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

Assessments aligned to 2026 AP Physics 1 standards

Mathematical Routines (MR)  |  MID-LEVEL  |  10 points  |  25 min

▤ 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. Figure 1 shows the rig.

Figure 1 — The bench-top rig. The dashed arrows show the sense of the motion; they are not forces.

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, air resistance is negligible, 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.

✎ Free Response Questions

Part A

i.    Consider the block at an instant while it slides toward the pulley.

On the following dot that represents the block, draw and label the forces (not components) that are exerted on the block. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.

Figure 2 — The dot representing the block. The dashed arrow is the block’s velocity, not a force.

ii.   Derive an expression for the magnitude a of the acceleration of the block while it slides, and then evaluate that expression for the quantities given in the Scenario.

Express your final answer in terms of M, m, μ_k, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

iii.  The class repeats the experiment several times with the same apparatus, holding M = 3.0 kg and μ_k = 0.20 fixed and changing only the hanging mass m. Figure 3 shows the acceleration measured in each trial. Air resistance on the block and the weight is too small to affect these measurements.

Figure 3 — The acceleration measured in each trial, with M = 3.0 kg and μ_k = 0.20 held fixed.

Identify one property of the real apparatus, left out of the model described in the Scenario, that could account for the difference between the acceleration measured at m = 2.0 kg and the value you calculated in part A (ii).

Justify your response.

Part B

The coefficient of static friction between the block and the bench is μ_s = 0.40. The class replaces the 2.0 kg hanging weight with a weight of mass 1.0 kg and again releases the system from rest. Use the model described in the Scenario.

Let f_1 be the magnitude of the friction force exerted on the block by the bench while the block slides with the 2.0 kg hanging weight, and let f_2 be the magnitude of the friction force exerted on the block by the bench after the release with the 1.0 kg weight.

Indicate whether f_2 is greater than, less than, or equal to f_1 by writing one of the following.

•   f_2 > f_1

•   f_2 < f_1

•   f_2 = f_1

Justify your response.

❖ Answer Key & Scoring Guide

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

Points are independent and there are no deductions. ▸ criteria follow College Board’s published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A (i) — Model Answer

The block touches the bench and the cord, and Earth exerts a gravitational force on it, so four forces are exerted on the block and no others. The cord pulls it horizontally toward the pulley with tension T. The bench exerts a kinetic friction force f_k on it, directed opposite to its velocity, 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 = 30 N on it, directed vertically downward.

The hanging weight does not touch the block, and the bench does not support it. It acts on the block only through the cord, so the block’s diagram carries the tension and no arrow for the hanging weight’s gravitational force.

Figure 4 — The sliding block’s four forces to one scale; T = 14.4 N is the part A (ii) check value.

Scoring (2 points):

▸ A1 — 1 point: For drawing the gravitational force directed downward and the normal force directed upward, as distinct labelled arrows starting on the dot, with no other vertical force. A downward arrow labelled mg is accepted as the gravitational force on the block.

▸ A2 — 1 point: For drawing the tension directed horizontally toward the pulley and the kinetic friction force directed horizontally opposite to the block’s velocity, as distinct labelled arrows starting on the dot, with no other horizontal force. The arrow toward the pulley may carry any label naming the cord’s force — T, F_T, F_cord or tension; a label m·g, or “weight of the hanging mass”, equates the cord’s pull with the hanging weight’s gravitational force and does not earn this point.

Scoring Note: Any label that names the force is accepted — F_g, mg, Mg, weight or gravitational force; F_N, N or normal force; T, F_T, F_cord or tension; f_k, f or friction. Equal lengths for the vertical pair are not required, no force scale having been supplied, and the response is not required to name the object exerting each force. A1 and A2 are earned on the response’s own diagram; there is no upstream result to carry from.

⚠︎ Common error (partial credit): Draws the four forces correctly and adds a fifth arrow, pointing down and labelled m·g, for the hanging weight — earns 1 of 2 points, A2. The hanging weight is not on the block and the bench does not support it, so no such force is exerted on the block.

⚠︎ Common error (partial credit): Draws all four arrows in the right directions but labels the arrow toward the pulley m·g, or “weight of the hanging mass”, instead of tension — earns 1 of 2 points, A1. The hanging weight’s gravitational force is exerted on the hanging weight; the block feels the cord’s pull, which is less than m·g while the system accelerates.

Part A (ii) — 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.

