AP Physics 1 Quiz: Kinetic And Static Friction

What this quiz covers

This quiz focuses on Kinetic And Static Friction, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $0\,\text{N}$
2. $6\,\text{N}$ (correct answer)
3. $mg$
4. $\mu_s N$

Explanation: This problem tests understanding of static friction and equilibrium. When the block remains at rest despite the applied force, the net force must be zero, meaning all forces balance. The student pulls with 6 N to the right, so static friction must exert 6 N to the left to maintain equilibrium. Static friction adjusts its magnitude (up to a maximum of μₛN) to exactly balance applied forces and prevent motion. Choice A (0 N) incorrectly assumes no friction acts when the object is at rest. The key strategy is: for objects at rest, static friction equals the applied force magnitude (not the maximum possible friction).

Question 2

1. It equals $\mu_s N$
2. It equals $4\,\text{N}$ (correct answer)
3. It is greater than $4\,\text{N}$
4. It is less than $4\,\text{N}$

Explanation: This question assesses understanding of static friction in AP Physics 1, distinguishing it from its maximum value. Static friction equals the applied force when it's less than μ_s N, keeping the object stationary. Kinetic friction would apply if the push exceeded the maximum, but here the block rests. Thus, friction must be exactly 4 N to balance the push. A common distractor is choice A, assuming it equals μ_s N, but that's only at the onset of motion, not necessarily here. A transferable strategy is to use equilibrium conditions to set friction equal to other parallel forces when motion doesn't occur.

Question 3

1. Leftward, because it opposes the cart's motion (correct answer)
2. Rightward, because it opposes the cart's slowing down
3. Rightward, because kinetic friction is greater than static friction
4. Zero, because the cart is already moving

Explanation: This question assesses understanding of kinetic friction in AP Physics 1 during deceleration. Static friction adjusts to prevent motion, but here kinetic friction is active since the cart is sliding. Kinetic friction opposes the velocity, pointing leftward against the rightward motion, contributing to slowing. Even with a rightward pull, if the cart slows, net force is leftward, consistent with leftward friction. A common distractor is choice B, thinking it opposes slowing by pointing right, but friction opposes velocity, not acceleration. A transferable strategy is to analyze net force direction from acceleration and ensure friction aligns with opposing motion.

Question 4

1. The normal force
2. The static friction force magnitude (correct answer)
3. The maximum possible static friction $\mu_s N$
4. The weight $mg$

Explanation: This problem tests static friction in equilibrium conditions. When the box remains at rest under an 8 N rightward push, static friction must provide an equal 8 N force leftward to maintain zero net force. Static friction adjusts its magnitude to match applied forces, up to its maximum value μₛN. The actual static friction (8 N) may be less than the maximum possible static friction. Choice C (μₛN) represents the maximum possible static friction, not necessarily the actual value. The strategy is: for stationary objects, static friction magnitude equals the applied force magnitude, not the maximum possible value.

Question 5

1. It must be $12\,\text{N}$
2. It must be greater than $12\,\text{N}$
3. It must be less than $12\,\text{N}$ (correct answer)
4. It must equal $\mu_s N$

Explanation: This question tests the concept of kinetic and static friction in AP Physics 1, analyzing accelerated motion with kinetic friction. Static friction adjusts up to μ_s N but switches to kinetic once motion begins, with constant magnitude μ_k N opposing the motion. Since the block accelerates to the right under a 12 N pull, the net force is rightward, meaning the applied force exceeds kinetic friction. Thus, kinetic friction must be less than 12 N. Choice B is a distractor, suggesting friction > 12 N, which would cause leftward acceleration, not rightward. For acceleration problems, apply Newton's second law: net force equals mass times acceleration, and isolate friction accordingly.

Question 6

1. It equals the component of gravity down the incline (correct answer)
2. It must be greater than the component of gravity down the incline
3. It equals $\mu_s N$
4. It is zero because the speed is constant

Explanation: This question tests the concept of kinetic and static friction in AP Physics 1, applied to constant-speed motion on an incline. Static friction varies up to μ_s N to prevent sliding, but here kinetic friction is involved since the block is already moving. Kinetic friction has a constant magnitude μ_k N and opposes the motion, balancing other forces for constant speed. Thus, its magnitude equals the downhill component of gravity, mg sin θ, to yield zero net force parallel to the incline. Choice D is a distractor, claiming friction is zero due to constant speed, but friction is necessary to counteract gravity's component. A key strategy is to set net force to zero for constant velocity and solve for the unknown friction force.

