis is much harder to spin.

68. A solid disk with a moment of inertia of $0.50\ \text{kg}\cdot\text{m}^2$ experiences a net torque of $2.0\ \text{N}\cdot\text{m}$ about its central axis. What is the magnitude of its angular acceleration?
    • $1.0\ \text{rad/s}^2$
    • $0.25\ \text{rad/s}^2$
    • $4.0\ \text{rad/s}^2$
    • $2.5\ \text{rad/s}^2$

    Correct answer: $4.0\ \text{rad/s}^2$

    The angular acceleration is $4.0\ \text{rad/s}^2$. The rotational form of Newton's second law states that net torque equals moment of inertia times angular acceleration, so $\alpha = \frac{\tau}{I} = \frac{2.0}{0.50} = 4.0\ \text{rad/s}^2$. This is the direct rotational counterpart of $a = \frac{F}{m}$.

69. A point on the rim of a rotating turntable of radius $0.20\ \text{m}$ moves with a tangential speed of $1.2\ \text{m/s}$. What is the angular velocity of the turntable?
    • $0.24\ \text{rad/s}$
    • $1.4\ \text{rad/s}$
    • $0.17\ \text{rad/s}$
    • $6.0\ \text{rad/s}$

    Correct answer: $6.0\ \text{rad/s}$

    The angular velocity is $6.0\ \text{rad/s}$. Tangential speed and angular velocity are related by $v = r\omega$, so $\omega = \frac{v}{r} = \frac{1.2}{0.20} = 6.0\ \text{rad/s}$. Every point on a rigid rotating object shares the same angular velocity even though points farther from the axis move with greater tangential speed.

70. A child pushes on a merry-go-round, but no matter how hard the push, it produces no rotation. Which situation would explain a zero net torque despite a nonzero applied force?
    • The force is directed straight toward the central axis of rotation
    • The force is applied perpendicular to the lever arm at the outer edge
    • The force is applied tangentially as far as possible from the axis
    • The force is applied at an angle of $90^\circ$ to the radius

    Correct answer: The force is directed straight toward the central axis of rotation

    A force directed straight toward the central axis produces zero torque. Torque equals $\tau = r F \sin\theta$, and when the force points along the radius toward the axis, the angle between the force and the radius is $0^\circ$ (or $180^\circ$), making $\sin\theta = 0$ and the lever arm zero. Only the component of force perpendicular to the radius produces rotation.

71. A rotating fan blade slows from $40\ \text{rad/s}$ to $10\ \text{rad/s}$ in $5.0\ \text{s}$ at a constant rate. What is the magnitude of its angular acceleration?
    • $8.0\ \text{rad/s}^2$
    • $6.0\ \text{rad/s}^2$
    • $30\ \text{rad/s}^2$
    • $2.0\ \text{rad/s}^2$

    Correct answer: $6.0\ \text{rad/s}^2$

    The magnitude of the angular acceleration is $6.0\ \text{rad/s}^2$. Using the rotational kinematics relation $\alpha = \frac{\Delta \omega}{\Delta t} = \frac{10 - 40}{5.0} = -6.0\ \text{rad/s}^2$, the negative sign indicates slowing down, so the magnitude is $6.0\ \text{rad/s}^2$.

72. A meter stick is pivoted at the $0\ \text{cm}$ end. A student can hang a single $10\ \text{N}$ weight anywhere along its length. To produce the greatest possible torque about the pivot, where should the weight be hung?
    • At the $50\ \text{cm}$ midpoint of the stick
    • At the $10\ \text{cm}$ mark, near the pivot
    • At the $100\ \text{cm}$ end, farthest from the pivot
    • Directly at the pivot at the $0\ \text{cm}$ end

    Correct answer: At the $100\ \text{cm}$ end, farthest from the pivot

    The weight should be hung at the $100\ \text{cm}$ end, farthest from the pivot. For a fixed force, torque increases with the lever arm, the perpendicular distance from the axis to the line of the force. Placing the weight as far from the pivot as possible maximizes that distance and therefore the torque, while hanging it at the pivot would give zero torque.

73. Two students sit on a balanced seesaw pivoted at its center. The heavier student weighs twice as much as the lighter student. For the seesaw to remain in rotational equilibrium, how must their distances from the pivot compare?
    • The heavier student must sit twice as far from the pivot as the lighter student
    • Both students must sit at equal distances from the pivot
    • The heavier student must sit four times as far from the pivot as the lighter student
    • The heavier student must sit half as far from the pivot as the lighter student

    Correct answer: The heavier student must sit half as far from the pivot as the lighter student

    The heavier student must sit half as far from the pivot as the lighter student. Rotational equilibrium requires the torques on each side to be equal, and torque is weight times distance. Because the heavier student exerts twice the weight, that student must sit at half the distance so that the products (torques) match and the net torque is zero.

74. Which expression correctly gives the rotational kinetic energy of a rigid object with moment of inertia $I$ rotating with angular velocity $\omega$?
    • $I\omega$
    • $\frac{1}{2} I \omega$
    • $\frac{1}{2} I \omega^2$
    • $I \omega^2$

    Correct answer: $\frac{1}{2} I \omega^2$

    The rotational kinetic energy is $\frac{1}{2} I \omega^2$. This is the rotational counterpart of translational kinetic energy $\frac{1}{2} m v^2$, with moment of inertia taking the role of mass and angular velocity taking the role of linear speed. The quantity $I\omega$ is the angular momentum, a different rotational quantity, so it does not describe the energy of spinning.

75. A figure skater spinning with her arms outstretched pulls her arms in close to her body. With no external torque acting, what happens to her rate of spin?
    • She spins faster because her moment of inertia decreases while angular momentum is conserved
    • She spins slower because pulling her arms in increases her moment of inertia
    • Her spin rate stays exactly the same because her mass does not change
    • She stops spinning because pulling her arms in removes angular momentum

    Correct answer: She spins faster because her moment of inertia decreases while angular momentum is conserved

    She spins faster because her moment of inertia decreases while her angular momentum stays the same. With no external torque, angular momentum $L = I\omega$ is conserved, so reducing the moment of inertia by bringing mass closer to the axis forces the angular velocity to rise. Her total mass is unchanged, but redistributing it nearer the spin axis is what lowers the moment of inertia and speeds up the spin.

76. A solid disk has a moment of inertia of $0.40\ \text{kg}\cdot\text{m}^2$ and spins about its central axis at an angular velocity of $5.0\ \text{rad/s}$. What is the magnitude of its angular momentum?
    • $0.08\ \text{kg}\cdot\text{m}^2/\text{s}$
    • $12.5\ \text{kg}\cdot\text{m}^2/\text{s}$
    • $5.0\ \text{kg}\cdot\text{m}^2/\text{s}$
    • $2.0\ \text{kg}\cdot\text{m}^2/\text{s}$

    Correct answer: $2.0\ \text{kg}\cdot\text{m}^2/\text{s}$

    The angular momentum is $2.0\ \text{kg}\cdot\text{m}^2/\text{s}$. Angular momentum equals the moment of inertia times the angular velocity, so $L = I\omega = 0.40 \times 5.0 = 2.0\ \text{kg}\cdot\text{m}^2/\text{s}$. This is the rotational analog of linear momentum $p = mv$, with moment of inertia and angular velocity replacing mass and linear speed.

77. A wheel of radius $0.30\ \text{m}$ rolls without slipping along level ground, and its center moves forward at $6.0\ \text{m/s}$. What is the angular velocity of the wheel about its center?
    • $1.8\ \text{rad/s}$
    • $20\ \text{rad/s}$
    • $6.0\ \text{rad/s}$
    • $0.05\ \text{rad/s}$

    Correct answer: $20\ \text{rad/s}$

    The angular velocity is $20\ \text{rad/s}$. For an object rolling without slipping, the speed of the center relates to the angular velocity by $v = r\omega$, so $\omega = \frac{v}{r} = \frac{6.0}{0.30} = 20\ \text{rad/s}$. The rolling-without-slipping condition is what links the translational motion of the center to the rotation about the center.

78. A satellite moves in a stable circular orbit around Earth at constant speed. What provides the centripetal force that keeps the satellite in its circular path?
    • The gravitational attraction between Earth and the satellite
    • The forward thrust continuously produced by the satellite's engines
    • The push of the surrounding air against the satellite's surface
    • The outward centrifugal force balancing the satellite's weight

    Correct answer: The gravitational attraction between Earth and the satellite

    The gravitational attraction between Earth and the satellite provides the centripetal force. Gravity continuously pulls the satellite toward the center of its orbit, supplying exactly the inward force needed to curve its path into a circle while it moves at constant speed. No engine thrust is required to maintain a stable orbit, and at orbital altitude there is essentially no air, so neither thrust nor air resistance is responsible.

79. A block on a frictionless surface attached to a spring undergoes simple harmonic motion. At which point in the motion is the magnitude of the block's acceleration the greatest?
    • At the equilibrium position, where the speed is greatest
    • At the points of maximum displacement from equilibrium
    • Exactly halfway between equilibrium and maximum displacement
    • The acceleration has the same magnitude everywhere in the motion

    Correct answer: At the points of maximum displacement from equilibrium

    The acceleration is greatest at the points of maximum displacement. In simple harmonic motion the restoring force is proportional to displacement ($F=-kx$), so the force, and therefore the acceleration, reaches its largest magnitude where displacement is largest, at the turning points. At equilibrium the displacement is zero, so the net force and acceleration are zero there even though the speed is maximum.

80. A simple pendulum has a period of $2.0\ \text{s}$ for small-angle swings. If the length of the pendulum is increased to four times its original length, what is the new period?
    • $1.0\ \text{s}$
    • $2.0\ \text{s}$
    • $4.0\ \text{s}$
    • $8.0\ \text{s}$

    Correct answer: $4.0\ \text{s}$

    The new period is $4.0\ \text{s}$. The period of a simple pendulum is $T=2\pi\sqrt{\frac{L}{g}}$, so the period is proportional to $\sqrt{L}$. Multiplying the length by 4 multiplies the period by $\sqrt{4}=2$, giving $2.0\ \text{s}\times 2=4.0\ \text{s}$.

81. A $0.50\ \text{kg}$ object oscillates in simple harmonic motion on a spring of spring constant $200\ \text{N/m}$. What is the period of the oscillation? (Use $\pi\approx 3.14$.)
    • $0.31\ \text{s}$
    • $0.63\ \text{s}$
    • $1.6\ \text{s}$
    • $6.3\ \text{s}$

    Correct answer: $0.31\ \text{s}$

    The period is about $0.31\ \text{s}$. For a mass on a spring the period is $T=2\pi\sqrt{\frac{m}{k}}$. The mass divided by the spring constant is $\frac{0.50}{200}=0.0025$, whose square root is $0.05$. Multiplying by $2\pi$ (about $6.28$) gives roughly $0.31\ \text{s}$.

82. An object in simple harmonic motion on a spring has total mechanical energy $E$. At the moment the object passes through the equilibrium position, how is its energy distributed?
    • All of the energy is elastic potential energy and the kinetic energy is zero
    • All of the energy is kinetic energy and the elastic potential energy is zero
    • The energy is split equally between kinetic and elastic potential energy
    • Both the kinetic energy and the elastic potential energy are momentarily zero

    Correct answer: All of the energy is kinetic energy and the elastic potential energy is zero

    At the equilibrium position the energy is entirely kinetic. The spring is at its natural length there, so the elastic potential energy is zero, while the object is moving at its maximum speed, so the kinetic energy equals the full total mechanical energy. As the object moves away from equilibrium, energy continuously transfers from kinetic to elastic potential energy.

83. Two identical mass-spring systems oscillate in simple harmonic motion. System X is given an amplitude that is twice the amplitude of system Y. How do the periods of the two systems compare?
    • System X has twice the period of system Y
    • System X has four times the period of system Y
    • System X has half the period of system Y
    • The two systems have the same period

    Correct answer: The two systems have the same period

    The two systems have the same period. For a mass on a spring the period $T=2\pi\sqrt{\frac{m}{k}}$ depends only on the mass and the spring constant, not on the amplitude. Because the systems are identical in mass and spring constant, doubling the amplitude of one leaves both periods equal.

84. A solid block has a mass of $480\ \text{g}$ and a volume of $600\ \text{cm}^3$. What is the density of the block?
    • $0.80\ \text{g/cm}^3$
    • $1.25\ \text{g/cm}^3$
    • $1080\ \text{g/cm}^3$
    • $0.0012\ \text{g/cm}^3$

    Correct answer: $0.80\ \text{g/cm}^3$

    The density is $0.80\ \text{g/cm}^3$. Density is mass divided by volume, so dividing the $480\ \text{g}$ mass by the $600\ \text{cm}^3$ volume gives $0.80\ \text{g/cm}^3$. Dividing volume by mass instead would give the reciprocal value of $1.25\ \text{g/cm}^3$, and adding the two quantities is meaningless because they have different units.

85. A diver descends in a lake of water with density $1000\ \text{kg/m}^3$. Taking $g=10\ \text{m/s}^2$, how much does the gauge pressure increase between a depth of $3.0\ \text{m}$ and a depth of $8.0\ \text{m}$?
    • $80000\ \text{Pa}$
    • $30000\ \text{Pa}$
    • $110000\ \text{Pa}$
    • $50000\ \text{Pa}$

    Correct answer: $50000\ \text{Pa}$

    The gauge pressure increases by $50000\ \text{Pa}$. The pressure added by a column of fluid equals $\rho g\,\Delta h$, so $1000\times 10\times 5.0\ \text{m}$ (the difference between $8.0\ \text{m}$ and $3.0\ \text{m}$) gives $50000\ \text{Pa}$. Using the full $8.0\ \text{m}$ depth instead of the $5.0\ \text{m}$ change would overstate the increase as $80000\ \text{Pa}$.

86. A block floats at rest on the surface of water, with part of it above the surface. What is the relationship between the buoyant force on the block and the block's weight?
    • The buoyant force is greater than the weight
    • The buoyant force equals the weight
    • The buoyant force is less than the weight
    • The buoyant force is zero because part of the block is above water

    Correct answer: The buoyant force equals the weight

    The buoyant force equals the block's weight. Because the block floats at rest, the net force on it is zero, so the upward buoyant force must exactly balance the downward weight. If the buoyant force were larger the block would accelerate upward, and if it were smaller the block would sink, so equilibrium requires the two to be equal.

87. An object with a volume of $2.0\times 10^{-3}\ \text{m}^3$ is fully submerged in water of density $1000\ \text{kg/m}^3$. Taking $g=10\ \text{m/s}^2$, what is the buoyant force on the object?
    • $20\ \text{N}$
    • $200\ \text{N}$
    • $0.02\ \text{N}$
    • $2.0\ \text{N}$

    Correct answer: $20\ \text{N}$

    The buoyant force is $20\ \text{N}$. By Archimedes' principle the buoyant force equals the weight of the displaced water, which is $\rho g V$, so $1000\times 10\times 2.0\times 10^{-3}$ gives $20\ \text{N}$. The buoyant force depends on the displaced fluid's properties and the object's volume, not on what the object itself is made of.

