AP Physics C Mechanics Quiz: Kinetic And Static Friction
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Kinetic And Static FrictionQuestion 1 of 20

A student wants to determine the coefficient of static friction μs\mu_s between a block and a wooden board. The student places the block on the board and slowly raises one end of the board, increasing the angle of inclination θ\theta until the block just begins to slide. Which of the following expressions represents μs\mu_s in terms of the angle θmax\theta_{max} at which the block starts to slide?

μs=sin⁡θmax\mu_s = \sin\theta_{max}
μs=cos⁡θmax\mu_s = \cos\theta_{max}
μs=tan⁡θmax\mu_s = \tan\theta_{max}
μs=1/tan⁡θmax\mu_s = 1 / \tan\theta_{max}
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Kinetic And Static Friction

Practice Kinetic And Static Friction in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 C Mechanics.

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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.

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Question 1

A student wants to determine the coefficient of static friction μs\mu_s between a block and a wooden board. The student places the block on the board and slowly raises one end of the board, increasing the angle of inclination θ\theta until the block just begins to slide. Which of the following expressions represents μs\mu_s in terms of the angle θmax\theta_{max} at which the block starts to slide?

  1. μs=sin⁡θmax\mu_s = \sin\theta_{max}
  2. μs=cos⁡θmax\mu_s = \cos\theta_{max}
  3. μs=tan⁡θmax\mu_s = \tan\theta_{max} (correct answer)
  4. μs=1/tan⁡θmax\mu_s = 1 / \tan\theta_{max}
Explanation: At the point where the block is about to slide, the static friction force has reached its maximum value, Fs,max=μsFNF_{s,max} = \mu_s F_N. This force must balance the component of gravity parallel to the incline, mgsin⁡θmaxmg \sin\theta_{max}. The normal force is FN=mgcos⁡θmaxF_N = mg \cos\theta_{max}. Setting the forces parallel to the incline equal: mgsin⁡θmax=μs(mgcos⁡θmax)mg \sin\theta_{max} = \mu_s (mg \cos\theta_{max}). The term mgmg cancels from both sides, leaving sin⁡θmax=μscos⁡θmax\sin\theta_{max} = \mu_s \cos\theta_{max}. Solving for μs\mu_s gives μs=sin⁡θmax/cos⁡θmax=tan⁡θmax\mu_s = \sin\theta_{max} / \cos\theta_{max} = \tan\theta_{max}.

Question 2

Consider a block of mass MM on a rough horizontal surface. It is acted upon by a horizontal force FF. If the static friction force acting on the block is fsf_s, which of the following inequalities must be true if the block remains at rest?

  1. fs>μsMgf_s > \mu_s Mg
  2. fs=μsMgf_s = \mu_s Mg
  3. fs=Ff_s = F (correct answer)
  4. fs>Ff_s > F
Explanation: If the block remains at rest, it is in static equilibrium, and the net force on it is zero. The applied horizontal force FF is balanced by the static friction force fsf_s. Therefore, the magnitude of the static friction force must be equal to the magnitude of the applied force, fs=Ff_s = F. This is true as long as FF does not exceed the maximum static friction, F≤μsMgF \le \mu_s Mg. The static friction force adjusts its magnitude to match the applied force.

Question 3

A time-varying horizontal force F(t)=ctF(t) = ct is applied to a block of mass mm resting on a horizontal surface with a coefficient of static friction μs\mu_s. Here, cc is a positive constant. At what time tt will the block begin to move?

  1. t=μsmg/ct = \mu_s mg / c (correct answer)
  2. t=mg/ct = mg / c
  3. t=c/(μsmg)t = c / (\mu_s mg)
  4. t=μsc/(mg)t = \mu_s c / (mg)
Explanation: The block will begin to move when the applied force F(t)F(t) becomes equal to the maximum static friction force, Fs,max=μsFNF_{s,max} = \mu_s F_N. For a block on a horizontal surface, FN=mgF_N = mg. So, the block starts to move when F(t)=μsmgF(t) = \mu_s mg. Since F(t)=ctF(t) = ct, we set ct=μsmgct = \mu_s mg. Solving for tt gives t=μsmg/ct = \mu_s mg / c.

