🧰FRQ: Sliding the Tool Chest

Assessments aligned to 2026 AP Physics 1 standards

📋Scenario

Two technicians need to reposition a 10 kg steel tool chest across a level concrete floor. They first notice it takes a hard initial shove to get the chest going, but 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. The chest’s base stays flat on the floor.

📝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, and sketch (describe) how the friction force on the chest changes as the push is raised from 0 N up through 70 N. 

(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, then drop.” Evaluate this claim. State precisely what the friction force is at the instant just before sliding and at the instant just after sliding begins, and explain whether the claim is right, partly right, or wrong. 

🗝️Answer Key & Scoring Guide

Part (a) — Model Answer

On level ground N = mg = (10)(10) = 100 N. At the verge of sliding the push equals the maximum static friction; at constant velocity while sliding the push equals the kinetic friction (zero acceleration in both cases):

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), consistent with μ_s > μ_k.

Scoring (3 points):

✔️1 point: N = mg = 100 N.

✔️1 point: μ_s = f_s,max / N = 60/100 = 0.60 (push at impending motion = maximum static friction).

✔️1 point: μ_k = f_k / N = 40/100 = 0.40 (steady-speed push = kinetic friction, a = 0).

Part (b) — Model Answer

To start the chest the push must overcome the maximum static friction f_s,max = μ_s N = 60 N. The moment it is sliding, the opposing force becomes kinetic friction f_k = μ_k N = 40 N. Because μ_k < μ_s for this surface pair, the kinetic friction (40 N) is smaller than the maximum static friction (60 N) that had to be beaten.

So less force is needed to balance kinetic friction and keep the chest moving at constant speed than was needed to break it loose.

Scoring (2 points):

✔️1 point: identifies that starting requires beating maximum static friction μ_s N, while keeping it moving only balances kinetic friction μ_k N.

✔️1 point: attributes the difference to μ_k < μ_s (so f_k < f_s,max) for the same surfaces.

⚠️Common error: attributing the difference to mass or to “inertia being harder to start” — no credit; the cause is the coefficient difference, not a change in N or m.

Part (c) — Model Answer

While sliding, kinetic friction is constant at f_k = μ_k N = 40 N. With a 70 N push:

F_push − f_k = ma

a = (F_push − f_k) / m = (70 − 40) / 10 = 30 / 10 = 3.0 m/s²

Friction vs push as the push rises from 0 to 70 N: while the chest is stationary, static friction rises linearly to match the push exactly (friction = push), climbing from 0 up to its maximum of 60 N at the verge of sliding. The instant sliding begins, the friction force drops to the constant kinetic value of 40 N and stays at 40 N for all larger pushes (it does not keep rising with the push).

Scoring (3 points):

✔️1 point: uses f_k = 40 N (constant kinetic friction) while sliding.

✔️1 point: correct acceleration a = (70 − 40)/10 = 3.0 m/s².

✔️1 point: describes the friction–push relationship correctly: rises with push up to 60 N (static), then drops to and stays at 40 N (kinetic).

⚠️Common error: stating friction equals 70 N or keeps increasing once sliding — no credit for the description point; kinetic friction is independent of the applied push.

Part (d) — Model Answer

The claim is partly right. Just before sliding, the chest is at the verge of motion, so static friction has reached its maximum value — exactly f_s = f_s,max = μ_s N = 60 N, balancing the 60 N push. So the “60 N at the start” part is correct, but only at the single instant of impending motion.

The instant sliding actually begins, the relevant force becomes kinetic friction, which is f_k = μ_k N = 40 N — not 60 N. So friction does drop, but it drops to 40 N, and the 60 N value applies only at the boundary instant, not during the slide. The claim is therefore correct about the value at impending motion and about the drop, but wrong if it implies friction is 60 N during any part of the actual sliding.

Scoring (3 points):

✔️ 1 point: Correctly evaluates the claim as partly right (must explicitly address both the valid and invalid aspects of the technician's statement).

✔️ 1 point: States that at the instant just before sliding (impending motion), the static friction force is exactly 60 N.

✔️ 1 point: States that at the instant just after sliding begins, the friction force immediately drops to the constant kinetic friction value of 40 N.

⚠️ Common error: Agreeing with the claim completely by assuming the 60 N force persists into the slide, or failing to explicitly state the 40 N kinetic friction value that takes over once motion begins.