88. Water flows steadily through a horizontal pipe that narrows from a wide section to a section with half the cross-sectional area. How does the water's speed in the narrow section compare with its speed in the wide section?
    • It is half as fast
    • It is the same
    • It is four times as fast
    • It is twice as fast

    Correct answer: It is twice as fast

    The water moves twice as fast in the narrow section. The continuity equation states that cross-sectional area times speed stays constant for an incompressible fluid, so halving the area must double the speed to keep the flow rate the same. The fluid speeds up where the pipe constricts, which is why a narrowed channel produces faster flow.

89. According to Bernoulli's equation for an ideal fluid flowing through a horizontal pipe, what happens to the fluid pressure in a region where the fluid speed increases?
    • The pressure increases
    • The pressure decreases
    • The pressure stays the same
    • The pressure drops to zero

    Correct answer: The pressure decreases

    The pressure decreases where the fluid speeds up. Bernoulli's equation says that along a horizontal flow the sum of pressure and the kinetic-energy-per-volume term is constant, so when the speed term rises the pressure term must fall to compensate. This inverse relationship between speed and pressure is the central idea of Bernoulli's principle.

90. Two identical balls have the same volume, but one is made of aluminum and the other of lead. Both are fully submerged and held underwater. How do the buoyant forces on the two balls compare?
    • The lead ball experiences a larger buoyant force
    • The aluminum ball experiences a larger buoyant force
    • The buoyant forces are equal
    • Neither ball experiences a buoyant force while held still

    Correct answer: The buoyant forces are equal

    The buoyant forces are equal. The buoyant force depends only on the weight of the fluid displaced, which is set by the submerged volume and the fluid's density, and both balls displace the same volume of water. The lead ball weighs more and is harder to hold up, but that extra difficulty comes from its greater weight, not from any difference in buoyant force.

91. A hydraulic lift has a small input piston and a large output piston connected by an enclosed fluid. The output piston has 10 times the area of the input piston. If a force of $50\ \text{N}$ is applied to the input piston, what output force is produced?
    • $5.0\ \text{N}$
    • $60\ \text{N}$
    • $500\ \text{N}$
    • $50\ \text{N}$

    Correct answer: $500\ \text{N}$

    The output force is $500\ \text{N}$. In an enclosed fluid the pressure applied at the input is transmitted undiminished to the output, and since pressure is force divided by area, the larger output area multiplies the force by the area ratio. Multiplying the $50\ \text{N}$ input by the area ratio of 10 gives $500\ \text{N}$ at the output piston.

92. A block of wood with density $600\ \text{kg/m}^3$ floats in water with density $1000\ \text{kg/m}^3$. What fraction of the block's volume is submerged below the surface?
    • $0.60$
    • $1.67$
    • $0.40$
    • $1.00$

    Correct answer: $0.60$

    Six-tenths of the block is submerged. For a floating object the fraction of volume below the surface equals the ratio of the object's density to the fluid's density, so $\frac{600}{1000}=0.60$. This means the denser the floating object is relative to the fluid, the more of it sits below the surface, with the remaining $0.40$ staying above.

93. An incompressible fluid flows at $4.0\ \text{m/s}$ through a pipe of cross-sectional area $0.030\ \text{m}^2$. What is the volume flow rate through the pipe?
    • $0.0075\ \text{m}^3\text{/s}$
    • $0.12\ \text{m}^3\text{/s}$
    • $1.2\ \text{m}^3\text{/s}$
    • $133\ \text{m}^3\text{/s}$

    Correct answer: $0.12\ \text{m}^3\text{/s}$

    The volume flow rate is $0.12\ \text{m}^3\text{/s}$. The volume flow rate equals the cross-sectional area multiplied by the flow speed, so $0.030\ \text{m}^2\times 4.0\ \text{m/s}$ gives $0.12\ \text{m}^3\text{/s}$. Dividing the area by the speed instead would give $0.0075$, but flow rate is a product of area and speed, not a quotient.

94. Two open containers sit on a table. One holds a tall, narrow column of water and the other holds a wide, shallow pool of water, but the water reaches the same height in both. How does the pressure at the bottom of each container compare?
    • It is greater in the narrow container because the column is taller-looking
    • It is greater in the wide container because it holds more water
    • It cannot be compared without knowing the container weights
    • It is the same in both because the depths are equal

    Correct answer: It is the same in both because the depths are equal

    The pressure at the bottom is the same in both containers. Pressure in a static fluid depends only on the fluid's density, $g$, and the depth below the surface, not on the total amount of fluid or the container's shape. Since the water reaches the same height in each, the depth is equal and therefore the bottom pressure is equal, even though the wide pool contains far more water.

95. A sealed metal can is gradually lowered into a deep lake. As the can sinks deeper, the surrounding water pressure squeezes it and slightly reduces its volume while its mass stays the same. How does the buoyant force on the can change as it descends?
    • The buoyant force increases because pressure increases with depth
    • The buoyant force decreases because the displaced volume decreases
    • The buoyant force stays exactly the same at all depths
    • The buoyant force becomes zero once the can is fully submerged

    Correct answer: The buoyant force decreases because the displaced volume decreases

    The buoyant force decreases as the can is compressed. The buoyant force equals the weight of the displaced water, which depends on the volume of water pushed aside, so when the can's volume shrinks it displaces less water and the buoyant force drops. Although the deeper water exerts higher pressure, what matters for buoyancy is the displaced volume, and that volume is getting smaller as the can is squeezed.

96. A runner jogs $300\ \text{m}$ east, then turns and jogs $100\ \text{m}$ west along the same straight road. What is the magnitude of the runner's total displacement?
    • $400\ \text{m}$
    • $200\ \text{m}$
    • $300\ \text{m}$
    • $100\ \text{m}$

    Correct answer: $200\ \text{m}$

    The displacement magnitude is $200\ \text{m}$. Displacement is the straight-line change in position with direction taken into account, so the $100\ \text{m}$ westward partly cancels the $300\ \text{m}$ eastward, leaving a net $200\ \text{m}$ east. The value $400\ \text{m}$ is the total path length traveled, which is the distance rather than the displacement.

97. A car traveling at a constant $25\ \text{m/s}$ passes a stationary billboard. How far does the car travel in $8.0\ \text{s}$?
    • $200\ \text{m}$
    • $3.1\ \text{m}$
    • $100\ \text{m}$
    • $33\ \text{m}$

    Correct answer: $200\ \text{m}$

    The car travels $200\ \text{m}$. With constant velocity there is no acceleration, so the distance is simply the speed of $25\ \text{m/s}$ multiplied by the time of $8.0\ \text{s}$, which equals $200\ \text{m}$. Dividing the values instead of multiplying would give the meaningless $3.1\ \text{m}$ figure, and halving the product would wrongly assume the car started from rest.

98. A bus speeds up from rest at a constant $2.5\ \text{m/s}^2$. What is its speed after $6.0\ \text{s}$?
    • $8.5\ \text{m/s}$
    • $0.42\ \text{m/s}$
    • $15\ \text{m/s}$
    • $90\ \text{m/s}$

    Correct answer: $15\ \text{m/s}$

    The bus reaches $15\ \text{m/s}$. Starting from rest, the final speed equals the acceleration of $2.5\ \text{m/s}^2$ multiplied by the time of $6.0\ \text{s}$, which gives $15\ \text{m/s}$. Adding the two numbers would incorrectly yield $8.5$, and multiplying acceleration by the square of the time would give a distance-like value rather than a speed.

99. An arrow is shot straight up and reaches a maximum height where its speed is momentarily zero. Taking $g=10\ \text{m/s}^2$, what is the arrow's acceleration at that highest point?
    • Zero, because the speed is zero
    • $10\ \text{m/s}^2$ directed downward
    • $10\ \text{m/s}^2$ directed upward
    • It cannot be determined without the launch speed

    Correct answer: $10\ \text{m/s}^2$ directed downward

    The acceleration is $10\ \text{m/s}^2$ directed downward at the highest point. Gravity acts continuously throughout the flight regardless of the arrow's instantaneous speed, so even when the velocity is momentarily zero at the top the downward acceleration remains $10\ \text{m/s}^2$. A common error is to assume zero acceleration whenever the speed is zero, but speed and acceleration are independent quantities.

100. A swimmer crosses a river by aiming straight toward the far bank at $1.2\ \text{m/s}$ while the current carries her downstream at $0.5\ \text{m/s}$. Treating the two perpendicular velocities together, what is her speed relative to the ground?
     • $1.3\ \text{m/s}$
     • $0.7\ \text{m/s}$
     • $1.7\ \text{m/s}$
     • $1.2\ \text{m/s}$

     Correct answer: $1.3\ \text{m/s}$

     Her speed relative to the ground is $1.3\ \text{m/s}$. Because the across-river velocity of $1.2\ \text{m/s}$ and the downstream current of $0.5\ \text{m/s}$ are perpendicular, they combine using the Pythagorean theorem, and $\sqrt{1.2^2+0.5^2}=\sqrt{1.69}=1.3\ \text{m/s}$. Simply adding the speeds to get $1.7\ \text{m/s}$ ignores that the two velocities point in different directions.

101. On a velocity versus time graph for an object moving in a straight line, what physical quantity is represented by the area between the line and the time axis?
     • The object's acceleration
     • The object's instantaneous speed
     • The object's displacement
     • The object's average force

     Correct answer: The object's displacement

     The area between a velocity versus time line and the time axis represents the object's displacement. Multiplying a velocity by a time interval yields a change in position, and summing those products over the graph is exactly what computing the enclosed area does. Acceleration is found from the slope of this graph rather than its area, and force does not appear on a kinematics graph at all.

102. A car uniformly accelerates from $10\ \text{m/s}$ to $18\ \text{m/s}$ over $4.0\ \text{s}$. What is its average velocity during this interval?
     • $14\ \text{m/s}$
     • $28\ \text{m/s}$
     • $8.0\ \text{m/s}$
     • $2.0\ \text{m/s}$

     Correct answer: $14\ \text{m/s}$

     The average velocity is $14\ \text{m/s}$. For uniform acceleration the average velocity equals the mean of the initial and final velocities, so $\frac{10+18}{2}=14\ \text{m/s}$. The value $28\ \text{m/s}$ is the sum of the speeds without averaging, and $2.0\ \text{m/s}$ is the acceleration rather than a velocity.

103. Which of the following pairs lists one scalar quantity and one vector quantity, in that exact order?
     • Acceleration, then mass
     • Speed, then acceleration
     • Velocity, then displacement
     • Distance, then temperature

     Correct answer: Speed, then acceleration

     The pair speed then acceleration correctly lists a scalar followed by a vector. Speed reports only how fast an object moves with no direction, making it a scalar, while acceleration includes a direction along with its magnitude, making it a vector. The other pairs either reverse this order or list two quantities of the same type.

104. A particle moving along a straight line has a positive velocity and a negative acceleration at a given instant. What is happening to the particle at that instant?
     • It is moving in the positive direction and speeding up
     • It is moving in the positive direction and slowing down
     • It is moving in the negative direction and speeding up
     • It is at rest and about to reverse

     Correct answer: It is moving in the positive direction and slowing down

     The particle is moving in the positive direction and slowing down. The positive velocity sets the direction of motion as positive, and because the acceleration points opposite to the velocity, it acts to reduce the speed. An object speeds up only when its velocity and acceleration share the same sign, which is not the case here.

105. A coin is dropped from rest at the top of a tall building and falls freely. Taking $g=10\ \text{m/s}^2$ and ignoring air resistance, what is its speed after falling for $3.0\ \text{s}$?
     • $3.3\ \text{m/s}$
     • $45\ \text{m/s}$
     • $30\ \text{m/s}$
     • $90\ \text{m/s}$

     Correct answer: $30\ \text{m/s}$

     The coin's speed after $3.0\ \text{s}$ is $30\ \text{m/s}$. Since it starts from rest, the speed equals the free-fall acceleration of $10\ \text{m/s}^2$ multiplied by the elapsed time of $3.0\ \text{s}$, giving $30\ \text{m/s}$. The value $45\ \text{m/s}$ would be the distance fallen, not the speed, and $90\ \text{m/s}$ incorrectly multiplies by the square of the time.

106. A projectile launched at an angle above the horizontal moves through the air with no air resistance. Which statement correctly describes its horizontal motion during the flight?
     • The horizontal velocity steadily decreases until it reaches zero at the peak
     • The horizontal velocity remains constant throughout the flight
     • The horizontal velocity increases as the projectile descends
     • The horizontal velocity reverses direction at the highest point

     Correct answer: The horizontal velocity remains constant throughout the flight

     The horizontal velocity remains constant throughout the flight. With air resistance ignored, gravity acts only vertically, so there is no horizontal force to change the horizontal velocity at any point in the path. This independence is why the horizontal component is the same at launch, at the peak, and on the way down.

107. A boat that can move at $4.0\ \text{m/s}$ in still water heads directly upstream against a current flowing at $4.0\ \text{m/s}$ relative to the shore. What is the boat's velocity relative to the shore?
     • $8.0\ \text{m/s}$ upstream
     • Zero
     • $4.0\ \text{m/s}$ upstream
     • $8.0\ \text{m/s}$ downstream

     Correct answer: Zero

     The boat's velocity relative to the shore is zero. Because the boat moves upstream at $4.0\ \text{m/s}$ while the current pushes it downstream at the same $4.0\ \text{m/s}$, the two opposite velocities cancel exactly and the boat stays in place relative to the shore. Adding the speeds to get $8.0\ \text{m/s}$ would apply only if the two velocities pointed the same way.

108. An object at rest at the origin has a velocity versus time graph that is a horizontal line at $+5\ \text{m/s}$ lasting $4.0\ \text{s}$, after which the velocity is shown. What is the object's displacement during those $4.0\ \text{s}$?
     • $1.25\ \text{m}$
     • $9.0\ \text{m}$
     • $20\ \text{m}$
     • $0\ \text{m}$

     Correct answer: $20\ \text{m}$

     The displacement is $20\ \text{m}$. On a velocity versus time graph the displacement equals the area under the line, and a constant $5\ \text{m/s}$ over $4.0\ \text{s}$ forms a rectangle of area $5\times 4=20\ \text{m}$. Dividing the velocity by the time would give $1.25$, which is neither a displacement nor a physically meaningful result here.

109. A ball is thrown straight up with an initial speed of $20\ \text{m/s}$. Taking $g=10\ \text{m/s}^2$ and ignoring air resistance, what maximum height above the launch point does it reach?
     • $40\ \text{m}$
     • $10\ \text{m}$
     • $20\ \text{m}$
     • $2.0\ \text{m}$

     Correct answer: $20\ \text{m}$

     The ball reaches a maximum height of $20\ \text{m}$. Using $v^2=v_0^2-2gh$ with the final velocity zero at the top, $20^2=2\times 10\times h$, so $h=\frac{400}{20}=20\ \text{m}$. The value $40\ \text{m}$ results from forgetting the factor of two in the denominator.