Question 7

1. To the right, because it opposes the applied force
2. To the left, because it always points in the direction of motion
3. To the right, because it opposes the sliding motion (correct answer)
4. Zero, because a force to the right cancels friction

Explanation: This question tests the concept of kinetic and static friction in AP Physics 1, highlighting kinetic friction's direction independent of applied forces. Static friction can adjust up to μ_s N to prevent motion, but once sliding occurs, kinetic friction is constant at μ_k N. Importantly, kinetic friction always opposes the direction of the object's velocity, not necessarily the applied force. Here, the block is sliding left, so friction points right to oppose that motion, even with a rightward applied force. Choice B is a distractor that misstates friction as pointing in the direction of motion, which would actually aid rather than resist it. To approach such questions, focus on the velocity vector to determine what friction opposes, regardless of other forces.

Question 8

1. Down the incline
2. Up the incline (correct answer)
3. Perpendicular to the incline
4. Zero because the box is not moving yet

Explanation: This question tests the concept of kinetic and static friction in AP Physics 1, particularly on inclined planes with impending motion. Static friction adjusts its magnitude up to the maximum μ_s N to counteract forces tending to cause motion, such as the component of gravity down the incline. When the box is about to slide down, static friction is at its maximum and directed up the incline to oppose the impending downward motion. Kinetic friction would take over once sliding begins, with a constant magnitude μ_k N opposing the actual motion. Choice A is a distractor, as it incorrectly suggests friction points down the incline, which would accelerate rather than oppose the motion. For these problems, always resolve forces parallel and perpendicular to the surface and consider the direction that opposes potential or actual sliding.

Question 9

1. It is zero because the speed is constant
2. It equals the horizontal pulling force magnitude (correct answer)
3. It equals $mg$
4. It is greater than the maximum static friction

Explanation: This problem tests kinetic friction during constant velocity motion. When an object moves at constant speed, the net force is zero, requiring all forces to balance. The rightward pulling force must equal the leftward kinetic friction force for horizontal equilibrium. Kinetic friction has magnitude μₖN regardless of speed, and this value must equal the applied force for constant velocity. Choice A (zero) incorrectly assumes no friction during constant speed motion, which would cause acceleration. The strategy is: constant velocity requires zero net force, so kinetic friction magnitude must equal the applied force magnitude.

Question 10

1. It must equal the pull's horizontal component (correct answer)
2. It must equal $\mu_s N$ because the sled is moving
3. It must be greater than the pull to keep speed constant
4. It must be zero because acceleration is zero

Explanation: This question tests understanding of kinetic friction at constant speed. When an object moves at constant velocity, the net force must be zero by Newton's first law. For horizontal motion, the horizontal component of the pull must exactly balance kinetic friction. Kinetic friction has magnitude μₖN regardless of speed, but this must equal the pull's horizontal component for equilibrium. Choice B incorrectly uses static friction coefficient, while C violates equilibrium conditions. At constant speed, kinetic friction magnitude equals the horizontal applied force.

Question 11

1. To the left, because friction opposes acceleration
2. To the right, because friction opposes the motion (correct answer)
3. To the left, because kinetic friction points with velocity when speeding up
4. Undetermined without $\mu_k$

Explanation: This question tests understanding of kinetic friction during acceleration. Kinetic friction always opposes the direction of sliding motion, regardless of whether the object accelerates or decelerates. Since the block slides left, kinetic friction points right, opposing this leftward motion. The block speeding up means there's a net leftward force, but kinetic friction still opposes motion by pointing right. Choice C incorrectly suggests friction direction depends on acceleration rather than velocity. Kinetic friction always points opposite to the sliding direction, independent of acceleration.

Question 12

1. $0\,\text{N}$
2. $3.0\,\text{N}$ (correct answer)
3. $mg$
4. $\mu_s N$

Explanation: This question tests understanding of static friction. When an object remains at rest despite an applied force, static friction adjusts to exactly balance that force. Since the block is in equilibrium with a 3.0 N pull to one direction, static friction must provide 3.0 N in the opposite direction to maintain zero net force. Static friction can vary from 0 up to μₛN, adjusting as needed to prevent motion. Choice A (0 N) incorrectly assumes no friction acts when at rest, while choices C and D give general expressions rather than the specific value. When an object is stationary under an applied force, static friction equals the applied force magnitude.

Question 13

1. $10\,\text{N}$ (correct answer)
2. $0\,\text{N}$
3. $\mu_s N$ (must equal the maximum)
4. $\mu_k N$ (since friction always uses $\mu_k$)

Explanation: This question tests understanding of static friction magnitude. When a block remains at rest under an applied force, static friction adjusts to exactly balance that force. With a 10 N pull to the right, static friction must be 10 N to the left for equilibrium. Static friction can range from 0 to μₛN, taking whatever value is needed to prevent motion. Choice C incorrectly assumes static friction always equals its maximum, while D confuses static with kinetic friction. For stationary objects, static friction equals the applied force magnitude, not necessarily its maximum value.

