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

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.

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.

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.

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.

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.

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.

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.

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.

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 $10\ \text{m}$.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $5.0\ \text{m/s}^2$.

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.

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.

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.

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 $0.25$.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $8.7$.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $6.4\ \text{N}$.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $40\ \text{N}$.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $\sqrt{2 \times 10 \times 1.8} = \sqrt{36} = 6.0\ \text{m/s}$.

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.