110. A hiker walks a winding $5\ \text{km}$ trail and ends exactly back at the starting point after $2\ \text{hours}$. What are the hiker's total distance traveled and the magnitude of the average velocity, respectively?
     • $5\ \text{km}$ and $2.5\ \text{km/h}$
     • $0\ \text{km}$ and $2.5\ \text{km/h}$
     • $5\ \text{km}$ and $0\ \text{km/h}$
     • $0\ \text{km}$ and $0\ \text{km/h}$

     Correct answer: $5\ \text{km}$ and $0\ \text{km/h}$

     The total distance is $5\ \text{km}$ and the average velocity magnitude is $0\ \text{km/h}$. Distance measures the full path length walked, which is the $5\ \text{km}$ of trail, while average velocity depends on net displacement, and returning to the starting point makes that displacement zero. A nonzero distance with zero average velocity is the signature of a round trip.

111. A sports car accelerates uniformly from rest and covers $80\ \text{m}$ in $4.0\ \text{s}$. What is the magnitude of its acceleration?
     • $20\ \text{m/s}^2$
     • $10\ \text{m/s}^2$
     • $5.0\ \text{m/s}^2$
     • $40\ \text{m/s}^2$

     Correct answer: $10\ \text{m/s}^2$

     The acceleration is $10\ \text{m/s}^2$. Starting from rest, $d=\frac{1}{2}at^2$, so $80=\frac{1}{2}a(4.0)^2$, giving $80=8a$ and an acceleration of $10\ \text{m/s}^2$. Forgetting the factor of one-half would wrongly yield $5.0\ \text{m/s}^2$.

112. A position versus time graph for an object moving in a straight line curves upward with an increasing slope. What does this shape indicate about the object's motion?
     • The object moves at constant velocity
     • The object is speeding up in the positive direction
     • The object is at rest
     • The object is moving in the negative direction

     Correct answer: The object is speeding up in the positive direction

     The upward-curving graph with increasing slope shows the object speeding up in the positive direction. Slope on a position versus time graph is velocity, so a slope that grows steeper means the velocity is increasing while staying positive. A straight sloped line would indicate constant velocity, and a horizontal line would indicate the object is at rest.

113. Two stones are thrown from the same point at the same instant with the same speed: one straight up and one straight down. Ignoring air resistance, how do their accelerations compare while both are in the air?
     • The downward-thrown stone has the greater acceleration
     • The upward-thrown stone has the greater acceleration
     • Both have the same downward acceleration of $g$
     • The upward-thrown stone has zero acceleration until it turns around

     Correct answer: Both have the same downward acceleration of $g$

     Both stones have the same downward acceleration equal to $g$ while in the air. Free-fall acceleration depends only on gravity, not on the direction or magnitude of an object's initial velocity, so each stone accelerates downward at the same rate the entire time. The initial direction affects the velocities but never the acceleration during free fall.

114. A passenger walks toward the front of a train at $1.5\ \text{m/s}$ relative to the train while the train moves forward at $20\ \text{m/s}$ relative to the ground. What is the passenger's speed relative to the ground?
     • $18.5\ \text{m/s}$
     • $20\ \text{m/s}$
     • $21.5\ \text{m/s}$
     • $1.5\ \text{m/s}$

     Correct answer: $21.5\ \text{m/s}$

     The passenger's speed relative to the ground is $21.5\ \text{m/s}$. Because the passenger walks in the same direction the train moves, the $1.5\ \text{m/s}$ adds to the train's $20\ \text{m/s}$, giving $21.5\ \text{m/s}$ relative to the ground. Subtracting to get $18.5\ \text{m/s}$ would be correct only if the passenger walked toward the back of the train.

115. A $4.0\ \text{kg}$ block slides down a frictionless ramp inclined at $30^\circ$ above the horizontal. Taking $g=10\ \text{m/s}^2$, what is the magnitude of the block's acceleration along the incline?
     • $10\ \text{m/s}^2$
     • $2.0\ \text{m/s}^2$
     • $8.7\ \text{m/s}^2$
     • $5.0\ \text{m/s}^2$

     Correct answer: $5.0\ \text{m/s}^2$

     The acceleration is $5.0\ \text{m/s}^2$. On a frictionless incline the only force along the slope is the component of gravity, which equals $g\sin\theta$, so $10\times\sin 30^\circ=10\times 0.50=5.0\ \text{m/s}^2$. The mass cancels out, and using the cosine instead of the sine would wrongly give about $8.7$.

116. A $6.0\ \text{kg}$ block rests on a frictionless horizontal surface and is connected by a light cord over a frictionless pulley to a hanging $4.0\ \text{kg}$ block. Taking $g=10\ \text{m/s}^2$, what is the acceleration of the system?
     • $10\ \text{m/s}^2$
     • $6.0\ \text{m/s}^2$
     • $4.0\ \text{m/s}^2$
     • $0.67\ \text{m/s}^2$

     Correct answer: $4.0\ \text{m/s}^2$

     The acceleration is $4.0\ \text{m/s}^2$. Treating the two blocks as one system, the only driving force is the weight of the hanging block, which is $4.0\times 10=40\ \text{N}$, and this acts on the combined mass of $6.0+4.0=10\ \text{kg}$, so $\frac{40\ \text{N}}{10\ \text{kg}}=4.0\ \text{m/s}^2$. Using only the hanging block's mass in the denominator would overstate the acceleration.

117. A $2.0\ \text{kg}$ object experiences a constant net force and its velocity increases from $3.0\ \text{m/s}$ to $11\ \text{m/s}$ over $4.0\ \text{s}$. What is the magnitude of the net force acting on it?
     • $4.0\ \text{N}$
     • $2.0\ \text{N}$
     • $16\ \text{N}$
     • $8.0\ \text{N}$

     Correct answer: $4.0\ \text{N}$

     The net force is $4.0\ \text{N}$. The acceleration is the change in velocity divided by time, which is $\frac{11-3.0}{4.0}=\frac{8.0}{4.0}=2.0\ \text{m/s}^2$, and Newton's second law then gives $F=ma=2.0\times 2.0=4.0\ \text{N}$. Skipping the acceleration step and multiplying mass by the velocity change would give the wrong value.

118. An object is in translational equilibrium. Which statement must be true about the forces acting on it?
     • The vector sum of all forces acting on the object is zero
     • No forces act on the object at all
     • The object must be at rest
     • Exactly two equal forces act on the object

     Correct answer: The vector sum of all forces acting on the object is zero

     Translational equilibrium means the vector sum of all forces on the object is zero. Equilibrium requires zero net force, but this can occur with many forces present as long as they cancel, and it includes objects moving at constant velocity, not just objects at rest. An object can be in equilibrium with three or more forces balancing one another.

119. A 1200 kg car traveling at $24\ \text{m/s}$ brakes to a stop in $6.0\ \text{s}$. What is the magnitude of the average net braking force?
     • $4800\ \text{N}$
     • $200\ \text{N}$
     • $28800\ \text{N}$
     • $2400\ \text{N}$

     Correct answer: $4800\ \text{N}$

     The braking force is $4800\ \text{N}$. The deceleration is the speed change of $24\ \text{m/s}$ divided by the $6.0\ \text{s}$ stopping time, giving $4.0\ \text{m/s}^2$, and multiplying this by the $1200\ \text{kg}$ mass yields $4800\ \text{N}$. Multiplying the mass by the initial speed instead of the acceleration would give an incorrect and far larger value.

120. A 3.0 kg block is held against a vertical wall by a horizontal push, and the coefficient of static friction between block and wall is 0.50. Taking $g = 10\ \text{m/s}^2$, what minimum horizontal push keeps the block from sliding down?
     • $15\ \text{N}$
     • $30\ \text{N}$
     • $60\ \text{N}$
     • $6.0\ \text{N}$

     Correct answer: $60\ \text{N}$

     The minimum push is $60\ \text{N}$. The block's weight of $3.0 \times 10$, or $30\ \text{N}$, must be held up by static friction, which can be at most the coefficient of 0.50 times the normal force; the normal force here equals the horizontal push, so setting $0.50$ times the push equal to $30\ \text{N}$ gives a required push of $60\ \text{N}$. A smaller push would not generate enough friction to support the weight.

121. Two blocks, one of mass 2.0 kg and one of mass 6.0 kg, are placed in contact on a frictionless surface and a $24\ \text{N}$ horizontal force is applied to the 2.0 kg block, pushing both. What is the magnitude of the contact force between the two blocks?
     • $24\ \text{N}$
     • $6.0\ \text{N}$
     • $18\ \text{N}$
     • $12\ \text{N}$

     Correct answer: $18\ \text{N}$

     The contact force is $18\ \text{N}$. The whole system of 8.0 kg accelerates at $24\ \text{N}$ divided by $8.0\ \text{kg}$, or $3.0\ \text{m/s}^2$, and the contact force is whatever pushes the 6.0 kg block at that acceleration, which is $6.0 \times 3.0$, giving $18\ \text{N}$. The full $24\ \text{N}$ is not transmitted because part of it accelerates the front block.

122. A satellite orbits a planet in a circular path. Which force serves as the centripetal force that keeps it in orbit?
     • The normal force from the atmosphere
     • An outward force balancing the satellite's inertia
     • The thrust of the satellite's engines
     • The gravitational attraction between the planet and the satellite

     Correct answer: The gravitational attraction between the planet and the satellite

     The centripetal force is the gravitational attraction between the planet and the satellite. For an orbiting satellite, gravity continuously pulls it toward the planet's center, supplying exactly the inward force needed to bend its path into a circle. No engine thrust or outward force is required; gravity alone plays the centripetal role.

123. A person stands on a bathroom scale inside an elevator. At what point will the scale read a value greater than the person's true weight?
     • When the elevator accelerates upward
     • When the elevator moves upward at constant speed
     • When the elevator accelerates downward
     • When the elevator is at rest

     Correct answer: When the elevator accelerates upward

     The scale reads more than the true weight when the elevator accelerates upward. The scale reads the normal force it pushes up with, and to accelerate the person upward this normal force must exceed gravity, so the reading climbs above the actual weight. At constant speed or at rest the net force is zero and the scale reads the true weight.

124. A 5.0 kg object is pushed up a frictionless incline angled at $37^\circ$ with a force directed up along the slope. If it moves up the incline at constant velocity, what is the magnitude of the applied force? Use $g = 10\ \text{m/s}^2$ and $\sin 37^\circ = 0.60$.
     • $50\ \text{N}$
     • $40\ \text{N}$
     • $30\ \text{N}$
     • $25\ \text{N}$

     Correct answer: $30\ \text{N}$

     The applied force is $30\ \text{N}$. Constant velocity means zero net force, so the up-the-slope push must balance the gravity component along the incline, which is the weight of $5.0 \times 10$, or $50\ \text{N}$, times the sine of $37^\circ$, giving $50 \times 0.60$, or $30\ \text{N}$. The full weight of $50\ \text{N}$ is not needed because only the slope component of gravity opposes the motion.

125. A 0.20 kg ball on a string is whirled in a vertical circle of radius 0.50 m. At the lowest point the ball moves at $4.0\ \text{m/s}$. Taking $g = 10\ \text{m/s}^2$, what is the tension in the string at that lowest point?
     • $6.4\ \text{N}$
     • $2.0\ \text{N}$
     • $4.4\ \text{N}$
     • $8.4\ \text{N}$

     Correct answer: $8.4\ \text{N}$

     The tension is $8.4\ \text{N}$. At the lowest point the net upward force must provide the centripetal force, so tension minus weight equals $\frac{mv^2}{r}$; the centripetal term is $\frac{0.20 \times 4.0^2}{0.50}$, or $6.4\ \text{N}$, and adding the weight of $0.20 \times 10$, or $2.0\ \text{N}$, gives a tension of $8.4\ \text{N}$. Forgetting to add the weight would leave only $6.4\ \text{N}$.

126. A constant horizontal force acts on a block resting on a frictionless surface. Which graph relationship correctly describes the block's velocity over time?
     • Velocity stays constant
     • Velocity decreases over time
     • Velocity increases at an ever-faster rate
     • Velocity increases at a steady, constant rate

     Correct answer: Velocity increases at a steady, constant rate

     A constant force produces a velocity that increases at a steady, constant rate. By Newton's second law a constant net force gives a constant acceleration, and constant acceleration means the velocity changes by the same amount each second, which is a steady linear increase. A constant velocity would require zero net force, not a steady push.

127. An object has a weight of $60\ \text{N}$ on Earth's surface. What is its mass? Use $g = 10\ \text{m/s}^2$.
     • $6.0\ \text{kg}$
     • $600\ \text{kg}$
     • $60\ \text{kg}$
     • $0.17\ \text{kg}$

     Correct answer: $6.0\ \text{kg}$

     The mass is $6.0\ \text{kg}$. Weight is the gravitational force on an object, equal to mass times $g$, so the mass is the weight divided by $g$, which is $60\ \text{N}$ divided by $10\ \text{m/s}^2$, giving $6.0\ \text{kg}$. Multiplying the weight by $g$ instead of dividing would wrongly give 600.

128. An astronaut takes a 10 kg object from Earth to the Moon, where the gravitational field strength is about one-sixth that on Earth. How do the object's mass and weight compare on the Moon to their values on Earth?
     • Both mass and weight are unchanged
     • Both mass and weight are one-sixth of their Earth values
     • Mass is unchanged but weight is about one-sixth as large
     • Weight is unchanged but mass is about one-sixth as large

     Correct answer: Mass is unchanged but weight is about one-sixth as large

     The mass stays the same while the weight drops to about one-sixth. Mass measures the amount of matter and does not depend on location, but weight is the gravitational force, which depends on the local gravitational field strength, so a weaker field on the Moon gives a smaller weight for the same mass. The 10 kg object is still 10 kg on the Moon but weighs much less.

129. A 1.5 kg block sits on top of a 3.5 kg block, and the lower block rests on a frictionless floor. A horizontal force is applied to the lower block, and the two move together without slipping. Which force is responsible for accelerating the upper block forward?
     • Gravity acting on the upper block
     • The normal force from the lower block
     • The applied force directly on the upper block
     • The static friction force from the lower block on the upper block

     Correct answer: The static friction force from the lower block on the upper block

     Static friction from the lower block on the upper block accelerates the upper block forward. The applied force acts only on the lower block, so the only horizontal force available to push the upper block along is the friction at the surface between them, which grips the upper block and carries it forward without slipping. Gravity and the normal force are vertical and cannot supply the horizontal acceleration.

130. Two forces act on a 5.0 kg object on a frictionless surface: $30\ \text{N}$ directed east and $40\ \text{N}$ directed north. What is the magnitude of the object's acceleration?
     • $14\ \text{m/s}^2$
     • $70\ \text{m/s}^2$
     • $10\ \text{m/s}^2$
     • $2.0\ \text{m/s}^2$

     Correct answer: $10\ \text{m/s}^2$

     The acceleration is $10\ \text{m/s}^2$. The two perpendicular forces combine by the Pythagorean theorem into a net force whose magnitude is $\sqrt{30^2 + 40^2}$, which is $\sqrt{2500}$, or $50\ \text{N}$, and dividing this by the $5.0\ \text{kg}$ mass gives $10\ \text{m/s}^2$. Simply adding the forces to get $70\ \text{N}$ ignores that they act at right angles.