The block has no vertical acceleration, so F_N = M·g, and the kinetic friction force on it is f_k = μ_k·F_N = μ_k·M·g, directed opposite to its motion:

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

Adding the two equations eliminates T:

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

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

a = (20 N − 6.0 N) / (5.0 kg) = 2.8 m/s²

The block and the weight each accelerate at 2.8 m/s². As a check beyond what part A (ii) asks, the tension is T = m(g − a) = (2.0 kg)(10 − 2.8) m/s² = 14.4 N, between the 6.0 N friction force and the 20 N weight, as it must be for the block to speed up and the weight to descend. The bench presses on the block alone, so kinetic friction on the block is μ_k·M·g however heavy the hanging weight is.

Scoring (3 points):

▸ A3 — 1 point: For a multistep derivation that includes Newton’s second law, applied to either object or to the two as one system, or an energy principle, such as the work–energy theorem or conservation of energy, applied to either object or to the system. This point is earned for the correct starting principle alone, even if the steps that follow are incorrect. A final expression for a on its own is not a multistep derivation.

▸ A4 — 1 point: For expressing the friction force on the block as μ_k·M·g, directed opposite to the block’s motion, from a normal force equal to M·g — or, on an energy route, the energy friction takes from the system over a distance d as μ_k·M·g·d, whether written as negative work or as thermal energy.

▸ A5 — 1 point: For a correct expression for a, a = (m − μ_k·M)·g/(M + m) or equivalent, together with its value for the given quantities, 2.8 m/s². Where A4 was not earned, an expression that follows correctly from the response’s own friction term, together with its value, earns this point; any other error prevents it.

Scoring Note: A correct, isolated, final expression for a earns point A4. A3, A4 and A5 may each be earned independently of the response in part A (i).

⚠︎ Common error (partial credit): Uses the wrong normal force for the friction — the hanging weight’s m·g, giving f_k = 4.0 N and a = 3.2 m/s², or the combined (M + m)·g, giving f_k = 10 N and a = 2.0 m/s² — earns 2 of 3 points, A3 and A5. The bench supports only the block, so the normal force on the block is M·g.

⚠︎ Common error (partial credit): Takes the friction force on the block to point toward the pulley, writing T + μ_k·M·g = M·a, and obtains a = (m + μ_k·M)·g/(M + m) = 5.2 m/s² — earns 2 of 3 points, A3 and A5. Friction on the sliding block points away from the pulley, against the cord’s pull.

⚠︎ Common error (partial credit): Sets the tension equal to the hanging weight, T = m·g = 20 N, and applies the second law to the block alone, obtaining a = (m·g − μ_k·M·g)/M = 4.7 m/s² — earns 2 of 3 points, A3 and A4. A tension equal to m·g would leave the hanging weight with no net force, contradicting its downward acceleration.

✗ Common error (no credit): States that the hanging weight falls freely, so a = g = 10 m/s², writing no force or energy equation for either object. A3 is not earned, no second-law or energy relation having been written; A4 is not earned, friction never appearing; and A5 is not earned, a = g leaving out the cord’s force as well as friction. The cord pulls up on the hanging weight throughout, so it cannot fall with acceleration g.

Part A (iii) — Model Answer

Part A (ii) 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 model treats the pulley as ideal, with negligible mass and turning on an axle with negligible friction. Neither is exactly true of a real pulley. Because it has mass, the pulley must itself be set turning faster and faster as the cord runs over it, so the system being accelerated has more inertia than the block and the weight alone; because its axle has friction, part of what drives the system goes into overcoming that friction. Either effect makes the real acceleration smaller than the model’s value, which is the direction observed.

The other trials tell the same story. For the five hanging masses the model predicts 2.00, 2.80, 3.45, 4.00 and 4.46 m/s², and every measured value lies below its prediction, by 0.15 to 0.26 m/s² (Figure 5). Random measurement error scatters readings to both sides of the true value, and a pulley’s mass and axle friction act the same way in every trial. A difference that every trial shows in the same direction points to a systematic cause rather than to random error.

Figure 5 — Every measured acceleration lies below the ideal-pulley model’s prediction.

Scoring (2 points):

▸ A6 — 1 point: For identifying a property of the real apparatus that the model leaves out and that could make the real acceleration smaller than the model predicts — the pulley has mass (rotational inertia), or friction acts at or in the pulley. A response that says only that the pulley is not ideal does not earn this point unless its justification identifies the mass or friction responsible. Air resistance, which part A (iii) states is too small to affect the measurements, and the mass of the light cord do not earn this point.

▸ A7 — 1 point: For a justification that explains correctly how the named property makes the real acceleration smaller than the ideal model predicts — for example, a pulley with mass must itself be set turning, which adds inertia, or takes energy, that the model does not count; a pulley with mass or axle friction needs a larger tension on the hanging side than on the block’s side, and that difference is taken out of the net force that accelerates the block and weight together; or friction at the axle opposes the pulley’s turning and reduces the force driving the system. The response need not restate the numerical comparison. A justification built on air resistance, on the mass of the light cord, or on timing error does not earn this point.