Question 4

A person pushes a 50 kg crate horizontally with a constant force of 200 N on a level floor. The crate accelerates at 1.0 m/s². What is the coefficient of kinetic friction between the crate and the floor?

  1. 0.10
  2. 0.20
  3. 0.31 (correct answer)
  4. 0.41
Explanation: First, apply Newton's second law in the horizontal direction: Fnet=Fapplied−Fk=maF_{net} = F_{applied} - F_k = ma. The kinetic friction force is Fk=Fapplied−ma=200 N−(50 kg)(1.0 m/s2)=150 NF_k = F_{applied} - ma = 200 \, \text{N} - (50 \, \text{kg})(1.0 \, \text{m/s}^2) = 150 \, \text{N}. The magnitude of the kinetic friction force is also given by Fk=μkFNF_k = \mu_k F_N. On a level floor, FN=mg=(50 kg)(9.8 m/s2)=490 NF_N = mg = (50 \, \text{kg})(9.8 \, \text{m/s}^2) = 490 \, \text{N}. Therefore, μk=Fk/FN=150 N/490 N≈0.306\mu_k = F_k / F_N = 150 \, \text{N} / 490 \, \text{N} \approx 0.306, which is closest to 0.31.

Question 5

A 10 kg block rests on a horizontal surface. The coefficient of static friction between the block and the surface is 0.50, and the coefficient of kinetic friction is 0.40. A horizontal force is applied to the block, and its magnitude is slowly increased from zero. At what applied force does the block begin to move?

  1. The block begins to move when the applied force just exceeds 39.2 N.
  2. The block begins to move when the applied force just exceeds 49.0 N. (correct answer)
  3. The block begins to move when the applied force just exceeds 88.2 N.
  4. The block begins to move at any applied force greater than zero, but its acceleration depends on the force.
Explanation: The block begins to move when the applied horizontal force exceeds the maximum force of static friction, Fs,max=μsFNF_{s,max} = \mu_s F_N. On a horizontal surface, the normal force FNF_N is equal to the weight of the block, mgmg. Using g≈9.8 m/s2g \approx 9.8 \, \text{m/s}^2, FN=(10 kg)(9.8 m/s2)=98 NF_N = (10 \, \text{kg})(9.8 \, \text{m/s}^2) = 98 \, \text{N}. Therefore, Fs,max=(0.50)(98 N)=49.0 NF_{s,max} = (0.50)(98 \, \text{N}) = 49.0 \, \text{N}. The block will start to move when the applied force is just greater than 49.0 N.

Question 6

A 10.0 kg10.0\,\text{kg} sled slides on packed snow (sled–snow contact) with μs=0.20\mu_s=0.20 and μk=0.10\mu_k=0.10. A student pulls horizontally with 30 N30\,\text{N} and the sled is already moving. Based on the scenario, calculate the force of kinetic friction acting on the sled.

  1. 9.8 N9.8\,\text{N} (correct answer)
  2. 19.6 N19.6\,\text{N}
  3. 29.4 N29.4\,\text{N}
  4. 3.0 N3.0\,\text{N}
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding of kinetic and static friction and their roles in force dynamics. Kinetic friction acts on moving objects with a constant magnitude equal to μk times the normal force, independent of the object's speed or applied force. In this scenario, a 10.0 kg sled is already moving on packed snow with μk = 0.10, and a student pulls with 30 N horizontally. Choice A (9.8 N) is correct because kinetic friction equals μk × N = μk × mg = 0.10 × 10.0 × 9.8 = 9.8 N, regardless of the 30 N applied force. Choice B (19.6 N) incorrectly uses the static coefficient or doubles the kinetic friction, while choice C uses the applied force magnitude. Students should understand that kinetic friction depends only on the coefficient and normal force, not on the applied force or velocity. Draw free-body diagrams showing all forces and emphasize that kinetic friction opposes motion with constant magnitude.

Question 7

A box is sliding to the right on a horizontal floor with an initial velocity v0v_0. The coefficient of kinetic friction between the box and the floor is μk\mu_k. Which of the following statements correctly describes the kinetic friction force?