Question 14

1. To the right, because friction always opposes motion
2. To the left, opposing the tendency to move right (correct answer)
3. Upward, to balance the normal force
4. Downward, to balance the weight

Explanation: This question tests understanding of static friction direction. Static friction opposes the tendency to move, not actual motion. With a 2 N force applied to the right, the box tends to move right, so static friction acts to the left to prevent this motion. Static friction adjusts both magnitude and direction to maintain equilibrium. Choice A incorrectly applies the rule for kinetic friction to a stationary object. When a force tries to move a stationary object, static friction points opposite to the applied force direction.

Question 15

1. To the right, because friction opposes the applied push
2. Upward, because friction balances weight
3. To the left, opposite the direction of motion (correct answer)
4. Downward, because friction increases the normal force

Explanation: This question tests understanding of kinetic friction direction. Kinetic friction always opposes the direction of sliding motion between surfaces. Since the crate moves to the right at constant speed, kinetic friction acts to the left, opposing this rightward motion. Unlike static friction which prevents motion, kinetic friction acts on moving objects with magnitude μₖN. Choice A incorrectly relates friction to the push rather than motion, while B and D confuse friction with vertical forces. When an object slides across a surface, kinetic friction points opposite to the velocity direction.

Question 16

1. $0\,\text{N}$
2. $6\,\text{N}$ (correct answer)
3. Greater than $6\,\text{N}$
4. $\mu_k N$

Explanation: This question tests understanding of maximum static friction. When an object is on the verge of slipping, static friction reaches its maximum value μₛN. At this critical point, static friction exactly equals the applied force of 6 N to maintain equilibrium. Below this threshold, static friction adjusts to match applied forces; at the threshold, it cannot increase further. Choice C incorrectly suggests friction exceeds the applied force, which would cause acceleration. When an object is about to slip, static friction equals the applied force and has reached its maximum possible value.

Question 17

1. zero because the book is not moving
2. less than the applied pull
3. equal to the applied pull (correct answer)
4. greater than the applied pull

Explanation: This question assesses understanding of static friction in AP Physics 1, particularly at the threshold of motion. Static friction increases to match the applied force, up to its maximum of μ_s N, to maintain equilibrium. Kinetic friction takes over once motion starts, with a constant value of μ_k N opposing the sliding. Just before the book moves, static friction equals the applied pull to keep the net force zero. A common distractor is choice B, suggesting it's less than the pull, but that would cause acceleration rather than rest. A transferable strategy is to apply Newton's first law, ensuring all forces balance for stationary objects.

Question 18

1. $0\,\text{N}$
2. $6.0\,\text{N}$ (correct answer)
3. Greater than $6.0\,\text{N}$
4. Equal to $\mu_s N$

Explanation: This question assesses understanding of static friction in AP Physics 1, focusing on its role in preventing motion. Static friction adjusts its magnitude to exactly match the applied force, up to a maximum value given by μ_s times the normal force, to keep the object at rest. In contrast, kinetic friction acts with a constant magnitude of μ_k times the normal force when the object is sliding. Here, since the block remains at rest under a 6.0 N pull, the static friction force must equal 6.0 N to balance the applied force. A common distractor is choice C, suggesting it's greater than 6.0 N, but that would imply a net force preventing equilibrium. A transferable strategy is to always draw a free-body diagram and set the net force to zero for objects in equilibrium.

Question 19

1. $f_s=F$
2. $f_s=\mu_k N$
3. $f_s=0$
4. $f_s=\mu_s N$ (correct answer)

Explanation: This problem tests understanding of maximum static friction. When an object is on the verge of sliding, static friction reaches its maximum value of μₛN. At this critical point, static friction no longer adjusts to match the applied force but instead equals its maximum possible value. Before this point, static friction would equal F, but at the threshold of motion, it specifically equals μₛN. Choice A (f = F) would be correct if the box weren't about to move, but misses that we're at the maximum static friction limit. When an object is about to slide, static friction equals μₛN regardless of the applied force magnitude.

Question 20

1. It decreases because friction matches the applied force
2. It increases because kinetic friction is greater than static friction
3. It stays the same because $f_k=\mu_k N$ (correct answer)
4. It becomes zero because the pull is smaller

Explanation: This problem tests understanding of kinetic friction's constant magnitude. Kinetic friction equals μₖN, depending only on the coefficient of kinetic friction and the normal force. As long as the block continues sliding, kinetic friction maintains this constant value regardless of changes in the applied force or sliding speed. The normal force remains unchanged (equal to weight on a horizontal surface), so kinetic friction stays constant. Choice A incorrectly applies static friction behavior to a kinetic situation. Remember that kinetic friction magnitude is independent of the applied force—it depends only on μₖ and N.