131. A child pulls a wagon by a handle that makes a $30^\circ$ angle above the horizontal with a force of $20\ \text{N}$. What is the magnitude of the horizontal component of this pulling force? Use $\cos 30^\circ = 0.87$.
     • $10\ \text{N}$
     • $17\ \text{N}$
     • $20\ \text{N}$
     • $23\ \text{N}$

     Correct answer: $17\ \text{N}$

     The horizontal component is about $17\ \text{N}$. The horizontal part of a force at an angle above the horizontal is the magnitude times the cosine of the angle, so $20 \times \cos 30^\circ$, or $20 \times 0.87$, gives about $17\ \text{N}$. Using the sine instead would give the vertical component of $10\ \text{N}$, not the horizontal one.

132. A book is pressed against a vertical wall and held in place by a horizontal push. Which direction does the normal force from the wall on the book point?
     • Vertically upward
     • Horizontally, away from the wall
     • Vertically downward
     • Diagonally, along the wall

     Correct answer: Horizontally, away from the wall

     The normal force points horizontally, away from the wall. A normal force is always perpendicular to the contact surface and directed away from that surface, so a vertical wall pushes the book straight out horizontally. It is the friction force, not the normal force, that acts vertically to help hold the book up.

133. Two stars of equal mass orbit their common center of mass. Where is the center of mass of this two-star system located?
     • At the position of the more massive star
     • Exactly at the midpoint between the two stars
     • Outside both stars
     • Constantly shifting toward whichever star is moving faster

     Correct answer: Exactly at the midpoint between the two stars

     The center of mass is exactly at the midpoint between the two stars. Because the stars have equal mass, the mass-weighted average position falls halfway between them by symmetry. If the masses were unequal the center of mass would shift toward the heavier star, but equal masses place it precisely in the middle.

134. A 2.0 kg block on a horizontal surface is pulled by a $10\ \text{N}$ horizontal force and experiences $4.0\ \text{N}$ of friction opposing its motion. What is the block's acceleration?
     • $5.0\ \text{m/s}^2$
     • $3.0\ \text{m/s}^2$
     • $7.0\ \text{m/s}^2$
     • $2.0\ \text{m/s}^2$

     Correct answer: $3.0\ \text{m/s}^2$

     The acceleration is $3.0\ \text{m/s}^2$. The net force is the $10\ \text{N}$ pull minus the $4.0\ \text{N}$ friction, giving $6.0\ \text{N}$, and dividing by the $2.0\ \text{kg}$ mass yields $3.0\ \text{m/s}^2$. Using the full $10\ \text{N}$ without subtracting friction would overstate the acceleration as 5.0.

135. The gravitational field strength at a point in space is defined as which quantity?
     • The gravitational force per unit mass placed at that point
     • The total gravitational force on any object at that point
     • The product of the two masses divided by their separation
     • The gravitational potential energy per unit volume

     Correct answer: The gravitational force per unit mass placed at that point

     Gravitational field strength is the gravitational force per unit mass at a point. It describes how strong gravity is at a location independent of the test object, so dividing the gravitational force on an object by that object's mass gives the field strength, which on Earth's surface is about $10\ \text{N/kg}$. The total force itself depends on the object's mass, so it is not the field by itself.

136. A 1000 kg car rounds a flat circular curve of radius 80 m. The coefficient of static friction between the tires and road is 0.50, and $g = 10\ \text{m/s}^2$. What is the maximum speed at which the car can round the curve without sliding?
     • $40\ \text{m/s}$
     • $20\ \text{m/s}$
     • $28\ \text{m/s}$
     • $14\ \text{m/s}$

     Correct answer: $20\ \text{m/s}$

     The maximum speed is $20\ \text{m/s}$. Friction supplies the centripetal force, so the maximum static friction of the coefficient times the weight, $0.50 \times 1000 \times 10$, or $5000\ \text{N}$, equals $\frac{mv^2}{r}$; solving gives $v^2 = \frac{5000 \times 80}{1000}$, or 400, so the speed is $20\ \text{m/s}$. The mass cancels, and forgetting to take the square root would leave an incorrect 400.

137. A 4.0 kg block hangs from a spring scale inside a stationary elevator. The elevator then accelerates downward at $3.0\ \text{m/s}^2$. Taking $g = 10\ \text{m/s}^2$, what does the spring scale read during the downward acceleration?
     • $40\ \text{N}$
     • $52\ \text{N}$
     • $28\ \text{N}$
     • $12\ \text{N}$

     Correct answer: $28\ \text{N}$

     The scale reads $28\ \text{N}$. When the elevator accelerates downward, the upward scale force is less than gravity, and Newton's second law gives weight minus scale reading equal to mass times downward acceleration, so the reading is the weight of $4.0 \times 10$, or $40\ \text{N}$, minus $4.0 \times 3.0$, or $12\ \text{N}$, leaving $28\ \text{N}$. A stationary elevator would instead read the full $40\ \text{N}$.

138. Two blocks are connected by a light string over a frictionless pulley at the edge of a table, with one block of mass 3.0 kg hanging and another of mass 3.0 kg on the frictionless tabletop. Taking $g = 10\ \text{m/s}^2$, what is the tension in the connecting string?
     • $30\ \text{N}$
     • $15\ \text{N}$
     • $5.0\ \text{m/s}^2$
     • $60\ \text{N}$

     Correct answer: $15\ \text{N}$

     The tension is $15\ \text{N}$. The system of 6.0 kg accelerates under the hanging block's weight of $30\ \text{N}$, giving an acceleration of $5.0\ \text{m/s}^2$, and the tension is the force that accelerates the tabletop block, which is $3.0 \times 5.0$, or $15\ \text{N}$. The full $30\ \text{N}$ weight is not the tension because the hanging block is itself accelerating downward.

139. A ball is thrown straight up and is momentarily at rest at the very top of its path. Ignoring air resistance, what is the net force on the ball at that highest point?
     • Zero, since the ball is momentarily at rest
     • Equal to the ball's weight, directed downward
     • Equal to the ball's weight, directed upward
     • Twice the ball's weight, directed downward

     Correct answer: Equal to the ball's weight, directed downward

     The net force at the top equals the ball's weight, directed downward. Even though the ball is instantaneously at rest, gravity still acts on it, so the net force is its weight pulling it down, which is why it immediately begins to fall. Zero velocity does not mean zero force; the acceleration of gravity continues throughout the motion.

140. A horizontal force of $25\ \text{N}$ is applied to a 5.0 kg box on a rough horizontal floor, but the box does not move. What is the magnitude of the static friction force acting on the box at that moment?
     • $25\ \text{N}$
     • $50\ \text{N}$
     • $5.0\ \text{N}$
     • $0\ \text{N}$

     Correct answer: $25\ \text{N}$

     The static friction force is $25\ \text{N}$. Because the box stays at rest, the net force must be zero, so static friction exactly matches and opposes the $25\ \text{N}$ applied force. Static friction adjusts itself up to a maximum value to prevent motion, so as long as the box is still it equals the applied force rather than some fixed coefficient times weight.

141. An object is launched and follows a projectile path through the air with negligible air resistance. While it is in flight, what is the direction of the net force acting on it?
     • In the direction of its velocity
     • Straight downward
     • Horizontal, in the direction of launch
     • Zero, since it is in free flight

     Correct answer: Straight downward

     The net force on a projectile in flight is straight downward. With air resistance ignored, the only force acting is gravity, which always points toward Earth, so the net force is downward throughout the entire arc regardless of which way the object is moving. The velocity changes direction along the path, but the force stays vertical.

142. Inertia is best described as which property of an object?
     • The tendency of an object to resist changes in its state of motion
     • The force that keeps a moving object moving
     • The gravitational pull an object experiences
     • The total momentum carried by a moving object

     Correct answer: The tendency of an object to resist changes in its state of motion

     Inertia is the tendency of an object to resist changes in its state of motion. It is the property described by Newton's first law, by which an object at rest stays at rest and a moving object keeps moving at constant velocity unless a net force acts. Inertia is measured by mass and is not itself a force.

143. A 50 kg crate is lowered by a cable that accelerates it downward at $2.0\ \text{m/s}^2$. Taking $g = 10\ \text{m/s}^2$, what is the tension in the cable?
     • $500\ \text{N}$
     • $600\ \text{N}$
     • $400\ \text{N}$
     • $100\ \text{N}$

     Correct answer: $400\ \text{N}$

     The cable tension is $400\ \text{N}$. For downward acceleration, Newton's second law gives weight minus tension equal to mass times acceleration, so the tension is the weight of $50 \times 10$, or $500\ \text{N}$, minus $50 \times 2.0$, or $100\ \text{N}$, leaving $400\ \text{N}$. A non-accelerating crate would require the full $500\ \text{N}$ of tension.

144. A spring is compressed 0.060 m and stores a restoring force of $18\ \text{N}$. What is the spring constant?
     • $300\ \text{N/m}$
     • $1.08\ \text{N/m}$
     • $108\ \text{N/m}$
     • $3.0\ \text{N/m}$

     Correct answer: $300\ \text{N/m}$

     The spring constant is $300\ \text{N/m}$. By Hooke's law the spring constant equals the restoring force divided by the displacement, so $18\ \text{N}$ divided by $0.060\ \text{m}$ gives $300\ \text{N/m}$. Multiplying the force by the displacement instead of dividing would give the much smaller and incorrect value of about 1.08.

145. Newton's first law tells us that an object moving at a constant velocity in a straight line has what net force acting on it?
     • A net force in the direction of motion
     • A net force opposite to the motion
     • A net force that grows with speed
     • Zero net force

     Correct answer: Zero net force

     An object moving at constant velocity has zero net force on it. Newton's first law states that an object continues at constant velocity unless acted on by a net force, so steady straight-line motion means the forces are balanced and the net force is zero. A net force in the direction of motion would instead speed the object up rather than keep its velocity constant.

146. A $60\ \text{N}$ horizontal force pushes a box 4.0 m in the same direction as the force across a level floor. How much work does this force do on the box?
     • $240\ \text{J}$
     • $64\ \text{J}$
     • $15\ \text{J}$
     • $120\ \text{J}$

     Correct answer: $240\ \text{J}$

     The force does $240\ \text{J}$ of work. When a constant force acts in the same direction as the displacement, the work equals the force times the distance, so $60\ \text{N} \times 4.0\ \text{m}$ gives $240\ \text{J}$. Adding the force and distance instead of multiplying them would give $64\ \text{J}$, which has the wrong units and ignores how work is actually defined.

147. A person carries a 10 kg suitcase horizontally across a level airport floor at constant speed for 25 m. Taking $g = 10\ \text{m/s}^2$, how much work does the upward carrying force do on the suitcase?
     • $2500\ \text{J}$
     • $0\ \text{J}$
     • $250\ \text{J}$
     • $1000\ \text{J}$

     Correct answer: $0\ \text{J}$

     The carrying force does $0\ \text{J}$ of work. Work depends on the component of the force along the direction of motion, and here the supporting force points straight up while the suitcase moves horizontally, so the force is perpendicular to the displacement. A force at a right angle to the motion does no work no matter how far the object travels.

148. A $25\ \text{N}$ force is applied to a sled at an angle of $60^\circ$ above the horizontal, and the sled moves 8.0 m horizontally. Given that $\cos 60^\circ = 0.50$, how much work does the applied force do on the sled?
     • $200\ \text{J}$
     • $173\ \text{J}$
     • $100\ \text{J}$
     • $50\ \text{J}$

     Correct answer: $100\ \text{J}$

     The applied force does $100\ \text{J}$ of work. For a force at an angle to the displacement, the work equals the force times the distance times the cosine of the angle between them, so $25\ \text{N} \times 8.0\ \text{m} \times 0.50$ gives $100\ \text{J}$. Forgetting the cosine factor would give $200\ \text{J}$, which overstates the work because only part of the force acts along the direction of motion.

149. A box is dragged across a floor and the force of kinetic friction acts opposite to the box's motion. What can be said about the work done by friction on the box?
     • It is positive because friction is a force
     • It is zero because friction is perpendicular to motion
     • It depends only on the mass, not the direction
     • It is negative because the force opposes the displacement

     Correct answer: It is negative because the force opposes the displacement

     The work done by friction is negative because the force opposes the displacement. When a force points opposite to the direction an object moves, the angle between force and displacement is 180 degrees, and its cosine is negative one, making the work negative. Friction removes kinetic energy from the box, which is consistent with the negative sign of the work it does.

150. The area under a force-versus-position graph, where force is plotted on the vertical axis and position on the horizontal axis, represents which physical quantity?
     • The work done by the force
     • The power delivered by the force
     • The impulse delivered by the force
     • The momentum of the object

     Correct answer: The work done by the force

     The area under a force-versus-position graph represents the work done by the force. Because work is force multiplied by displacement, summing force times small position steps across the graph gives the total work as the enclosed area. Impulse would instead come from the area under a force-versus-time graph, which is a different quantity entirely.

151. A variable force acts on a cart along a straight line. On a force-versus-position graph the force is constant at $10\ \text{N}$ from 0 m to 3.0 m. How much work does the force do over this interval?
     • $13\ \text{J}$
     • $30\ \text{J}$
     • $3.3\ \text{J}$
     • $10\ \text{J}$

     Correct answer: $30\ \text{J}$

     The force does $30\ \text{J}$ of work over this interval. The work equals the area under the force-versus-position graph, and a constant $10\ \text{N}$ force over a 3.0 m span forms a rectangle of area $10\ \text{N} \times 3.0\ \text{m}$, or $30\ \text{J}$. Dividing the force by the distance would give the wrong quantity, since work comes from multiplying force by displacement, not dividing.

152. An object's kinetic energy is doubled while its mass stays the same. By what factor does its speed change?
     • It doubles
     • It quadruples
     • It increases by $\sqrt{2}$
     • It stays the same

     Correct answer: It increases by $\sqrt{2}$

     The speed increases by a factor of $\sqrt{2}$. Kinetic energy is $\frac{1}{2}mv^2$, so the energy depends on the square of the speed; doubling the energy requires the speed squared to double, which means the speed itself grows by $\sqrt{2}$. The speed does not simply double, because that would make the kinetic energy four times as large.

153. A 4.0 kg object has a kinetic energy of $50\ \text{J}$. What is its speed?
     • $12.5\ \text{m/s}$
     • $25\ \text{m/s}$
     • $3.5\ \text{m/s}$
     • $5.0\ \text{m/s}$

     Correct answer: $5.0\ \text{m/s}$

     The object's speed is $5.0\ \text{m/s}$. Setting kinetic energy equal to $\frac{1}{2}mv^2$ gives $50\ \text{J} = \frac{1}{2} \times 4.0 \times v^2$, so $v^2 = 25$ and the speed is $5.0\ \text{m/s}$. Skipping the square root and stopping at the speed squared would wrongly suggest $25\ \text{m/s}$.

154. Two objects move with the same kinetic energy, but object A has four times the mass of object B. How does the speed of object A compare with the speed of object B?
     • Object A moves at half the speed of object B
     • Object A moves at the same speed as object B
     • Object A moves at four times the speed of object B
     • Object A moves at twice the speed of object B

     Correct answer: Object A moves at half the speed of object B

     Object A moves at half the speed of object B. For a fixed kinetic energy, the speed squared is inversely proportional to the mass, so four times the mass means one-fourth the speed squared, and the speed is therefore $\sqrt{\frac{1}{4}}$, or one-half. The heavier object must move more slowly to carry the same energy of motion.

