Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.

FRQ: Kinetic and Static Friction — The Bench-Top Pulley Rig

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Mathematical Routines (MR)  |  MID-LEVEL  |  10 points

▤ Scenario

A physics class sets up a bench-top rig to measure friction. A wooden block of mass M = 3.0 kg rests on a horizontal laboratory bench. A light cord runs horizontally from the block to a pulley mounted at the edge of the bench, passes over the pulley, and hangs vertically down to a metal weight of mass m = 2.0 kg. The coefficient of kinetic friction between the block and the bench is μ_k = 0.20.

The class releases the system from rest. The hanging weight descends and the block slides toward the pulley. No one touches either object after release.

Model the cord as massless and unstretchable and the pulley as ideal, so the tension is the same at every point in the cord. All motion is one-dimensional, and all quantities are measured in the reference frame of the laboratory bench, which is inertial. For each object, take its own direction of motion as positive. Use g = 10 N/kg.

The bench-top rig. The dashed arrows show the sense of the motion after release; they are not forces.

✎ Free Response Questions

(a) Draw a free-body diagram of the sliding block, labelling each force by the object that exerts it, and state which force opposes the block's motion. Then write an expression for the magnitude of the kinetic friction force exerted on the block and evaluate it.

(b) Starting from a fundamental physics principle, derive an expression for the magnitude of the acceleration of the system in terms of M, m, μ_k and g, and evaluate it. Then determine the tension in the cord.

(c) The class repeats the experiment several times, holding M = 3.0 kg and μ_k = 0.20 fixed and changing only the hanging mass m. The acceleration measured in each trial is shown below.

Compare the acceleration measured at m = 2.0 kg with the value you predicted in part (b). Identify one physical effect that is present in the real apparatus but absent from the idealized model, and explain how that effect accounts for a discrepancy in the direction you found.

(d) The coefficient of static friction between the block and the bench is μ_s = 0.40. Determine the largest hanging mass m for which the block remains at rest when the system is released, justifying your answer with an equilibrium argument. Then explain how the same apparatus can be in equilibrium for a small hanging mass and yet accelerate for a large one.

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

Four forces are exerted on the block, and no others. The cord pulls it horizontally toward the pulley with tension T. The bench exerts kinetic friction f_k on it, directed opposite to the block's motion — that is, away from the pulley. The bench also exerts a normal force F_N on it, directed vertically upward. Earth exerts the gravitational force F_g = M·g on it, directed vertically downward. Kinetic friction is the force that opposes the block's motion.

The block does not accelerate vertically, so the vertical forces on it are balanced and F_N = M·g. The magnitude of the kinetic friction force is therefore

f_k = μ_k·F_N = μ_k·M·g = (0.20)(3.0 kg)(10 N/kg) = 6.0 N

Free-body diagram of the sliding block, each force labelled by the object that exerts it. Arrow lengths show direction only; the magnitude of T is determined in part (b).

The bench exerts 6.0 N of kinetic friction on the block, directed away from the pulley.

Scoring (2 points):

▸ 1 point: The free-body diagram shows exactly four forces on the block — tension toward the pulley, kinetic friction opposite the motion, the normal force upward and the gravitational force downward — with no additional forward force, and kinetic friction is named as the force opposing the motion.

▸ 1 point: Writes f_k = μ_k·F_N, justifies F_N = M·g from the absence of vertical acceleration, and evaluates f_k = 6.0 N.

⚠︎ Common error (partial credit): Uses the hanging weight for the normal force, F_N = m·g, giving f_k = 4.0 N. The friction point is lost; the diagram point is still available. Earns 1 of 2.

✗ Common error (no credit): Adds the hanging weight m·g to the forces on the block, or sets F_N = (M + m)·g and finds f_k = 10 N. The bench supports only the block, so the diagram carries a force no object exerts on it and the friction magnitude that follows is wrong. Earns 0 of 2.

Part (b) — Model Answer

Apply Newton's second law to each object separately, taking each object's own direction of motion as positive. The cord does not stretch, so both objects have the same magnitude of acceleration a, and the pulley is ideal, so the tension T is the same at both ends of the cord.

hanging weight:  m·g − T = m·a
block:  T − f_k = M·a,  with  f_k = μ_k·M·g

Adding the two equations eliminates T and leaves one equation for the whole system:

m·g − μ_k·M·g = (M + m)·a
a = (m·g − μ_k·M·g) / (M + m)

Substituting the given values:

a = (20 N − 6.0 N) / (5.0 kg) = 2.8 m/s²
T = m(g − a) = (2.0 kg)(10 − 2.8) N/kg = 14.4 N

The two objects isolated, each with its own positive direction marked in amber. Adding the two equations eliminates the tension.

The system accelerates at 2.8 m/s² and the cord tension is 14.4 N — between the 6.0 N friction force and the 20 N hanging weight, as it must be for the block to speed up and the weight to descend.