  1. The friction force is directed to the right and has a magnitude of μkmg\mu_k mg.
  2. The friction force is directed to the left and has a magnitude of μkmg\mu_k mg. (correct answer)
  3. The friction force is directed to the left and its magnitude decreases as the box slows down.
  4. The friction force is directed to the right and its magnitude decreases as the box slows down.
Explanation: The kinetic friction force always opposes the direction of relative motion. Since the box is sliding to the right, the friction force is directed to the left. The magnitude of the kinetic friction force is given by Fk=μkFNF_k = \mu_k F_N. For an object on a horizontal surface, the normal force FNF_N equals the weight mgmg. Thus, the magnitude of the friction force is μkmg\mu_k mg and is constant as long as the object is in motion and the coefficient of kinetic friction is constant.

Question 8

A 20 kg crate is pulled across a rough horizontal floor by a rope that exerts a force of 100 N at an angle of 37° above the horizontal. If the coefficient of kinetic friction is 0.20, what is the magnitude of the friction force on the crate?

  1. 27.2 N (correct answer)
  2. 39.2 N
  3. 51.2 N
  4. 60.0 N
Explanation: The friction force is Fk=μkFNF_k = \mu_k F_N. First, we must find the normal force FNF_N. The vertical forces must sum to zero. Let the applied force be TT. The upward forces are the normal force and the vertical component of the tension (Tsin⁡θT \sin\theta). The downward force is gravity (mgmg). So, FN+Tsin⁡θ−mg=0F_N + T \sin\theta - mg = 0, which means FN=mg−Tsin⁡θF_N = mg - T \sin\theta. Using g≈9.8 m/s2g \approx 9.8 \, \text{m/s}^2, sin⁡37°≈0.6\sin 37° \approx 0.6: FN=(20)(9.8)−(100)(0.6)=196−60=136 NF_N = (20)(9.8) - (100)(0.6) = 196 - 60 = 136 \, \text{N}. Now, calculate the friction force: Fk=(0.20)(136 N)=27.2 NF_k = (0.20)(136 \, \text{N}) = 27.2 \, \text{N}.

Question 9

Two blocks, A and B, are made of the same material and have the same mass. Block A has a square base with side length LL, and Block B has a rectangular base of length 2L2L and width L/2L/2. Both are placed on the same rough, horizontal surface. How does the maximum static friction force Fs,max,AF_{s,max,A} on Block A compare to the maximum static friction force Fs,max,BF_{s,max,B} on Block B?

  1. Fs,max,A=2Fs,max,BF_{s,max,A} = 2 F_{s,max,B}
  2. Fs,max,A=Fs,max,BF_{s,max,A} = F_{s,max,B} (correct answer)
  3. Fs,max,A=(1/2)Fs,max,BF_{s,max,A} = (1/2) F_{s,max,B}
  4. Fs,max,A=4Fs,max,BF_{s,max,A} = 4 F_{s,max,B}
Explanation: The maximum static friction force is given by Fs,max=μsFNF_{s,max} = \mu_s F_N. The area of the base of Block A is L2L^2 and for Block B is (2L)(L/2)=L2(2L)(L/2) = L^2. Although the areas are the same, the friction force is independent of the contact area. Since both blocks have the same mass, their weight mgmg is the same, and the normal force FNF_N from the horizontal surface is the same for both. They are made of the same material and are on the same surface, so μs\mu_s is the same. Therefore, the maximum static friction force is the same for both blocks.

Question 10

A block of mass mm is pushed against a vertical wall by a horizontal force of magnitude PP. The coefficient of static friction between the block and the wall is μs\mu_s. What is the minimum magnitude of PP required to prevent the block from sliding down the wall?

  1. mgmg
  2. μsmg\mu_s mg
  3. mg/μsmg / \mu_s (correct answer)
  4. mg(1+μs2)1/2mg (1 + \mu_s^2)^{1/2}
Explanation: For the block to be held in place, the upward static friction force must balance the downward gravitational force, Fs=mgF_s = mg. The static friction force cannot exceed its maximum value, Fs≤μsFNF_s \leq \mu_s F_N. The normal force FNF_N is provided by the wall and is equal in magnitude to the applied horizontal force PP, so FN=PF_N = P. The condition for the block not to slide is mg≤μsPmg \leq \mu_s P. Therefore, the minimum force PP required is when mg=μsPmg = \mu_s P, which gives P=mg/μsP = mg / \mu_s.