155. A 3.0 kg object speeds up from $2.0\ \text{m/s}$ to $6.0\ \text{m/s}$. By how much does its kinetic energy increase?
     • $12\ \text{J}$
     • $48\ \text{J}$
     • $54\ \text{J}$
     • $24\ \text{J}$

     Correct answer: $48\ \text{J}$

     The kinetic energy increases by $48\ \text{J}$. The final kinetic energy is $\frac{1}{2} \times 3.0 \times 6.0^2$, or $54\ \text{J}$, and the initial is $\frac{1}{2} \times 3.0 \times 2.0^2$, or $6.0\ \text{J}$, so the change is $54 - 6.0$, which is $48\ \text{J}$. Squaring the change in speed first would be incorrect, because kinetic energy depends on the square of each speed separately, not on the square of the difference.

156. A net force does $120\ \text{J}$ of positive work on a 2.0 kg cart that starts from rest on a frictionless track. What is the cart's final speed?
     • $7.7\ \text{m/s}$
     • $60\ \text{m/s}$
     • $11\ \text{m/s}$
     • $120\ \text{m/s}$

     Correct answer: $11\ \text{m/s}$

     The cart's final speed is about $11\ \text{m/s}$. By the work-energy theorem the $120\ \text{J}$ of net work equals the cart's final kinetic energy since it started from rest, so $120\ \text{J} = \frac{1}{2} \times 2.0 \times v^2$, giving $v^2 = 120$ and a speed of about $11\ \text{m/s}$. Dividing the work by the mass without the kinetic energy relationship would not give a correct speed.

157. A $1000\ \text{kg}$ car traveling at $15\ \text{m/s}$ comes to a complete stop. How much net work was done on the car while it stopped?
     • Positive $112500\ \text{J}$
     • Negative $15000\ \text{J}$
     • Zero $\text{J}$
     • Negative $112500\ \text{J}$

     Correct answer: Negative $112500\ \text{J}$

     The net work done on the car is $-112500\ \text{J}$. By the work-energy theorem the net work equals the change in kinetic energy, and the kinetic energy drops from $\frac{1}{2}\times 1000\times 15^{2}=112500\ \text{J}$ down to zero, a change of $-112500\ \text{J}$. The work is negative because the forces that stop the car act opposite to its motion, removing energy.

158. A child throws a $0.20\ \text{kg}$ ball straight up, and at the moment it leaves the hand it moves at $12\ \text{m/s}$. Ignoring air resistance and taking $g=10\ \text{m/s}^2$, how high above the release point does the ball rise?
     • $7.2\ \text{m}$
     • $1.2\ \text{m}$
     • $14.4\ \text{m}$
     • $24\ \text{m}$

     Correct answer: $7.2\ \text{m}$

     The ball rises $7.2\ \text{m}$ above the release point. Using conservation of energy, the kinetic energy at release equals the gravitational potential energy at the top, so the height equals $\frac{v^{2}}{2g}=\frac{12^{2}}{2\times 10}=\frac{144}{20}=7.2\ \text{m}$. The mass cancels out, so it does not affect the maximum height reached.

159. A $0.50\ \text{kg}$ ball is dropped from a height of $2.0\ \text{m}$ and rebounds to a height of $1.5\ \text{m}$. Taking $g=10\ \text{m/s}^2$, how much mechanical energy was lost in the bounce?
     • $7.5\ \text{J}$
     • $2.5\ \text{J}$
     • $10\ \text{J}$
     • $0.5\ \text{J}$

     Correct answer: $2.5\ \text{J}$

     The ball loses $2.5\ \text{J}$ of mechanical energy. The gravitational potential energy before the bounce is $0.50\times 10\times 2.0=10\ \text{J}$, and after the bounce it is $0.50\times 10\times 1.5=7.5\ \text{J}$, so the lost energy is $10-7.5=2.5\ \text{J}$. The missing energy is converted to heat and sound during the collision with the floor.

160. A skier of mass $70\ \text{kg}$ starts from rest at the top of a frictionless slope and reaches the bottom at $20\ \text{m/s}$. Taking $g=10\ \text{m/s}^2$, what was the vertical height of the slope?
     • $10\ \text{m}$
     • $40\ \text{m}$
     • $20\ \text{m}$
     • $2.0\ \text{m}$

     Correct answer: $20\ \text{m}$

     The slope had a vertical height of $20\ \text{m}$. Conservation of energy on a frictionless slope means the gravitational potential energy at the top equals the kinetic energy at the bottom, so the height equals $\frac{v^{2}}{2g}=\frac{20^{2}}{2\times 10}=\frac{400}{20}=20\ \text{m}$. The skier's mass cancels and so does not change the required height.

161. A $0.40\ \text{kg}$ ball is launched horizontally from a tabletop and leaves the edge moving at $5.0\ \text{m/s}$. The tabletop is $1.25\ \text{m}$ above the floor. Ignoring air resistance and taking $g=10\ \text{m/s}^2$, what is the ball's total kinetic energy just before it lands?
     • $2.5\ \text{J}$
     • $5.0\ \text{J}$
     • $15\ \text{J}$
     • $10\ \text{J}$

     Correct answer: $10\ \text{J}$

     The ball's kinetic energy just before landing is $10\ \text{J}$. Its initial kinetic energy is $\frac{1}{2}\times 0.40\times 5.0^{2}=5.0\ \text{J}$, and the gravitational potential energy converted during the fall is $0.40\times 10\times 1.25=5.0\ \text{J}$, so the total kinetic energy at the floor is $5.0+5.0=10\ \text{J}$. Energy is conserved, so all the lost potential energy adds to the kinetic energy regardless of the horizontal motion.

162. A spring stores $18\ \text{J}$ of elastic potential energy when compressed a distance $x$. How much elastic potential energy does it store when compressed only half that distance?
     • $4.5\ \text{J}$
     • $9.0\ \text{J}$
     • $18\ \text{J}$
     • $36\ \text{J}$

     Correct answer: $4.5\ \text{J}$

     The spring stores $4.5\ \text{J}$ when compressed half as far. Elastic potential energy is proportional to the square of the compression, so halving the compression multiplies the energy by $\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$, giving $\frac{18}{4}=4.5\ \text{J}$. Simply halving the energy to $9.0\ \text{J}$ would ignore the squared dependence on displacement.

163. A spring stores $8.0\ \text{J}$ of elastic potential energy when stretched $0.20\ \text{m}$ from its natural length. What is the spring constant?
     • $40\ \text{N/m}$
     • $400\ \text{N/m}$
     • $200\ \text{N/m}$
     • $80\ \text{N/m}$

     Correct answer: $400\ \text{N/m}$

     The spring constant is $400\ \text{N/m}$. Elastic potential energy equals $\frac{1}{2}kx^{2}$, so $8.0=\frac{1}{2}\times k\times 0.20^{2}=\frac{1}{2}\times k\times 0.04$. Solving gives $k=\frac{16}{0.04}=400\ \text{N/m}$.

164. A $2.0\ \text{kg}$ block rests on a frictionless surface against a spring compressed $0.30\ \text{m}$. The spring has a spring constant of $500\ \text{N/m}$. When the block is released, what speed does it reach as it leaves the spring at the spring's natural length?
     • $11\ \text{m/s}$
     • $2.3\ \text{m/s}$
     • $4.7\ \text{m/s}$
     • $22\ \text{m/s}$

     Correct answer: $4.7\ \text{m/s}$

     The block leaves the spring at about $4.7\ \text{m/s}$. The elastic potential energy stored in the spring, $\frac{1}{2}\times 500\times 0.30^{2}=22.5\ \text{J}$, all converts to the block's kinetic energy on a frictionless surface. Setting $22.5\ \text{J}=\frac{1}{2}\times 2.0\times v^{2}$ gives $v^{2}=22.5$ and $v\approx 4.7\ \text{m/s}$.

165. An elevator motor lifts a total load of $800\ \text{kg}$ a height of $20\ \text{m}$ in $40\ \text{s}$ at a steady speed. Taking $g=10\ \text{m/s}^2$, what average power does the motor supply?
     • $160000\ \text{W}$
     • $16000\ \text{W}$
     • $400\ \text{W}$
     • $4000\ \text{W}$

     Correct answer: $4000\ \text{W}$

     The motor supplies an average power of $4000\ \text{W}$. The work done is the weight times the height, $800\times 10\times 20=160000\ \text{J}$, and dividing this work by the $40\ \text{s}$ time gives $4000\ \text{W}$. Forgetting to divide by the time would leave the total work of $160000\ \text{J}$ rather than the rate at which it is delivered.

166. A $1500\ \text{W}$ motor operates for $30\ \text{s}$. How much energy does it deliver in that time?
     • $45000\ \text{J}$
     • $50\ \text{J}$
     • $1500\ \text{J}$
     • $4500\ \text{J}$

     Correct answer: $45000\ \text{J}$

     The motor delivers $45000\ \text{J}$ of energy. Since power is energy per unit time, the energy delivered equals the power times the time, so $1500\ \text{W}\times 30\ \text{s}=45000\ \text{J}$. Dividing the power by the time would give the wrong quantity, because energy is the product of power and time, not their quotient.

167. Two engines do the same amount of work, but engine A takes $10\ \text{s}$ and engine B takes $20\ \text{s}$. How does the average power of engine A compare with that of engine B?
     • Engine A has half the power of engine B
     • Engine A has twice the power of engine B
     • Engine A has the same power as engine B
     • Engine A has four times the power of engine B

     Correct answer: Engine A has twice the power of engine B

     Engine A has twice the power of engine B. Power is work divided by time, and since both engines do equal work, the engine taking less time delivers more power; engine A finishes in half the time, so it has twice the power. Doing the same work faster always means producing greater power.

168. A locomotive provides a constant forward force of $5000\ \text{N}$ to a train moving at a steady $12\ \text{m/s}$. What power is the locomotive delivering?
     • $5000\ \text{W}$
     • $417\ \text{W}$
     • $60000\ \text{W}$
     • $30000\ \text{W}$

     Correct answer: $60000\ \text{W}$

     The locomotive delivers $60000\ \text{W}$ of power. When a constant force acts on an object moving at a steady speed, the power equals the force times the speed, so $5000\ \text{N}\times 12\ \text{m/s}=60000\ \text{W}$. Dividing the force by the speed would give the wrong quantity, because power is found by multiplying force and velocity together.

169. A $3.0\ \text{kg}$ block is pushed $2.0\ \text{m}$ up a frictionless incline by a force that does $90\ \text{J}$ of work, and the block rises a vertical height of $2.0\ \text{m}$. Taking $g=10\ \text{m/s}^2$, what is the block's kinetic energy at the top if it started from rest?
     • $150\ \text{J}$
     • $90\ \text{J}$
     • $60\ \text{J}$
     • $30\ \text{J}$

     Correct answer: $30\ \text{J}$

     The block has $30\ \text{J}$ of kinetic energy at the top. Of the $90\ \text{J}$ of work supplied, the gain in gravitational potential energy is $3.0\times 10\times 2.0=60\ \text{J}$, and on a frictionless incline the remaining energy becomes kinetic energy, so $90-60=30\ \text{J}$. The work input is shared between raising the block and speeding it up.

170. An object is raised so that its gravitational potential energy relative to the ground increases by $250\ \text{J}$. The object has a mass of $5.0\ \text{kg}$, and $g=10\ \text{m/s}^2$. Through what vertical height was it raised?
     • $5.0\ \text{m}$
     • $50\ \text{m}$
     • $2.5\ \text{m}$
     • $25\ \text{m}$

     Correct answer: $5.0\ \text{m}$

     The object was raised through a vertical height of $5.0\ \text{m}$. Gravitational potential energy gained equals $mgh$, so $250=5.0\times 10\times h$, meaning $h=\frac{250}{50}=5.0\ \text{m}$. Only the vertical change in height matters, regardless of any horizontal motion involved.

171. A $0.10\ \text{kg}$ arrow is shot straight up and momentarily stops at a height of $40\ \text{m}$. Ignoring air resistance and taking $g=10\ \text{m/s}^2$, what was the arrow's speed when it left the bow?
     • $40\ \text{m/s}$
     • $28\ \text{m/s}$
     • $20\ \text{m/s}$
     • $800\ \text{m/s}$

     Correct answer: $28\ \text{m/s}$

     The arrow left the bow at about $28\ \text{m/s}$. By conservation of energy the launch kinetic energy equals the gravitational potential energy at the top, so the speed equals $\sqrt{2gh}=\sqrt{2\times 10\times 40}=\sqrt{800}\approx 28\ \text{m/s}$. The mass cancels out, so it does not affect the launch speed needed to reach that height.

172. On a frictionless track a cart at the top of a hill has only gravitational potential energy, and at the bottom it has only kinetic energy. Which statement correctly compares the cart's total mechanical energy at the two points?
     • The total mechanical energy is greater at the top
     • The total mechanical energy is greater at the bottom
     • The total mechanical energy is the same at both points
     • The total mechanical energy is zero at the bottom

     Correct answer: The total mechanical energy is the same at both points

     The total mechanical energy is the same at both points. On a frictionless track no energy is lost to heat, so the sum of kinetic and potential energy stays constant; the energy simply changes form from potential at the top to kinetic at the bottom. This constant total is the defining feature of conservation of mechanical energy.

173. A $2.0\ \text{kg}$ object is moving at $3.0\ \text{m/s}$ when a constant net force does an additional $16\ \text{J}$ of positive work on it along its direction of motion. What is its new speed?
     • $8.0\ \text{m/s}$
     • $7.0\ \text{m/s}$
     • $4.0\ \text{m/s}$
     • $5.0\ \text{m/s}$

     Correct answer: $5.0\ \text{m/s}$

     The object's new speed is $5.0\ \text{m/s}$. Its initial kinetic energy is $\frac{1}{2}\times 2.0\times 3.0^{2}=9.0\ \text{J}$, and adding the $16\ \text{J}$ of work gives a final kinetic energy of $25\ \text{J}$ by the work-energy theorem. Setting $25\ \text{J}=\frac{1}{2}\times 2.0\times v^{2}$ gives $v^{2}=25$ and a final speed of $5.0\ \text{m/s}$.

174. A heavy crate is pushed at constant velocity across a rough horizontal floor. Comparing the work done by the applied push to the work done by friction over the same displacement, which statement is correct?
     • The push and friction do equal magnitudes of work with opposite signs
     • The push does more positive work than friction does negative work
     • Friction does no work because the crate moves at constant velocity
     • The push does negative work and friction does positive work

     Correct answer: The push and friction do equal magnitudes of work with opposite signs

     The push and friction do equal magnitudes of work with opposite signs. Because the crate moves at constant velocity, its kinetic energy does not change, so by the work-energy theorem the net work is zero; this requires the positive work of the push to be exactly canceled by the negative work of friction. The two contributions therefore have equal magnitude but opposite sign.

175. A $1.0\ \text{kg}$ ball is held at rest $5.0\ \text{m}$ above the ground and then released. Taking $g=10\ \text{m/s}^2$ and the ground as the reference level, what is the ball's kinetic energy after it has fallen $2.0\ \text{m}$?
     • $50\ \text{J}$
     • $20\ \text{J}$
     • $30\ \text{J}$
     • $10\ \text{J}$

     Correct answer: $20\ \text{J}$

     After falling $2.0\ \text{m}$ the ball has $20\ \text{J}$ of kinetic energy. As it falls, the gravitational potential energy lost converts to kinetic energy, and the energy lost over a $2.0\ \text{m}$ drop is $1.0\times 10\times 2.0=20\ \text{J}$. Only the distance actually fallen matters here, so the full $5.0\ \text{m}$ starting height is not needed to find the kinetic energy at this point.

