FRQ: Static and Kinetic Friction — Sliding the Tool Chest

Assessments aligned to 2026 AP Physics 1 standards — 12 points

Question Type: Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  

▤ Scenario

Two technicians need to reposition a 10 kg steel tool chest across a level concrete floor. They first notice that it takes a hard initial shove to get the chest going, but that once it is sliding it is much easier to keep it moving at a steady speed.

By experiment they establish that the chest is on the verge of sliding when the horizontal push reaches 60 N, and that a steady horizontal push of 40 N keeps it sliding at constant velocity.

Take g = 10 m/s². All motion is one-dimensional and horizontal; take the direction of motion as positive and the floor as the reference frame. The chest’s base stays flat on the floor throughout.

✎ Free Response Questions

(a) Using the two experimental observations, determine the coefficient of static friction and the coefficient of kinetic friction between the tool chest and the concrete floor.

(b) Explain, in terms of the forces involved, why a larger push is needed to start the chest moving than to keep it moving at constant speed. Refer to the specific friction forces and coefficients in your answer.

(c) Once the chest is sliding, the technicians increase their steady push to 70 N. Calculate the chest’s acceleration. Then sketch a graph of the magnitude of the friction force on the chest as a function of the magnitude of the applied horizontal push, for pushes from 0 N up to 70 N. Label the value of the friction force at every point where it changes behaviour.

(d) A technician argues: “Since it took 60 N to get the chest moving, the friction force on the chest must be 60 N at the very start of the slide, and then drop.” In a clear, paragraph-length response, evaluate this claim. State precisely what the friction force is at the instant just before sliding and at the instant just after sliding begins, use those values as evidence, and conclude whether the claim is right, partly right, or wrong.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit

Part (a) — Model Answer

On level ground the chest’s vertical forces balance, so the normal force is N = mg. At the verge of sliding the push equals the maximum static friction; while sliding at constant velocity the push equals the kinetic friction (the acceleration is zero in both cases).

N = mg = (10)(10) = 100 N

f_s,max = 60 N = μ_s N  ⇒  μ_s = 60 / 100 = 0.60

f_k = 40 N = μ_k N  ⇒  μ_k = 40 / 100 = 0.40

Result: μ_s = 0.60 and μ_k = 0.40, both dimensionless, and consistent with the expected ordering μ_s > μ_k for a single pair of surfaces.

Scoring (3 points):

▸ 1 point: correctly obtains the normal force from vertical force balance, N = mg = 100 N.

▸ 1 point: μ_s = f_s,max / N = 60 / 100 = 0.60, recognising that the push at impending motion equals the maximum static friction.

▸ 1 point: μ_k = f_k / N = 40 / 100 = 0.40, recognising that a steady-speed push equals the kinetic friction because a = 0.

⚠︎ Common error (partial credit): correct method and correct ratios but N taken as 10 N (mass used in place of weight) — earns the two ratio points if the substitution is otherwise consistent, loses the normal-force point.

✗ Common error (no credit): assigning units such as N or kg to μ — a coefficient of friction is a dimensionless ratio of two forces.

Part (b) — Model Answer

To start the chest moving, the push must exceed the maximum static friction, f_s,max = μ_s N = 60 N. The moment the chest is sliding, the opposing force is no longer static friction but kinetic friction, f_k = μ_k N = 40 N. The normal force has not changed — the chest still rests flat on level ground, so N = mg = 100 N in both situations — and the mass has not changed either. What changes is the coefficient: because μ_k < μ_s for this pair of surfaces, the kinetic friction of 40 N is smaller than the 60 N of maximum static friction that had to be broken. A smaller push is therefore enough to balance kinetic friction and hold the chest at constant speed.

Scoring (2 points):

▸ 1 point: identifies that starting the chest requires exceeding the maximum static friction μ_s N, whereas keeping it moving at constant speed only requires balancing the kinetic friction μ_k N.

▸ 1 point: attributes the difference to μ_k < μ_s for the same pair of surfaces, so f_k < f_s,max at the same normal force.

⚠︎ Common error (partial credit): correctly names static and kinetic friction but never states the coefficient inequality — earns the first point only.

✗ Common error (no credit): attributing the difference to the chest’s mass, to “inertia being harder to start”, or to a change in the normal force — m and N are identical in the two situations, so neither can be the cause.

Part (c) — Model Answer

While the chest is sliding, the kinetic friction is constant at f_k = μ_k N = 40 N, independent of how hard the technicians push. Applying Newton’s second law along the direction of motion:

F_push − f_k = ma

a = (F_push − f_k) / m = (70 − 40) / 10 = 3.0 m/s²

The chest accelerates at 3.0 m/s² in the direction of the push.

Graph of friction force against applied push: while the chest is stationary, static friction rises linearly to match the push exactly (f_s = F_push), climbing along a line of slope 1 from 0 N up to its maximum of 60 N at the verge of sliding. The instant sliding begins, the friction force drops discontinuously to the constant kinetic value of 40 N and stays at 40 N for every larger push — it does not continue to rise with the applied force.

Friction force on the chest versus applied push: static friction tracks the push to its maximum of 60 N, then falls to the constant kinetic value of 40 N once the chest slides.

Scoring (4 points):

▸ 1 point: uses the constant kinetic friction f_k = μ_k N = 40 N while the chest is sliding.

▸ 1 point: applies Newton’s second law to the sliding chest as F_push − f_k = ma.

▸ 1 point: correct acceleration a = (70 − 40) / 10 = 3.0 m/s², with units, in the direction of the push.

▸ 1 point: graph shows friction rising along a slope-1 line to 60 N while stationary, then dropping to and remaining at the constant value of 40 N for all larger pushes, with both 60 N and 40 N labelled.

⚠︎ Common error (partial credit): correct acceleration but a graph in which friction keeps rising after sliding begins — earns the three quantitative points, loses the graph point.

✗ Common error (no credit): setting the friction force equal to 70 N while sliding, or scaling friction with the applied push — kinetic friction depends on μ_k and N only, never on the applied force.

Part (d) — Model Answer

The claim is partly right. Just before sliding, the chest is at the verge of motion, so static friction has risen to its maximum value and is exactly f_s = f_s,max = μ_s N = 60 N, balancing the 60 N push and keeping the chest in equilibrium. To that extent the technician is correct — but the 60 N figure applies only at that single instant of impending motion, while the chest is still at rest. The instant sliding actually begins, static friction is no longer the relevant force at all; the chest experiences kinetic friction, f_k = μ_k N = (0.40)(100) = 40 N. So the friction force does drop, exactly as the technician expects, but it drops immediately to 40 N — it never has the value 60 N at any moment during the slide. The claim is therefore right about the value at impending motion and right that a drop occurs, but wrong if it is read as saying friction is 60 N during any part of the actual sliding.

Scoring (3 points):

▸ 1 point: evaluates the claim as partly right, explicitly addressing both the valid and the invalid part of the technician’s statement.

▸ 1 point: states that at the instant just before sliding the static friction is exactly its maximum value, f_s,max = μ_s N = 60 N, and that this holds only at impending motion.

▸ 1 point: states that at the instant just after sliding begins the friction force is the kinetic value f_k = μ_k N = 40 N, and uses that value as the evidence for the conclusion.

⚠︎ Common error (partial credit): correct 60 N and 40 N values but no explicit verdict on the claim, or a verdict with no supporting values — earns the value points or the verdict point, not both.

✗ Common error (no credit): agreeing with the claim in full by treating 60 N as the friction force at the start of the slide — this confuses the maximum static friction, which acts only at rest, with the kinetic friction that acts once the surfaces are moving relative to one another.