Question 11

A puck is given an initial speed v0v_0 and slides up a ramp inclined at an angle θ\theta. The coefficient of kinetic friction between the puck and the ramp is μk\mu_k. What is the magnitude of the puck's acceleration as it slides up the ramp?

  1. g(sin⁡θ+μkcos⁡θ)g(\sin\theta + \mu_k \cos\theta) (correct answer)
  2. g(sin⁡θ−μkcos⁡θ)g(\sin\theta - \mu_k \cos\theta)
  3. g(cos⁡θ+μksin⁡θ)g(\cos\theta + \mu_k \sin\theta)
  4. gsin⁡θg\sin\theta
Explanation: When the puck slides up the ramp, two forces act on it parallel to the ramp's surface and directed down the ramp: the component of gravity mgsin⁡θmg\sin\theta and the kinetic friction force FkF_k. The normal force is FN=mgcos⁡θF_N = mg\cos\theta, so the friction force is Fk=μkFN=μkmgcos⁡θF_k = \mu_k F_N = \mu_k mg\cos\theta. The net force down the ramp is Fnet=mgsin⁡θ+μkmgcos⁡θF_{net} = mg\sin\theta + \mu_k mg\cos\theta. According to Newton's second law, Fnet=maF_{net} = ma. Therefore, ma=mgsin⁡θ+μkmgcos⁡θma = mg\sin\theta + \mu_k mg\cos\theta. Dividing by mm gives the acceleration a=g(sin⁡θ+μkcos⁡θ)a = g(\sin\theta + \mu_k \cos\theta).

Question 12

A block of mass mm rests on a larger block of mass MM, which rests on a frictionless horizontal table. The coefficient of static friction between the blocks is μs\mu_s. A horizontal force FF is applied to the lower block MM. What is the maximum force FF that can be applied such that the block mm does not slip relative to MM?

  1. μsmg\mu_s mg
  2. μs(m+M)g\mu_s (m+M)g (correct answer)
  3. μsg(m+M)/M\mu_s g (m+M)/M
  4. μsg(m+M)2/M\mu_s g (m+M)^2/M
Explanation: The top block mm accelerates due to the static friction force from the bottom block, FsF_s. The maximum acceleration of the top block is amax=Fs,max/m=μsFN/m=μsmg/m=μsga_{max} = F_{s,max}/m = \mu_s F_N / m = \mu_s mg / m = \mu_s g. For the top block not to slip, the entire system (both blocks) must have this as its maximum acceleration. Applying Newton's second law to the combined system of mass (m+M)(m+M), the maximum applied force is Fmax=(m+M)amax=(m+M)μsgF_{max} = (m+M)a_{max} = (m+M)\mu_s g.

Question 13

A coin of mass mm is placed on a horizontal turntable at a distance rr from the center. The turntable rotates with a constant angular speed ω\omega. The coefficient of static friction between the coin and the turntable is μs\mu_s. What is the maximum angular speed ωmax\omega_{max} at which the coin can rotate without slipping?

  1. (μsg/r)1/2(\mu_s g / r)^{1/2} (correct answer)
  2. μsg/r\mu_s g / r
  3. (μsgr)1/2(\mu_s g r)^{1/2}
  4. r/(μsg)1/2r / (\mu_s g)^{1/2}
Explanation: The centripetal force required to keep the coin in a circular path is provided by the static friction force. The required centripetal force is Fc=mac=mω2rF_c = m a_c = m \omega^2 r. The maximum available static friction force is Fs,max=μsFN=μsmgF_{s,max} = \mu_s F_N = \mu_s mg. The coin will not slip as long as mω2r≤μsmgm \omega^2 r \leq \mu_s mg. To find the maximum angular speed, we set these equal: mωmax2r=μsmgm \omega_{max}^2 r = \mu_s mg. Solving for ωmax\omega_{max} gives ωmax2=μsg/r\omega_{max}^2 = \mu_s g / r, so ωmax=(μsg/r)1/2\omega_{max} = (\mu_s g / r)^{1/2}.