176. A force of $200\ \text{N}$ is applied to lift a load, and the force acts through a distance of $3.0\ \text{m}$ in the direction of the load's motion over a time of $5.0\ \text{s}$. What average power does the force deliver?
     • $3000\ \text{W}$
     • $600\ \text{W}$
     • $120\ \text{W}$
     • $40\ \text{W}$

     Correct answer: $120\ \text{W}$

     The force delivers an average power of $120\ \text{W}$. The work done is the force times the distance, $200\ \text{N}\times 3.0\ \text{m}=600\ \text{J}$, and dividing this work by the $5.0\ \text{s}$ time gives an average power of $120\ \text{W}$. Stopping at the $600\ \text{J}$ of work would report the total energy rather than the rate at which it is delivered.

177. A $2.0\ \text{kg}$ object falls freely from rest, and at one instant it is moving downward at $8.0\ \text{m/s}$. What is its kinetic energy at that instant?
     • $16\ \text{J}$
     • $32\ \text{J}$
     • $128\ \text{J}$
     • $64\ \text{J}$

     Correct answer: $64\ \text{J}$

     The object's kinetic energy at that instant is $64\ \text{J}$. Kinetic energy equals $\frac{1}{2}mv^{2}$, so $\frac{1}{2}\times 2.0\times 8.0^{2}=\frac{1}{2}\times 2.0\times 64=64\ \text{J}$. The fact that the object is in free fall does not change the calculation, since kinetic energy depends only on the mass and the current speed.

178. Linear momentum is best classified as which type of physical quantity?
     • A vector quantity having both magnitude and direction
     • A scalar quantity having only magnitude
     • A quantity that has direction but no magnitude
     • A unitless ratio describing how fast mass changes

     Correct answer: A vector quantity having both magnitude and direction

     Linear momentum is a vector quantity having both magnitude and direction. Because momentum is mass times velocity and velocity is itself a vector, momentum points in the same direction as the object's velocity and must be combined using vector rules. Kinetic energy, by contrast, is a scalar with only magnitude, which is why momentum and energy behave differently when motions point in opposite directions.

179. What are the SI units of linear momentum?
     • Newtons per second
     • Joules
     • Kilogram-meters per second
     • Kilogram-meters per second squared

     Correct answer: Kilogram-meters per second

     The SI units of momentum are kilogram-meters per second, $\text{kg}\cdot\text{m/s}$. Momentum equals mass times velocity, so multiplying the kilogram unit of mass by the meters-per-second unit of velocity gives $\text{kg}\cdot\text{m/s}$. Joules are units of energy and $\text{kg}\cdot\text{m/s}^2$ is the newton, the unit of force, so neither describes momentum.

180. Two objects have the same kinetic energy, but object X has twice the mass of object Y. Which object has the greater magnitude of momentum?
     • Object Y, because lighter objects always carry more momentum
     • Object X, the more massive object
     • They have equal momentum because their kinetic energies are equal
     • Neither, momentum cannot be compared from kinetic energy

     Correct answer: Object X, the more massive object

     Object X, the more massive object, has the greater momentum. For a given kinetic energy, momentum magnitude increases with $\sqrt{m}$ because momentum can be written as $p=\sqrt{2m\,KE}$. Since object X has twice the mass at the same kinetic energy, its momentum is larger by a factor of $\sqrt{2}$ even though its speed is lower.

181. The area under a force versus time graph for a single object represents which physical quantity?
     • The work done on the object
     • The impulse delivered to the object
     • The average power supplied to the object
     • The kinetic energy of the object

     Correct answer: The impulse delivered to the object

     The area under a force versus time graph represents the impulse delivered to the object. Impulse is defined as force accumulated over a time interval, which is exactly what the area beneath the curve measures. Work is the area under a force versus position graph instead, so it would require distance on the horizontal axis rather than time.

182. A constant net force of $12\ \text{N}$ acts on a $3.0\ \text{kg}$ object initially at rest for $4.0\ \text{s}$ along a frictionless surface. What is the object's final speed?
     • $16\ \text{m/s}$
     • $1.0\ \text{m/s}$
     • $9.0\ \text{m/s}$
     • $48\ \text{m/s}$

     Correct answer: $16\ \text{m/s}$

     The final speed is $16\ \text{m/s}$. The impulse equals the force of $12\ \text{N}$ times the $4.0\ \text{s}$ interval, giving $48\ \text{kg}\cdot\text{m/s}$, and by the impulse-momentum theorem this equals the change in momentum; dividing $48\ \text{kg}\cdot\text{m/s}$ by the $3.0\ \text{kg}$ mass gives a speed of $16\ \text{m/s}$ from rest. Forgetting to divide by mass would wrongly leave 48 as the answer.

183. An airbag protects a passenger during a crash primarily because, compared with hitting the rigid dashboard, it does which of the following?
     • Decreases the passenger's total change in momentum
     • Increases the contact time, reducing the average force on the passenger
     • Increases the impulse delivered to the passenger
     • Converts the passenger's momentum directly into heat

     Correct answer: Increases the contact time, reducing the average force on the passenger

     An airbag works by increasing the contact time, which reduces the average force on the passenger. The passenger must undergo the same change in momentum to stop regardless of what they hit, and because impulse equals average force times time, stretching that momentum change over a longer collision time lowers the force. The total momentum change is fixed by the crash speed, so it is the longer time, not a smaller momentum change, that protects the passenger.

184. A $0.30\ \text{kg}$ ball strikes a wall horizontally at $6.0\ \text{m/s}$ and bounces straight back at $4.0\ \text{m/s}$. Taking the incoming direction as positive, what is the impulse delivered to the ball by the wall?
     • $0.60\ \text{kg}\cdot\text{m/s}$ in the incoming direction
     • $3.0\ \text{kg}\cdot\text{m/s}$ in the incoming direction
     • $3.0\ \text{kg}\cdot\text{m/s}$ opposite the incoming direction
     • $0.60\ \text{kg}\cdot\text{m/s}$ opposite the incoming direction

     Correct answer: $3.0\ \text{kg}\cdot\text{m/s}$ opposite the incoming direction

     The impulse is $3.0\ \text{kg}\cdot\text{m/s}$ directed opposite the incoming direction. The ball's velocity changes from $+6.0\ \text{m/s}$ to $-4.0\ \text{m/s}$, a change of $-10\ \text{m/s}$, and multiplying this by the $0.30\ \text{kg}$ mass gives an impulse of $-3.0\ \text{kg}\cdot\text{m/s}$, meaning it points back the way the ball came. Treating the bounce as only a $2.0\ \text{m/s}$ change would incorrectly give $0.60\ \text{kg}\cdot\text{m/s}$.

185. An astronaut floating at rest in deep space throws a wrench away from herself. Which statement best describes what happens to the astronaut?
     • She remains at rest because she is much more massive than the wrench
     • She drifts in the same direction as the thrown wrench
     • She drifts in the direction opposite the thrown wrench
     • She accelerates continuously after releasing the wrench

     Correct answer: She drifts in the direction opposite the thrown wrench

     The astronaut drifts in the direction opposite the thrown wrench. The astronaut-wrench system starts with zero total momentum and no external force acts, so the total momentum must stay zero; giving the wrench forward momentum requires the astronaut to gain equal and opposite momentum. She moves slowly because of her larger mass, but once the wrench leaves her hand there is no further force, so she drifts at constant velocity rather than continuing to accelerate.

186. A $60\ \text{kg}$ child runs at $4.0\ \text{m/s}$ and jumps onto a stationary $20\ \text{kg}$ sled on frictionless ice. What is the speed of the child and sled together immediately after?
     • $4.0\ \text{m/s}$
     • $3.0\ \text{m/s}$
     • $1.0\ \text{m/s}$
     • $12\ \text{m/s}$

     Correct answer: $3.0\ \text{m/s}$

     The child and sled move at $3.0\ \text{m/s}$ together. Momentum is conserved, so the initial momentum of $60\ \text{kg}\times 4.0\ \text{m/s}=240\ \text{kg}\cdot\text{m/s}$ must equal the final momentum carried by the combined $80\ \text{kg}$ mass; dividing $\frac{240}{80}=3.0\ \text{m/s}$. The speed is below the original $4.0\ \text{m/s}$ because the same momentum is now shared by a greater total mass.

187. In an isolated system of two interacting carts, the total momentum is conserved fundamentally because of which principle?
     • Newton's first law, since objects at rest stay at rest
     • Newton's third law, since the carts exert equal and opposite forces on each other
     • The work-energy theorem applied to each cart
     • Conservation of kinetic energy during the interaction

     Correct answer: Newton's third law, since the carts exert equal and opposite forces on each other

     Momentum conservation follows from Newton's third law, since the carts exert equal and opposite forces on each other. These internal forces produce equal and opposite impulses over the same contact time, so the momentum gained by one cart is exactly the momentum lost by the other and the total stays constant. Kinetic energy need not be conserved during the interaction, so that principle does not explain why momentum is preserved.

188. A firecracker initially at rest on a frictionless surface explodes into exactly two unequal pieces. Which statement about the two fragments is correct?
     • Both fragments fly off in the same direction
     • The fragments have equal and opposite momenta
     • The fragments have equal speeds in opposite directions
     • The heavier fragment carries more momentum than the lighter one

     Correct answer: The fragments have equal and opposite momenta

     The two fragments have equal and opposite momenta. Because the firecracker started at rest with zero total momentum and the explosion forces are internal, the total momentum afterward must also be zero, requiring the fragments to carry equal momentum in opposite directions. Their speeds are not equal, however, since the lighter fragment must move faster to match the momentum of the heavier one.

189. A $1.0\ \text{kg}$ ball moving at $5.0\ \text{m/s}$ makes a head-on elastic collision with an identical $1.0\ \text{kg}$ ball at rest. After the collision, the incoming ball does what?
     • Continues forward at $5.0\ \text{m/s}$, pushing the second ball ahead
     • Stops, and the second ball moves forward at $5.0\ \text{m/s}$
     • Bounces straight back at $5.0\ \text{m/s}$
     • Moves forward at $2.5\ \text{m/s}$ while the second ball moves at $2.5\ \text{m/s}$

     Correct answer: Stops, and the second ball moves forward at $5.0\ \text{m/s}$

     The incoming ball stops, and the second ball moves forward at $5.0\ \text{m/s}$. In a head-on elastic collision between equal masses where one is initially at rest, the two objects simply exchange velocities, which conserves both momentum and kinetic energy. If the balls instead shared the speed equally at $2.5\ \text{m/s}$ each, momentum would be conserved but kinetic energy would not, so that outcome cannot occur in an elastic collision.

190. Two carts collide on a frictionless track and afterward move off as separate objects with different velocities. Total momentum is conserved and total kinetic energy is also exactly conserved. This collision is best described as which type?
     • Perfectly inelastic
     • Partially inelastic
     • Elastic
     • An explosion

     Correct answer: Elastic

     This collision is elastic. A collision in which both total momentum and total kinetic energy are conserved is by definition perfectly elastic. A perfectly inelastic collision would have the carts stick together and lose kinetic energy, and a partially inelastic collision would conserve momentum but lose some kinetic energy, so neither matches a collision that preserves all the kinetic energy.

191. A $3.0\ \text{kg}$ cart moving right at $2.0\ \text{m/s}$ collides with and sticks to a $1.0\ \text{kg}$ cart moving left at $2.0\ \text{m/s}$ on a frictionless track. What is the velocity of the combined carts after the collision?
     • $1.0\ \text{m/s}$ to the right
     • $2.0\ \text{m/s}$ to the right
     • $1.0\ \text{m/s}$ to the left
     • Zero

     Correct answer: $1.0\ \text{m/s}$ to the right

     The combined carts move at $1.0\ \text{m/s}$ to the right. Taking right as positive, the total momentum before is $3.0\times 2.0+1.0\times(-2.0)=6.0-2.0=4.0\ \text{kg}\cdot\text{m/s}$; dividing this by the combined $4.0\ \text{kg}$ mass gives $1.0\ \text{m/s}$ in the positive, rightward, direction. The result points right because the heavier cart's momentum dominates.

192. On a horizontal frictionless surface, a $2.0\ \text{kg}$ puck moving north at $3.0\ \text{m/s}$ strikes and sticks to a $2.0\ \text{kg}$ puck moving east at $3.0\ \text{m/s}$. What is the direction of the combined pucks immediately afterward?
     • Due north
     • Due east
     • Toward the northeast, halfway between north and east
     • Toward the southwest

     Correct answer: Toward the northeast, halfway between north and east

     The combined pucks move toward the northeast, halfway between north and east. Momentum is conserved separately in the north and east directions, and since the two equal-mass pucks contribute equal momentum components of $6.0\ \text{kg}\cdot\text{m/s}$ each, the resultant momentum points at a $45^{\circ}$ angle. That equal mix of northward and eastward momentum aims the combined object exactly northeast.

193. Two skaters of unequal mass push off from rest on frictionless ice. Which quantity is the same in magnitude for both skaters immediately after they separate?
     • Their speeds
     • Their kinetic energies
     • Their accelerations after separation
     • The magnitudes of their momenta

     Correct answer: The magnitudes of their momenta

     The magnitudes of their momenta are the same for both skaters. Starting from rest with zero total momentum, conservation of momentum forces the two skaters to acquire equal and opposite momenta regardless of their masses. Their speeds differ because the lighter skater must move faster to carry the same momentum, and once separated neither skater experiences a force, so neither accelerates.

194. Two particles with no external forces acting on them form a system whose total momentum is constant. What does this imply about the velocity of the system's center of mass?
     • It is constant, since constant total momentum means a constant center-of-mass velocity
     • It steadily increases as the particles interact
     • It is always zero
     • It depends only on the heavier particle's motion

     Correct answer: It is constant, since constant total momentum means a constant center-of-mass velocity

     The center-of-mass velocity is constant, since constant total momentum means a constant center-of-mass velocity. The total momentum of a system equals its total mass times the velocity of the center of mass, so if total momentum does not change and the total mass is fixed, the center-of-mass velocity cannot change either. The center of mass need not be at rest; it simply moves at a steady velocity when no external force acts.

195. During a collision between two objects, the momentum of object A decreases by $8.0\ \text{kg}\cdot\text{m/s}$. If no external forces act on the two-object system, what happens to the momentum of object B?
     • It also decreases by $8.0\ \text{kg}\cdot\text{m/s}$
     • It increases by $8.0\ \text{kg}\cdot\text{m/s}$
     • It remains unchanged
     • It increases by $16\ \text{kg}\cdot\text{m/s}$

     Correct answer: It increases by $8.0\ \text{kg}\cdot\text{m/s}$

     Object B's momentum increases by $8.0\ \text{kg}\cdot\text{m/s}$. With no external forces, the total momentum of the system is conserved, so any momentum lost by object A must be gained by object B in equal amount. This balance reflects the equal and opposite impulses the objects exert on each other during the collision.