Scoring Note: A6 and A7 may each be earned independently of the response in part A (ii). No numerical comparison is required, at m = 2.0 kg or at any other mass.

⚠︎ Common error (partial credit): Names the pulley’s mass or its axle friction but gives no mechanism, or argues that the property would make the acceleration larger — earns 1 of 2 points, A6. Either property can only make the real acceleration smaller than the ideal model’s.

⚠︎ Common error (partial credit): Says only that the pulley is not ideal, but explains correctly that the tension is then larger on the hanging side than on the block’s side, a difference taken out of the net force that accelerates the block and weight together — earns 1 of 2 points, A7. The explanation is right, but it never names the pulley’s mass or axle friction, the property that makes the tensions differ.

✗ Common error (no credit): Attributes the difference to human error, reaction time or imprecise timing, naming no property of the apparatus. Neither point is earned. Timing error is a flaw in the measurement, not a property of the apparatus that the model leaves out, and it does not change how the block and weight actually move.

Part B — Model Answer

f_2 > f_1

While the block slides with the 2.0 kg weight, the friction force on it is kinetic: f_1 = μ_k·M·g = 6.0 N, fixed by the normal force whatever the hanging mass.

With the 1.0 kg weight, first decide whether the block moves at all. If it stays at rest, the hanging weight is at rest too and the cord holds it, so the cord pulls on the block with T = (1.0 kg)(10 N/kg) = 10 N, and static friction must balance that pull. The largest force static friction can supply is

f_s,max = μ_s·F_N = μ_s·M·g = (0.40)(3.0 kg)(10 N/kg) = 12 N

The required 10 N does not exceed 12 N, so the block remains at rest. Equivalently, the block stays at rest for any hanging mass up to μ_s·M = 1.2 kg, a threshold in which g cancels. With a hanging mass above 1.2 kg the required friction would exceed 12 N, the block would start to slide, and the friction on it would drop to its kinetic value, 6.0 N, the situation of part A.

Static friction is not fixed at its maximum. It takes the value needed to keep the block from sliding, so here it balances the cord’s pull exactly:

f_2 = T = 10 N, which is greater than f_1 = 6.0 N

The lighter hanging weight produces the larger friction force. With the 2.0 kg weight the cord pulled the accelerating weight with 14.4 N, less than its 20 N weight, and the sliding block felt only the kinetic 6.0 N; here nothing moves, and the bench must supply the full 10 N the cord exerts. Static friction takes whatever value keeps the surfaces from slipping, up to μ_s·F_N, so a resting block can feel more friction than a sliding one.

Figure 6 — Horizontal forces on the block with each hanging weight, drawn to one force scale.

Scoring (3 points):

▸ B1 — 1 point: For indicating f_2 > f_1.

▸ B2 — 1 point: For establishing that the block remains at rest, by comparing the cord’s pull, (1.0 kg)(10 N/kg) = 10 N, with the largest force static friction can supply, μ_s·M·g = 12 N — or, equivalently, the hanging mass, 1.0 kg, with μ_s·M = 1.2 kg.

▸ B3 — 1 point: For indicating that the friction force on the resting block equals the cord’s pull, f_2 = 10 N, and is greater than the kinetic friction force on the sliding block, f_1 = μ_k·M·g = 6.0 N. A response that finds f_2 = 10 N for the resting block and compares that value correctly with its own friction force from part A (ii) earns this point.

Scoring Note: B1 is scored independently of the response in part A (ii), and B2 does not depend on part A. B1, B2 and B3 may each be earned independently of one another.

⚠︎ Common error (partial credit): Compares the 10 N pull with μ_s·M·g = 12 N and concludes that the block stays at rest, but then takes the static friction force to be that maximum, f_2 = 12 N, and indicates f_2 > f_1 — earns 2 of 3 points, B1 and B2. A 12 N friction force against a 10 N pull would leave a 2 N net force pointing away from the pulley, and a block at rest does not start sliding away from the pulley.

⚠︎ Common error (partial credit): Assumes the block stays at rest without comparing the pull with μ_s·M·g, then correctly sets f_2 = 10 N, compares it with f_1 = μ_k·M·g = 6.0 N, and indicates f_2 > f_1 — earns 2 of 3 points, B1 and B3. Whether the block moves at all is exactly what μ_s decides, and a 1.3 kg weight would have made it slide.

✗ Common error (no credit): Uses μ_k in place of μ_s for the largest friction force, finds μ_k·M·g = 6.0 N less than the 10 N pull, concludes the block slides, and indicates f_2 = f_1. None of the three points is earned. Whether a resting block starts to slide is decided by the coefficient of static friction, which here allows up to 12 N.

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