Question 14

A block is projected with an initial velocity v0v_0 across a horizontal surface with a coefficient of kinetic friction μk\mu_k. How much work is done by the friction force as the block slides to a stop?

  1. Wf=−12mv02W_f = -\frac{1}{2}mv_0^2 (correct answer)
  2. Wf=12mv02W_f = \frac{1}{2}mv_0^2
  3. Wf=−μkmgv0W_f = -\mu_k mgv_0
  4. Wf=−μkmgW_f = -\mu_k mg
Explanation: According to the work-energy theorem, the net work done on an object is equal to its change in kinetic energy: Wnet=ΔKW_{net} = \Delta K. In this case, the net work is done by the friction force, Wnet=WfW_{net} = W_f. The change in kinetic energy is ΔK=Kf−Ki\Delta K = K_f - K_i. Since the block comes to a stop, Kf=0K_f = 0. The initial kinetic energy is Ki=12mv02K_i = \frac{1}{2}mv_0^2. Therefore, Wf=0−12mv02=−12mv02W_f = 0 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mv_0^2. The work is negative because the friction force opposes the displacement.

Question 15

A crate is at rest on a rough horizontal surface. A person can either push the crate with a force PP directed at an angle θ\theta below the horizontal, or pull the crate with a force PP directed at the same angle θ\theta above the horizontal. To make the crate start moving, which method requires a smaller force PP, and why?

  1. Pushing, because the downward component of the push increases the friction force.
  2. Pulling, because the upward component of the pull reduces the normal force and thus the friction. (correct answer)
  3. Pushing, because the horizontal component of the force is larger when pushing.
  4. Both methods require the same force because the horizontal component of the force is Pcos⁡θP\cos\theta in both cases.
Explanation: To start the crate moving, the horizontal component of the applied force, Pcos⁡θP\cos\theta, must overcome the maximum static friction, μsFN\mu_s F_N. When pushing down at an angle, the normal force is FN=mg+Psin⁡θF_N = mg + P\sin\theta. When pulling up at an angle, the normal force is FN=mg−Psin⁡θF_N = mg - P\sin\theta. Since the normal force is smaller when pulling, the maximum static friction force is also smaller. Therefore, a smaller applied force PP is required to overcome friction when pulling.

Question 16

A car of mass mm travels at a constant speed vv around a flat, circular track of radius RR. The coefficient of static friction between the tires and the road is μs\mu_s. What is the magnitude of the friction force acting on the car?

  1. μsmg\mu_s mg
  2. Zero, because the speed is constant.
  3. mv2/Rm v^2 / R (correct answer)
  4. μsmv2/R\mu_s m v^2 / R
Explanation: For the car to move in a circle, there must be a net force directed towards the center of the circle (a centripetal force). This force is provided by static friction between the tires and the road. The required centripetal force has a magnitude of Fc=mac=mv2/RF_c = m a_c = m v^2 / R. Since static friction is the only horizontal force, its magnitude must be equal to this value. Note that this force must be less than or equal to the maximum static friction, μsmg\mu_s mg, but it is not necessarily equal to it.

Question 17

A block is placed on an inclined plane. The coefficient of static friction is 0.80 and the coefficient of kinetic friction is 0.60. The angle of the incline is slowly increased. The block first starts to slide, and then it continues to slide down the plane. What happens to its speed?

  1. It slides at a constant speed.
  2. It slows down and comes to a stop.
  3. It speeds up. (correct answer)
  4. Its motion cannot be determined without knowing the mass of the block.
Explanation: The block starts to slide when the component of gravity down the incline equals the maximum static friction: mgsin⁡θ=μsmgcos⁡θmg \sin\theta = \mu_s mg \cos\theta, so tan⁡θ=μs=0.80\tan\theta = \mu_s = 0.80. Once it starts sliding, the friction force becomes kinetic friction, Fk=μkmgcos⁡θF_k = \mu_k mg \cos\theta. The net force down the incline is Fnet=mgsin⁡θ−μkmgcos⁡θF_{net} = mg \sin\theta - \mu_k mg \cos\theta. Since μk(0.60)<μs(0.80)\mu_k (0.60) < \mu_s (0.80), it follows that μk<tan⁡θ\mu_k < \tan\theta. This means μkmgcos⁡θ<mgsin⁡θ\mu_k mg \cos\theta < mg \sin\theta, so there is a net force down the incline. This net force causes the block to accelerate, so it speeds up.