196. A net force is applied to an object for a fixed time interval. Doubling the magnitude of that net force while keeping the time the same has what effect on the impulse delivered?
     • The impulse is unchanged
     • The impulse is halved
     • The impulse is doubled
     • The impulse is quadrupled

     Correct answer: The impulse is doubled

     Doubling the force doubles the impulse. Impulse equals force multiplied by the time over which it acts, $J = F \Delta t$, so with the time interval held constant the impulse is directly proportional to the force. Quadrupling would only occur if both the force and the time were doubled, not the force alone.

197. A grinding wheel rotating at $50\ \text{rad/s}$ is brought to rest with a constant angular acceleration of magnitude $10\ \text{rad/s}^2$. Through what angular displacement does the wheel turn before stopping?
     • $250\ \text{rad}$
     • $5.0\ \text{rad}$
     • $500\ \text{rad}$
     • $125\ \text{rad}$

     Correct answer: $125\ \text{rad}$

     The wheel turns through $125\ \text{rad}$. Using the rotational kinematics equation $\omega^2 = \omega_0^2 + 2\alpha\theta$ with a final $\omega$ of 0, $\theta = -\frac{\omega_0^2}{2\alpha} = -\frac{(50)^2}{2 \times (-10)} = \frac{2500}{20} = 125\ \text{rad}$. This is the rotational analog of $v^2 = v_0^2 + 2ax$.

198. A $5.0\ \text{N}$ force is applied at the end of a $0.40\ \text{m}$ lever arm, directed at an angle of $30^\circ$ to the lever arm. What is the magnitude of the torque produced about the pivot?
     • $2.0\ \text{N}\cdot\text{m}$
     • $1.0\ \text{N}\cdot\text{m}$
     • $1.7\ \text{N}\cdot\text{m}$
     • $0.50\ \text{N}\cdot\text{m}$

     Correct answer: $1.0\ \text{N}\cdot\text{m}$

     The torque is $1.0\ \text{N}\cdot\text{m}$. Torque is $\tau = r F \sin\theta = 0.40\ \text{m} \times 5.0\ \text{N} \times \sin(30^\circ) = 0.40 \times 5.0 \times 0.50 = 1.0\ \text{N}\cdot\text{m}$. Only the component of the force perpendicular to the lever arm contributes to the torque, which is why the $\sin(30^\circ)$ factor of $0.50$ reduces the result.

199. A solid sphere and a hollow sphere have the same mass and the same radius and rotate about an axis through their centers. Which statement correctly compares their moments of inertia?
     • The solid sphere has the larger moment of inertia
     • Both have exactly the same moment of inertia
     • The moment of inertia depends only on the mass, so they are equal
     • The hollow sphere has the larger moment of inertia

     Correct answer: The hollow sphere has the larger moment of inertia

     The hollow sphere has the larger moment of inertia. For the same mass and radius, a hollow sphere concentrates its mass farther from the rotation axis than a solid sphere, and moment of inertia increases as mass is placed farther from the axis. A solid sphere has $I = \frac{2}{5}MR^2$ while a hollow sphere has $I = \frac{2}{3}MR^2$, and $\frac{2}{3}$ is larger than $\frac{2}{5}$.

200. A net torque of $6.0\ \text{N}\cdot\text{m}$ acts on a flywheel, giving it an angular acceleration of $3.0\ \text{rad/s}^2$. What is the moment of inertia of the flywheel?
     • $18\ \text{kg}\cdot\text{m}^2$
     • $0.50\ \text{kg}\cdot\text{m}^2$
     • $9.0\ \text{kg}\cdot\text{m}^2$
     • $2.0\ \text{kg}\cdot\text{m}^2$

     Correct answer: $2.0\ \text{kg}\cdot\text{m}^2$

     The moment of inertia is $2.0\ \text{kg}\cdot\text{m}^2$. Rearranging the rotational form of Newton's second law, $\tau = I\alpha$, gives $I = \frac{\tau}{\alpha} = \frac{6.0\ \text{N}\cdot\text{m}}{3.0\ \text{rad/s}^2} = 2.0\ \text{kg}\cdot\text{m}^2$. This parallels solving $m = \frac{F}{a}$ in linear dynamics.

201. A disk rotating at $8.0\ \text{rad/s}$ undergoes a constant angular acceleration of $2.0\ \text{rad/s}^2$ for $4.0\ \text{s}$. What is its angular velocity at the end of that interval?
     • $10\ \text{rad/s}$
     • $16\ \text{rad/s}$
     • $32\ \text{rad/s}$
     • $4.0\ \text{rad/s}$

     Correct answer: $16\ \text{rad/s}$

     The final angular velocity is $16\ \text{rad/s}$. Using $\omega = \omega_0 + \alpha t = 8.0\ \text{rad/s} + (2.0\ \text{rad/s}^2)(4.0\ \text{s}) = 8.0 + 8.0 = 16\ \text{rad/s}$. This is the rotational analog of $v = v_0 + at$.

202. A uniform horizontal beam of weight $200\ \text{N}$ and length $3.0\ \text{m}$ is supported at its left end by a hinge. A single upward support force at the right end holds it level. Treating the beam's weight as acting at its center, what upward force at the right end keeps the beam in rotational equilibrium about the hinge?
     • $200\ \text{N}$
     • $50\ \text{N}$
     • $100\ \text{N}$
     • $400\ \text{N}$

     Correct answer: $100\ \text{N}$

     The upward support force must be $100\ \text{N}$. Taking torques about the hinge, the beam's weight acts at the center ($1.5\ \text{m}$ from the hinge) producing a torque of $200\ \text{N} \times 1.5\ \text{m} = 300\ \text{N}\cdot\text{m}$, and the support force at $3.0\ \text{m}$ must balance it: $F \times 3.0\ \text{m} = 300\ \text{N}\cdot\text{m}$, so $F = 100\ \text{N}$. Choosing the hinge as the pivot eliminates the unknown hinge force from the torque equation.

203. Two children sit on opposite ends of a balanced uniform seesaw. The mass and weight of the seesaw itself are placed symmetrically about the central pivot. Why does the weight of the seesaw produce no net torque about that pivot?
     • Its weight acts at the center of mass, located at the pivot, giving zero lever arm
     • Its weight is canceled by the normal force from the ground
     • The seesaw is too rigid to rotate under its own weight
     • Weight never produces torque on a horizontal object

     Correct answer: Its weight acts at the center of mass, located at the pivot, giving zero lever arm

     The seesaw's weight produces no net torque because its weight acts at its center of mass, which lies at the pivot, giving a lever arm of zero. Torque equals force times the perpendicular distance from the axis, so a force whose line of action passes through the pivot has zero lever arm and zero torque, regardless of how large the weight is.

204. A point on a rotating wheel is located $0.30\ \text{m}$ from the axis. If the wheel has an angular acceleration of $4.0\ \text{rad/s}^2$, what is the tangential (linear) acceleration of that point?
     • $13\ \text{m/s}^2$
     • $0.075\ \text{m/s}^2$
     • $1.2\ \text{m/s}^2$
     • $4.0\ \text{m/s}^2$

     Correct answer: $1.2\ \text{m/s}^2$

     The tangential acceleration is $1.2\ \text{m/s}^2$. Tangential acceleration relates to angular acceleration by $a_t = r\alpha = 0.30\ \text{m} \times 4.0\ \text{rad/s}^2 = 1.2\ \text{m/s}^2$. This is distinct from centripetal acceleration, and it describes how fast the point's tangential speed is changing.

205. A massless rod has two point masses attached: a $2.0\ \text{kg}$ mass at $1.0\ \text{m}$ from the axis and a $3.0\ \text{kg}$ mass at $2.0\ \text{m}$ from the same axis. What is the total moment of inertia of the system about that axis?
     • $14\ \text{kg}\cdot\text{m}^2$
     • $10\ \text{kg}\cdot\text{m}^2$
     • $8.0\ \text{kg}\cdot\text{m}^2$
     • $5.0\ \text{kg}\cdot\text{m}^2$

     Correct answer: $14\ \text{kg}\cdot\text{m}^2$

     The total moment of inertia is $14\ \text{kg}\cdot\text{m}^2$. For point masses, $I = \sum m r^2$, so $I = (2.0\ \text{kg})(1.0\ \text{m})^2 + (3.0\ \text{kg})(2.0\ \text{m})^2 = 2.0 + 12 = 14\ \text{kg}\cdot\text{m}^2$. Each mass contributes according to the square of its distance, so the more distant mass dominates the total.

206. Two forces act on a rod that can rotate about a fixed pivot. One produces a clockwise torque of $8.0\ \text{N}\cdot\text{m}$ and the other produces a counterclockwise torque of $5.0\ \text{N}\cdot\text{m}$. What is the magnitude and sense of the net torque on the rod?
     • $13\ \text{N}\cdot\text{m}$ counterclockwise
     • $3.0\ \text{N}\cdot\text{m}$ counterclockwise
     • $40\ \text{N}\cdot\text{m}$ clockwise
     • $3.0\ \text{N}\cdot\text{m}$ clockwise

     Correct answer: $3.0\ \text{N}\cdot\text{m}$ clockwise

     The net torque is $3.0\ \text{N}\cdot\text{m}$ clockwise. Torques in opposite rotational senses subtract, so the net torque is $8.0\ \text{N}\cdot\text{m} - 5.0\ \text{N}\cdot\text{m} = 3.0\ \text{N}\cdot\text{m}$, and because the clockwise torque is larger, the net torque acts clockwise. The rod will angularly accelerate in the clockwise direction.

207. A solid cylinder of mass $4.0\ \text{kg}$ and radius $0.50\ \text{m}$ rotates about its central axis. Using $I = \frac{1}{2}MR^2$ for a solid cylinder, what is its moment of inertia?
     • $1.0\ \text{kg}\cdot\text{m}^2$
     • $0.50\ \text{kg}\cdot\text{m}^2$
     • $2.0\ \text{kg}\cdot\text{m}^2$
     • $4.0\ \text{kg}\cdot\text{m}^2$

     Correct answer: $0.50\ \text{kg}\cdot\text{m}^2$

     The moment of inertia is $0.50\ \text{kg}\cdot\text{m}^2$. Substituting into $I = \frac{1}{2}MR^2 = \frac{1}{2}(4.0\ \text{kg})(0.50\ \text{m})^2 = \frac{1}{2}(4.0)(0.25) = 0.50\ \text{kg}\cdot\text{m}^2$. The radius is squared, so the $0.50\ \text{m}$ radius contributes a factor of $0.25$ to the result.

208. A wheel turns through $100\ \text{rad}$ while accelerating uniformly from rest. If it reaches an angular velocity of $20\ \text{rad/s}$, what is the magnitude of its angular acceleration?
     • $0.20\ \text{rad/s}^2$
     • $4.0\ \text{rad/s}^2$
     • $2.0\ \text{rad/s}^2$
     • $10\ \text{rad/s}^2$

     Correct answer: $2.0\ \text{rad/s}^2$

     The angular acceleration is $2.0\ \text{rad/s}^2$. Using $\omega^2 = \omega_0^2 + 2\alpha\theta$ with $\omega_0 = 0$, $\alpha = \frac{\omega^2}{2\theta} = \frac{(20)^2}{2 \times 100} = \frac{400}{200} = 2.0\ \text{rad/s}^2$. This rotational equation mirrors the linear kinematics relation $v^2 = v_0^2 + 2ax$.

209. A solid disk of moment of inertia $0.80\ \text{kg}\cdot\text{m}^2$ is at rest. A constant net torque of $1.6\ \text{N}\cdot\text{m}$ is applied for $3.0\ \text{s}$. What is the angular velocity of the disk at the end of that time?
     • $6.0\ \text{rad/s}$
     • $2.0\ \text{rad/s}$
     • $1.5\ \text{rad/s}$
     • $4.8\ \text{rad/s}$

     Correct answer: $6.0\ \text{rad/s}$

     The angular velocity is $6.0\ \text{rad/s}$. First find the angular acceleration from $\tau = I\alpha$, so $\alpha = \frac{1.6\ \text{N}\cdot\text{m}}{0.80\ \text{kg}\cdot\text{m}^2} = 2.0\ \text{rad/s}^2$. Then apply $\omega = \omega_0 + \alpha t = 0 + (2.0\ \text{rad/s}^2)(3.0\ \text{s}) = 6.0\ \text{rad/s}$. This combines the rotational second law with rotational kinematics.

210. A bicycle wheel of radius $0.35\ \text{m}$ rolls without slipping. If its angular velocity is $10\ \text{rad/s}$, what is the linear speed of the wheel's center across the ground?
     • $0.035\ \text{m/s}$
     • $29\ \text{m/s}$
     • $3.5\ \text{m/s}$
     • $10\ \text{m/s}$

     Correct answer: $3.5\ \text{m/s}$

     The center moves at $3.5\ \text{m/s}$. For rolling without slipping, the linear speed of the center equals $v = r\omega = 0.35\ \text{m} \times 10\ \text{rad/s} = 3.5\ \text{m/s}$. The rolling-without-slipping condition links the wheel's rotation directly to its translation across the ground.

211. When a figure skater pulls her outstretched arms inward while spinning, her rate of spin increases noticeably. In terms of rotational quantities, what change explains this within the same physical system?
     • Her moment of inertia decreases as mass moves closer to the rotation axis
     • Her moment of inertia increases as mass moves closer to the rotation axis
     • The net torque on her body increases sharply
     • Her total mass decreases when she pulls her arms in

     Correct answer: Her moment of inertia decreases as mass moves closer to the rotation axis

     Her moment of inertia decreases because pulling her arms inward moves mass closer to the rotation axis. Moment of inertia depends on how mass is distributed relative to the axis, and concentrating mass nearer the axis lowers it. With less rotational inertia, conservation of angular momentum produces a faster spin. Her total mass does not change, and no external torque is needed.

212. A uniform ladder leans against a frictionless wall and rests on the ground. Which condition, in addition to zero net force, must be satisfied for the ladder to remain in static equilibrium?
     • The sum of all torques about any chosen axis must be zero
     • The torque from gravity must be greater than all other torques
     • The ladder's angular velocity must be constant and nonzero
     • The moment of inertia of the ladder must equal zero

     Correct answer: The sum of all torques about any chosen axis must be zero

     The sum of all torques about any chosen axis must be zero. Static equilibrium requires both that the net force is zero (translational equilibrium) and that the net torque about any axis is zero (rotational equilibrium). If the net torque were not zero, the ladder would begin to rotate, so balancing the torques is essential.

213. Two identical forces are applied to a door to push it open. Force 1 is applied at the outer edge of the door, far from the hinges, and Force 2 is applied near the hinges. Which force produces the greater torque about the hinge axis, and why?
     • Force 1, because it acts with a longer lever arm from the hinge axis
     • Force 2, because it acts closer to the rotation axis
     • Both produce equal torque because the forces are identical
     • Neither produces torque because the door is rigid

     Correct answer: Force 1, because it acts with a longer lever arm from the hinge axis

     Force 1 produces the greater torque because it acts with a longer lever arm from the hinge axis. Torque equals force times the perpendicular distance from the axis, so for equal forces the one applied farther from the hinges yields more torque. This is why door handles are placed on the edge opposite the hinges.

