FRQ: Friction Joins the Problem

Assessments aligned to 2026 AP Physics 1 standards

Scenario

A wooden block of mass M = 3.0 kg sits on a horizontal table whose surface is NOT frictionless. A light, unstretchable cord runs from the block, over an ideal pulley, to a hanging mass m = 2.0 kg. The coefficient of kinetic friction between the block and the table is μₖ = 0.20. When released, the hanging mass descends and the block slides. Use g ≈ 10 N/kg.

All motion is one-dimensional. Take each object’s direction of motion as positive for that object. The pulley and cord are ideal.

(a) On a free-body diagram of the sliding block, identify every force and state which force opposes the motion. Write the expression for the magnitude of the kinetic friction force and evaluate it.

(b) Determine the acceleration of the system and the tension in the cord.

(c) A lab group keeps M = 3.0 kg and μₖ = 0.20 fixed and varies the hanging mass m, recording the measured acceleration each time. Their data are shown. Using the row for m = 2.0 kg, determine whether their measured value is consistent with your prediction in part (b), and identify one physical effect not in the idealized model that could explain any discrepancy. 

Plaintext

m (kg):           1.0   1.5   2.0   2.5   3.0

a measured (m/s²): 0.9   1.8   2.6   3.3   3.8

(d) Below what hanging mass m would the block remain stationary, assuming the coefficient of static friction is μₛ = 0.40? Justify your reasoning with an equilibrium argument, and explain why a system can be in equilibrium for small m but accelerate for large m.

Answer Key & Scoring Guide

Part (a) — Model Answer

Forces on the block: tension T (toward the pulley, by the cord); kinetic friction f (opposite the motion, by the table); normal force N (up, by the table); weight Mg (down, by Earth). Friction opposes the block’s motion. Because the block does not accelerate vertically, N = Mg, so:

f = μₖN = μₖMg = (0.20)(3.0)(10) = 6.0 N

Scoring (2 points):

✔️1 point: Free-body diagram includes T, f, N, and Mg with correct directions; friction identified as opposing the motion.

✔️ 1 point: Correct friction magnitude f = μₖMg = 6.0 N, using N = Mg.

✔️ Common error: using N = mg (the hanging mass) instead of N = Mg — earns 0 for the friction point.

Part (b) — Model Answer

Hanging mass (down positive): mg − T = ma. Block (toward pulley positive): T − f = Ma. Adding eliminates T:

mg − f = (M + m)a

a = (mg − μₖMg) / (M + m) = (20 − 6.0) / (5.0) = 2.8 m/s²

T = m(g − a) = (2.0)(10 − 2.8) = 14.4 N

The system accelerates at 2.8 m/s², and the tension is 14.4 N.

Scoring (3 points):

✔️1 point: Correct system equation mg − μₖMg = (M + m)a (friction subtracted).

✔️ 1 point: Correct acceleration a = 2.8 m/s².

✔️ 1 point: Correct tension T = 14.4 N (consistent with the computed a).

✔️ Common error: adding friction instead of subtracting it (a = (mg + f)/(M+m)) — earns 0 for the acceleration point.

Part (c) — Model Answer

Predicted acceleration for m = 2.0 kg is 2.8 m/s² (part b). The measured value is 2.6 m/s², which is slightly lower — close but not identical. The measured acceleration is smaller, consistent with real effects the idealized model ignores: friction or small mass in the pulley (so the cord must also do work to spin it up), a cord of non-negligible mass, or air resistance on the falling mass. Any of these removes energy or adds inertia the model leaves out, lowering the observed acceleration below the prediction.

Scoring (3 points):

✔️ 1 point: Correctly compares measured 2.6 m/s² to predicted 2.8 m/s² and notes the measured value is lower.

✔️ 1 point: States the comparison quantitatively (a difference of about 0.2 m/s²) rather than just “close.”

✔️ 1 point: Names one plausible non-ideal effect (pulley mass/friction, cord mass, or air resistance) that would reduce a.

✔️ Common error: claiming the data prove the model wrong — earns at most 1 point; a small discrepancy is expected from non-ideal effects.

Part (d) — Model Answer

The block stays at rest only if static friction can balance the cord tension, and at the verge of motion T = mg (the hanging mass is in equilibrium) while friction is at its maximum, μₛMg. Setting the pull equal to the maximum static friction:

mg ≤ μₛMg  ⟹  m ≤ μₛM = (0.40)(3.0) = 1.2 kg

So for m below 1.2 kg the system remains stationary. The reason a system can be in equilibrium for small m but accelerate for large m is that static friction is self-adjusting up to a maximum: for small hanging weights it exactly cancels the pull (net force zero, translational equilibrium), but once mg exceeds μₛMg the forces become unbalanced and the system accelerates.

Scoring (2 points):

✔️ 1 point: Correct threshold m = μₛM = 1.2 kg from an equilibrium/maximum-static-friction argument.

✔️ 1 point: Explains that static friction adjusts up to a maximum, so equilibrium holds until mg exceeds μₛMg.

✔️ Common error: using μₖ = 0.20 instead of μₛ = 0.40 — earns 0 for the threshold point.