Question 18

A heavy sled is pulled by a rope. The work required to pull the sled at a constant speed vv over a distance dd on a rough horizontal snowfield is WW. What is the work required to pull the same sled at a constant speed 2v2v over the same distance dd? Assume the coefficient of kinetic friction is independent of speed.

  1. W/2W/2
  2. WW (correct answer)
  3. 2W2W
  4. 4W4W
Explanation: The work done against friction is given by W=Fpull⋅dW = F_{pull} \cdot d. Since the sled is pulled at a constant speed, the pulling force must be equal in magnitude to the kinetic friction force, Fpull=Fk=μkFNF_{pull} = F_k = \mu_k F_N. The coefficient of kinetic friction is assumed to be independent of speed, and the normal force is constant. Therefore, the friction force FkF_k is the same at speed vv and speed 2v2v. Since the pulling force and the distance are the same in both cases, the work done is also the same, WW.

Question 19

A block slides down an inclined plane at constant velocity when the angle is θ1=30°\theta_1 = 30°. The same block is then placed on an inclined plane at angle θ2=45°\theta_2 = 45°. What is the acceleration of the block down the 45°45° incline?

  1. g(sin⁡45°−cos⁡45°tan⁡30°)g(\sin 45° - \cos 45° \tan 30°) (correct answer)
  2. g(sin⁡45°−cos⁡45°/tan⁡30°)g(\sin 45° - \cos 45°/\tan 30°)
  3. g(sin⁡45°−sin⁡30°)g(\sin 45° - \sin 30°)
  4. g(cos⁡45°−cos⁡30°)g(\cos 45° - \cos 30°)
Explanation: At constant velocity on the 30° incline: mgsin⁡30°=μkmgcos⁡30°mg\sin 30° = \mu_k mg\cos 30°, so μk=tan⁡30°\mu_k = \tan 30°. On the 45° incline: ma=mgsin⁡45°−μkmgcos⁡45°=mg(sin⁡45°−cos⁡45°tan⁡30°)ma = mg\sin 45° - \mu_k mg\cos 45° = mg(\sin 45° - \cos 45° \tan 30°). Therefore a=g(sin⁡45°−cos⁡45°tan⁡30°)a = g(\sin 45° - \cos 45° \tan 30°). Option B incorrectly uses 1/tan⁡30°=cot⁡30°1/\tan 30° = \cot 30°. Options C and D don't account for the friction coefficient properly.

Question 20

A 5.0 kg block is placed on a rough inclined plane that makes an angle of 30° with the horizontal. The block remains at rest. What is the magnitude of the static friction force acting on the block?

  1. The static friction force is zero because the block is not moving.
  2. The static friction force is approximately 42.4 N, acting up the incline.
  3. The static friction force is approximately 24.5 N, acting up the incline. (correct answer)
  4. The static friction force cannot be determined without the coefficient of static friction.
Explanation: Since the block is at rest, it is in static equilibrium. The net force on the block is zero. The forces acting parallel to the incline are the component of gravity pulling the block down the incline (mgsin⁡θmg \sin\theta) and the static friction force (FsF_s) acting up the incline. Therefore, Fs=mgsin⁡θF_s = mg \sin\theta. Using g≈9.8 m/s2g \approx 9.8 \, \text{m/s}^2, Fs=(5.0 kg)(9.8 m/s2)(sin⁡30°)=(49 N)(0.5)=24.5 NF_s = (5.0 \, \text{kg})(9.8 \, \text{m/s}^2)(\sin 30°) = (49 \, \text{N})(0.5) = 24.5 \, \text{N}. The static friction force adjusts its magnitude to be equal and opposite to the net force that would cause motion, as long as that force does not exceed the maximum static friction.