214. A potter's wheel decelerates uniformly from $30\ \text{rad/s}$ to rest in $12\ \text{s}$. How many radians does it rotate during this deceleration?
     • $360\ \text{rad}$
     • $180\ \text{rad}$
     • $90\ \text{rad}$
     • $2.5\ \text{rad}$

     Correct answer: $180\ \text{rad}$

     The wheel rotates through $180\ \text{rad}$. With constant angular acceleration, the angular displacement equals the average angular velocity times time, $\theta = \frac{\omega_0 + \omega}{2} t = \frac{30 + 0}{2}(12) = 15 \times 12 = 180\ \text{rad}$. Using the average angular velocity avoids needing to compute the angular acceleration first.

215. A net torque is applied to a wheel, but the wheel's angular velocity remains exactly constant. What can be correctly concluded about the situation?
     • The net torque on the wheel must actually be zero
     • The wheel must have an infinite moment of inertia
     • The wheel is accelerating angularly at a constant rate
     • The applied torque is balancing the wheel's angular momentum

     Correct answer: The net torque on the wheel must actually be zero

     The net torque must actually be zero. By the rotational form of Newton's second law, $\tau = I\alpha$, a constant angular velocity means zero angular acceleration, which requires the net torque to be zero. If a torque is applied yet the spin stays constant, an opposing torque (such as friction) must be canceling it so that the net torque is zero.

216. A liquid has a density of $800\ \text{kg/m}^3$. What volume of this liquid has a mass of $2.0\ \text{kg}$?
     • $2.5 \times 10^{-3}\ \text{m}^3$
     • $1600\ \text{m}^3$
     • $0.40\ \text{m}^3$
     • $400\ \text{m}^3$

     Correct answer: $2.5 \times 10^{-3}\ \text{m}^3$

     The volume is $2.5 \times 10^{-3}\ \text{m}^3$. Since density equals mass divided by volume, volume equals mass divided by density, $V = \frac{m}{\rho} = \frac{2.0\ \text{kg}}{800\ \text{kg/m}^3} = 0.0025\ \text{m}^3$. Multiplying mass by density instead would wrongly give $1600$, but volume comes from dividing, not multiplying.

217. The pressure exerted by a fluid on a surface is defined as which of the following?
     • The total force the fluid exerts on the surface
     • The force the fluid exerts per unit area of the surface
     • The force the fluid exerts per unit volume
     • The energy the fluid stores per unit area

     Correct answer: The force the fluid exerts per unit area of the surface

     Pressure is the force the fluid exerts per unit area of the surface. By definition, $P = \frac{F}{A}$, the perpendicular force divided by the area over which it acts, which is why the SI unit is the pascal, equal to one newton per square meter. Total force alone is not pressure because the same force spread over a larger area gives a smaller pressure.

218. A tank holds oil of density $900\ \text{kg/m}^3$. Taking $g = 10\ \text{m/s}^2$, what is the gauge pressure due to the oil at a depth of $2.0\ \text{m}$ below its surface?
     • $180\ \text{Pa}$
     • $1800\ \text{Pa}$
     • $18000\ \text{Pa}$
     • $45000\ \text{Pa}$

     Correct answer: $18000\ \text{Pa}$

     The gauge pressure is $18000\ \text{Pa}$. The pressure from a column of fluid equals density times $g$ times depth, $P = \rho g h = 900 \times 10 \times 2.0 = 18000\ \text{Pa}$. The smaller values come from dropping factors of ten, but keeping all three factors gives the correct $18000\ \text{Pa}$.

219. A flat plate of area $0.50\ \text{m}^2$ lies horizontally at a depth where the surrounding fluid pressure is $40000\ \text{Pa}$. What is the magnitude of the force the fluid exerts on one face of the plate due to this pressure?
     • $80000\ \text{N}$
     • $20000\ \text{N}$
     • $40000\ \text{N}$
     • $0.50\ \text{N}$

     Correct answer: $20000\ \text{N}$

     The force is $20000\ \text{N}$. Since pressure is force per unit area, the force equals pressure times area, $F = PA = 40000\ \text{Pa} \times 0.50\ \text{m}^2 = 20000\ \text{N}$. Dividing pressure by area instead would be incorrect because force is the product of pressure and the area it acts on.

220. An object is fully submerged and floats in equilibrium in the middle of a fluid, neither rising nor sinking. How does the object's average density compare with the fluid's density?
     • The object's density is greater than the fluid's density
     • The object's density is less than the fluid's density
     • The object's density equals the fluid's density
     • The object's density is zero

     Correct answer: The object's density equals the fluid's density

     The object's density equals the fluid's density. For an object suspended in equilibrium while fully submerged, the buoyant force exactly balances its weight, and since both depend on the same volume, this balance requires the object and fluid to share the same density. If the object were denser it would sink, and if it were less dense it would rise to the surface.

221. A wooden raft of volume $0.30\ \text{m}^3$ floats in water of density $1000\ \text{kg/m}^3$ with $0.20\ \text{m}^3$ of its volume below the surface. Taking $g = 10\ \text{m/s}^2$, what is the buoyant force on the raft?
     • $3000\ \text{N}$
     • $2000\ \text{N}$
     • $5000\ \text{N}$
     • $200\ \text{N}$

     Correct answer: $2000\ \text{N}$

     The buoyant force is $2000\ \text{N}$. The buoyant force equals the weight of the displaced water, which depends only on the submerged volume, $F_b = \rho g V_{sub} = 1000 \times 10 \times 0.20 = 2000\ \text{N}$. Using the raft's full $0.30\ \text{m}^3$ would overstate the displaced water, since only the submerged part displaces fluid.

222. A solid object weighs $60\ \text{N}$ in air and has an apparent weight of $45\ \text{N}$ when fully submerged in water of density $1000\ \text{kg/m}^3$. Taking $g = 10\ \text{m/s}^2$, what is the volume of the object?
     • $1.5 \times 10^{-3}\ \text{m}^3$
     • $6.0 \times 10^{-3}\ \text{m}^3$
     • $4.5 \times 10^{-3}\ \text{m}^3$
     • $15 \times 10^{-3}\ \text{m}^3$

     Correct answer: $1.5 \times 10^{-3}\ \text{m}^3$

     The volume is $1.5 \times 10^{-3}\ \text{m}^3$. The buoyant force equals the apparent weight loss, $60\ \text{N} - 45\ \text{N} = 15\ \text{N}$, and since $F_b = \rho g V$, the volume is $V = \frac{15}{1000 \times 10} = 0.0015\ \text{m}^3$. Using the full $60\ \text{N}$ weight instead of the $15\ \text{N}$ buoyant force would give a volume four times too large.

223. A beaker of water sits on a scale. A metal ball hanging from a string is then lowered until it is fully submerged in the water without touching the bottom or sides. What happens to the scale reading?
     • It increases
     • It decreases
     • It stays the same
     • It drops to zero

     Correct answer: It increases

     The scale reading increases. By Newton's third law, the water pushes up on the ball with the buoyant force, so the ball pushes down on the water with an equal and opposite force, and this extra downward force is transmitted to the scale. The added reading equals the buoyant force, which is the weight of the water the ball displaces.

224. A piece of ice floats in a glass of water with part of it above the surface. As the ice melts completely, what happens to the water level in the glass, assuming no spilling?
     • It rises noticeably
     • It drops noticeably
     • It stays the same
     • It rises then falls below the start

     Correct answer: It stays the same

     The water level stays the same. While floating, the ice displaces a volume of water whose weight equals the ice's weight, and when the ice melts it produces exactly that same mass of water, which fills precisely the volume that had been displaced. Because the melted water occupies the displaced volume, the overall level does not change.

225. A hydraulic system has an input piston of area $0.0020\ \text{m}^2$ and an output piston of area $0.080\ \text{m}^2$. A $30\ \text{N}$ force is applied to the input piston. What force is exerted by the output piston?
     • $0.75\ \text{N}$
     • $30\ \text{N}$
     • $120\ \text{N}$
     • $1200\ \text{N}$

     Correct answer: $1200\ \text{N}$

     The output force is $1200\ \text{N}$. Pascal's principle says the pressure is transmitted equally, so the output force equals the input force times the ratio of output area to input area, $F_{out} = 30 \times \frac{0.080}{0.0020} = 30 \times 40 = 1200\ \text{N}$. The large area ratio of $40$ is what multiplies the modest input force into a much larger output force.

226. In a hydraulic lift, the input piston is pushed down a distance of $0.40\ \text{m}$ and the output piston rises only $0.010\ \text{m}$. Ignoring losses, how does the output force compare with the input force?
     • The output force is 40 times the input force
     • The output force equals the input force
     • The output force is $\frac{1}{40}$ of the input force
     • The output force is 0.40 times the input force

     Correct answer: The output force is 40 times the input force

     The output force is 40 times the input force. Because an ideal hydraulic system conserves energy, the work done on the input equals the work done by the output, so force times distance is the same on both sides. Since the output moves only $\frac{1}{40}$ of the input distance ($\frac{0.40}{0.010} = 40$), its force must be 40 times larger to keep the work equal.

227. Water flows steadily through a horizontal pipe that branches and then rejoins. The volume flow rate entering the pipe is $0.060\ \text{m}^3\text{/s}$. If one branch carries $0.025\ \text{m}^3\text{/s}$, what is the flow rate in the other branch?
     • $0.085\ \text{m}^3\text{/s}$
     • $0.035\ \text{m}^3\text{/s}$
     • $0.025\ \text{m}^3\text{/s}$
     • $0.060\ \text{m}^3\text{/s}$

     Correct answer: $0.035\ \text{m}^3\text{/s}$

     The other branch carries $0.035\ \text{m}^3\text{/s}$. By conservation of mass for an incompressible fluid, the total flow rate entering must equal the sum of the flow rates in the two branches, $0.060 - 0.025 = 0.035\ \text{m}^3\text{/s}$. Adding the branch flow to the inlet would violate continuity, since the flow splits rather than accumulates.

228. An incompressible fluid flows through a horizontal pipe whose radius doubles from a narrow section to a wide section. How does the fluid's speed in the wide section compare with its speed in the narrow section?
     • It is twice as fast
     • It is half as fast
     • It is one-fourth as fast
     • It is four times as fast

     Correct answer: It is one-fourth as fast

     The speed is one-fourth as fast in the wide section. The cross-sectional area depends on the square of the radius ($A = \pi r^2$), so doubling the radius makes the area four times larger, and the continuity equation $A_1 v_1 = A_2 v_2$ requires speed to drop by the same factor that area grows. With four times the area, the fluid slows to one-fourth of its original speed.

229. Air moves rapidly across the top of an open chimney while the air inside the room below is nearly still. Based on Bernoulli's principle, what effect does the fast-moving air at the top have?
     • It raises the pressure at the top, pushing air down the chimney
     • It lowers the pressure at the top, helping draw air up the chimney
     • It has no effect on pressure because air is so light
     • It increases the pressure equally at the top and bottom

     Correct answer: It lowers the pressure at the top, helping draw air up the chimney

     The fast-moving air lowers the pressure at the top, helping draw air up the chimney. Bernoulli's principle states that faster-moving fluid has lower pressure, so the rapid air across the opening creates a low-pressure region above. The higher pressure of the still air below then pushes upward toward this low-pressure region, drawing air up and out.

230. An ideal fluid of density $1000\ \text{kg/m}^3$ flows through a horizontal pipe. At a point where the speed is $2.0\ \text{m/s}$ the pressure is $150000\ \text{Pa}$. At a wider point the speed drops to $0\ \text{m/s}$. Using Bernoulli's equation, what is the pressure at the wider point?
     • $148000\ \text{Pa}$
     • $152000\ \text{Pa}$
     • $150000\ \text{Pa}$
     • $100000\ \text{Pa}$

     Correct answer: $152000\ \text{Pa}$

     The pressure at the wider point is $152000\ \text{Pa}$. For a horizontal flow, Bernoulli's equation makes the sum of pressure and the kinetic-energy term per volume constant, and the kinetic term at the first point is $\frac{1}{2}\rho v^2 = \frac{1}{2}(1000)(2.0)^2 = 2000\ \text{Pa}$. Since the speed drops to zero, all of that $2000\ \text{Pa}$ converts to pressure, raising it from $150000\ \text{Pa}$ to $152000\ \text{Pa}$.

231. A large open tank of water has a small hole in its side at a depth of $5.0\ \text{m}$ below the water surface. Taking $g = 10\ \text{m/s}^2$ and treating the water as an ideal fluid, what is the speed of the water as it exits the hole?
     • $5.0\ \text{m/s}$
     • $10\ \text{m/s}$
     • $50\ \text{m/s}$
     • $100\ \text{m/s}$

     Correct answer: $10\ \text{m/s}$

     The exit speed is $10\ \text{m/s}$. Applying Bernoulli's equation between the open surface and the hole gives an exit speed of $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5.0} = \sqrt{100} = 10\ \text{m/s}$. This result, known as Torricelli's theorem, shows the water exits as if it had fallen freely from the surface.

232. Two liquids that do not mix are poured into a container, and the less dense liquid settles above the denser one. Considering the pressure within the static liquids, why does the denser liquid end up at the bottom?
     • Pressure pushes denser fluid downward against gravity
     • A given height of denser fluid produces more pressure, so it sinks beneath the lighter fluid
     • Density has no effect on which liquid is on top
     • The denser liquid is pushed up by the larger buoyant force

     Correct answer: A given height of denser fluid produces more pressure, so it sinks beneath the lighter fluid

     The denser liquid settles at the bottom because a given height of it produces more pressure, so it sinks beneath the lighter fluid. Since fluid weight per unit volume increases with density, the heavier fluid is not held up by the lighter fluid above it and settles below. The system reaches its lowest energy state with the denser fluid underneath the less dense one.

233. A balloon filled with a gas is fully submerged and released at the bottom of a deep tank of water. As it rises toward the surface, the water pressure on it decreases and the gas expands, increasing the balloon's volume while its mass stays constant. How does the buoyant force on the balloon change as it rises?
     • It increases because the displaced volume increases
     • It decreases because the water pressure decreases
     • It stays exactly the same at all depths
     • It becomes zero as soon as the balloon starts moving

     Correct answer: It increases because the displaced volume increases

     The buoyant force increases as the balloon rises. The buoyant force equals the weight of the displaced water, which depends on the volume of water pushed aside, and as the gas expands the balloon displaces more water. Even though the surrounding pressure is dropping, what determines buoyancy is the displaced volume, and that volume grows as the balloon swells.

234. A solid sphere is placed in oil and floats with exactly half its volume submerged. The density of the oil is $800\ \text{kg/m}^3$. What is the density of the sphere?
     • $1600\ \text{kg/m}^3$
     • $800\ \text{kg/m}^3$
     • $400\ \text{kg/m}^3$
     • $200\ \text{kg/m}^3$

     Correct answer: $400\ \text{kg/m}^3$

     The sphere's density is $400\ \text{kg/m}^3$. For a floating object, the fraction of volume submerged equals the ratio of the object's density to the fluid's density, so with half submerged the sphere's density is half the oil's density, $0.5 \times 800 = 400\ \text{kg/m}^3$. The sphere must be less dense than the oil to float, which rules out the larger values.