Scoring (3 points):

▸ 1 point: Begins from Newton's second law applied to each object, or to the two-object system, with kinetic friction included as the only horizontal resistive force. This point is awarded for the correct starting principle alone, before any algebra is performed.

▸ 1 point: Eliminates T and reaches a = (m·g − μ_k·M·g) / (M + m) symbolically before substituting, then evaluates a = 2.8 m/s².

▸ 1 point: Returns to a single-object equation for the tension, T = m(g − a) = 14.4 N, or equivalently T = M·a + μ_k·M·g = 14.4 N. Awarded for a tension computed consistently from the student's own value of a.

⚠︎ Common error (partial credit): Adds the friction force instead of subtracting it, giving a = (m·g + μ_k·M·g) / (M + m) = 5.2 m/s². The starting-principle point still stands, and the tension point remains available if T follows consistently from that acceleration. Earns up to 2 of 3.

✗ Common error (no credit): Sets T = m·g = 20 N and uses that value throughout. A tension equal to the hanging weight would leave the weight in equilibrium, contradicting the premise that it descends, so Newton's second law was never applied to it. Earns 0 of 3.

Part (c) — Model Answer

Part (b) predicts a = 2.8 m/s² at m = 2.0 kg. The measured value is 2.6 m/s², smaller than the prediction by 0.2 m/s², about 7% of it. The discrepancy is small and one-sided: every trial in the table falls below the value the idealized model gives.

The model treats the pulley as ideal — negligible mass, turning on an axle with negligible friction. A real pulley has neither property. Because it has mass, the cord must spin it up as well as accelerate the block and the weight, so the driving force acts on more inertia than M + m alone; because its axle has friction, part of that force is spent turning it. Either effect lowers the measured acceleration below the prediction, which is the direction observed.

Scoring (3 points):

▸ 1 point: Compares the measured 2.6 m/s² with the predicted 2.8 m/s² numerically and states that the measured value is the smaller of the two.

▸ 1 point: Names one specific effect present in the apparatus but absent from the model — the mass of the pulley, friction in the pulley's axle, the mass of the cord, or air resistance on the descending weight.

▸ 1 point: Explains why that effect lowers the acceleration: it either adds inertia the model does not count or removes part of the driving force.

⚠︎ Common error (partial credit): Names a legitimate non-ideal effect but reasons it backwards — arguing, for instance, that air resistance makes the weight fall faster. The comparison and the named effect still score; the explanation point is lost. Earns 2 of 3.

✗ Common error (no credit): Reports only that the values are close, or attributes the gap to human error, with no numerical comparison and no named physical effect. Earns 0 of 3.

Part (d) — Model Answer

While the block stays at rest, both objects are in translational equilibrium. The hanging weight is held by the cord, so T = m·g. The block is held by static friction, so f_s = T = m·g. Static friction is not a fixed quantity: it takes whatever value is needed to prevent slipping, up to a maximum set by the normal force,

f_s ≤ f_s,max = μ_s·F_N = μ_s·M·g

The block therefore remains at rest as long as the pull does not exceed that maximum:

m·g ≤ μ_s·M·g,  so  m ≤ μ_s·M = (0.40)(3.0 kg) = 1.2 kg

The largest hanging mass for which the block remains at rest is 1.2 kg. Notice that g cancels: the threshold depends only on μ_s and M, so the same 1.2 kg would be found on the Moon.

The apparatus can be in equilibrium for a small hanging mass and accelerate for a large one because static friction adjusts itself up to a ceiling. Below the threshold it takes exactly the value m·g, the forces on the block are balanced, and its velocity stays zero. Above the threshold the required value exceeds μ_s·M·g and friction cannot supply it, so the forces become unbalanced and the block accelerates — at which moment friction drops to its kinetic value μ_k·M·g = 6.0 N, smaller still, which is why a block on the verge of moving starts abruptly.

Scoring (2 points):

▸ 1 point: Sets the cord's pull equal to the maximum static friction force, m·g = μ_s·M·g, from the equilibrium of both objects, and obtains m = μ_s·M = 1.2 kg.

▸ 1 point: Explains that static friction takes whatever value is required to prevent slipping, up to a maximum of μ_s·F_N, so the forces stay balanced below that ceiling and become unbalanced above it.

⚠︎ Common error (partial credit): Uses μ_k = 0.20 in place of μ_s = 0.40 and reports 0.6 kg. The threshold point is lost, but the self-adjusting-friction explanation is unaffected and still earns its point. Earns 1 of 2.

✗ Common error (no credit): Applies the friction condition to the hanging weight, writing m·g ≤ μ_s·m·g. The hanging weight touches no surface, so no normal force and no friction act on it, and the argument is not about the block at all — neither point can be earned. Earns 0 of 2.