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FRQ: Kinetic Friction and Inertial Mass — The Two Sliding Crates

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

Two crates rest on the level concrete floor of a warehouse. Crate A has mass m_A = 12 kg and crate B has mass m_B = 3.0 kg. A worker gives each crate a brief horizontal push and then releases it, so that both crates leave the worker’s hands at the same speed v_0 = 2.0 m/s, travelling in the same direction along a straight line. After release no one touches either crate again, and each crate slides across the floor until it stops. Figure 1 shows the two crates at the instant of release.

Figure 1 — Crate A and crate B at the instant of release, both moving at 2.0 m/s.

The coefficient of kinetic friction between the floor and each crate is the same, μ_k = 0.20. Use g = 10 m/s², and treat air resistance as negligible.

All motion is one-dimensional and is measured in the warehouse (ground) frame, which is inertial. Model each crate as a particle. Take the direction of the initial push as positive, and take t = 0 for each crate to be the instant at which the worker releases it.

✎ Free Response Questions

Part A

i.    Consider crate A at an instant while it is sliding across the floor.

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

Figure 2 — The dot representing crate A, for the free-body diagram in part A (i).

ii.   Derive an expression for the acceleration of a crate while it is sliding across the floor, and then evaluate that expression for the quantities given in the Scenario. Your derivation must show explicitly the step at which the crate’s mass divides out.

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

iii.  Calculate the time each crate slides before coming to rest. A worker claims:

“Crate A is four times as heavy as crate B, so crate A has more inertia and will keep sliding for a longer time before it stops.”

State whether your results support the worker’s claim, and what, if anything, the worker’s reasoning does not take into account.

Justify your response.

Part B

The worker now repeats the experiment in the same warehouse, on a long, level, frictionless horizontal surface, releasing each crate at the same speed v_0 = 2.0 m/s as before. The crates and their masses are unchanged, and after release no one touches either crate again. Both crates remain on this surface throughout the interval considered below.

Indicate whether the speed v_A of crate A at t = 3.0 s is greater than, less than, or equal to the speed v_B of crate B at t = 3.0 s by writing one of the following.

•   v_A > v_B

•   v_A < v_B

•   v_A = v_B

Justify your response.

Your justification should identify the condition under which the crates’ inertial masses would affect their motion, and state whether that condition is met here.

❖ 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

Crate A is in contact with the floor and is sliding along it. Three forces are exerted on the crate: the gravitational force F_g = m_A·g = 120 N directed straight down; the normal force F_N from the floor, directed straight up and perpendicular to the surface; and the kinetic friction force f_k from the floor, directed horizontally and opposite to the crate’s velocity. The worker’s push ended at release, so no forward force is exerted on the crate while it slides.

The crate has no vertical acceleration, so the vertical forces balance, and the friction force follows from the normal force they fix:

ΣF_y = F_N − m_A·g = 0   so   F_N = m_A·g = 120 N
f_k = μ_k·F_N = (0.20)(120 N) = 24 N

Three arrows: F_g down and F_N up, the same length because they balance, and f_k horizontal and pointing opposite to the velocity, one-fifth as long as either vertical arrow. A diagram of a sliding crate carries no forward arrow, because the push ended at release.

Figure 3 — Free-body diagram of crate A while sliding; the vertical pair balances.

Scoring (2 points):

▸ A1 — 1 point: For drawing the gravitational force and the normal force as distinct arrows starting on the dot and pointing in opposite vertical directions, both labelled.

▸ A2 — 1 point: For drawing kinetic friction as a distinct arrow starting on the dot, horizontal and opposite to the crate’s velocity, and labelled, and for showing no forward horizontal arrow.

Scoring Note: A forward horizontal arrow — “applied force”, “force of the push”, “force of motion” — costs A2 only, being an error about which horizontal forces are exerted on a crate no one is touching; A1 is unaffected by it. Equal lengths for the vertical pair are not required, no force scale having been supplied; the balance is shown in Figure 3 and is scored as physics in A4. Any label that names the force is accepted — F_g, mg, weight, gravitational force, and likewise F_N, N, normal force, f_k, friction. The response is not required to name the object exerting each force, and numerical magnitudes are not required. A1 and A2 are earned on the response’s own diagram and carry no consistency credit, there being no upstream result to carry from.

⚠︎ Common error (partial credit): All three forces drawn, labelled and correctly directed, with a fourth arrow labelled “force of the push” added pointing forward — earns 1 of 2 points, A1. The push ended at release and exerts nothing on the sliding crate.

✗ Common error (no credit): Draws a forward arrow labelled “force of motion” and a backward friction arrow, with no vertical forces at all. A1 is not earned, neither vertical arrow being present; A2 is not earned, the forward arrow being present alongside the friction arrow. This is the misconception the part is built on — that a moving object must have something pushing it along.

Part A (ii) — Model Answer

Begin from Newton’s second law applied along the direction of motion, together with the kinetic friction relationship from the reference information:

ΣF_x = m·a       f_k = μ_k·F_N

The crate does not accelerate vertically, so the vertical forces balance and F_N = m·g. The friction force is therefore f_k = μ_k·m·g. It is the only horizontal force on the sliding crate, and it is directed opposite to the velocity, so:

−μ_k·m·g = m·a

The mass stands on both sides of this equation, so it divides out:

a = −μ_k·g
a = −(0.20)(10 m/s²) = −2.0 m/s²

The acceleration has magnitude 2.0 m/s² and is directed opposite to the velocity. An acceleration expression with no m in it predicts the same motion for every crate on this floor, whatever each one weighs.

Scoring (3 points):

▸ A3 — 1 point: For beginning from a fundamental principle — Newton’s second law ΣF = m·a written for the horizontal direction, or the relationship f_k = μ_k·F_N from the reference information. This point is earned for the correct starting principle alone, even if the algebra that follows is incorrect.

▸ A4 — 1 point: For a symbolic derivation that obtains F_N = m·g from the vertical forces balancing, uses it in a horizontal equation in which kinetic friction is the only horizontal force, and cancels the m standing on both sides to reach an expression for the acceleration that does not contain m. The expected expression is a = −μ_k·g, or μ_k·g with the direction stated as opposite to the velocity. A derivation reaching the same mass-free expression through impulse and momentum, or through work and energy, with friction as the only horizontal interaction with the floor, satisfies this criterion equally; a horizontal force written explicitly as zero does not count against “the only horizontal force”.

▸ A5 — 1 point: For evaluating the acceleration correctly from the response’s own work and stating its sign, or its direction relative to the velocity. The expected value is a = −2.0 m/s², directed opposite to the velocity. Award this point for a value correctly obtained from an expression the response derived, or from the response’s own force analysis where no symbolic expression was written, whatever that expression or analysis is. An acceleration asserted with no force analysis behind it is not the response’s own derivation, and evaluating it does not earn A5.

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

⚠︎ Common error (partial credit): Reaches −2.0 m/s² by substituting m = 12 kg and m = 3.0 kg separately and observing that the two answers agree — earns 2 of 3 points, A3 and A5. Two worked cases are evidence that the result is the same; they are not the symbolic cancellation A4 asks for, and m is present in every line of the working.

⚠︎ Common error (partial credit): Writes a = −μ_k·m·g, leaving the acceleration proportional to mass, and evaluates it for crate A as −24 m/s² — earns 2 of 3 points, A3 and A5. A4 is lost: the m on the right-hand side was cancelled and the m inside the friction force was not. A deceleration of 24 m/s² should itself prompt a check.

⚠︎ Common error (partial credit): Treats m·g as a horizontal force, writing ΣF_x = m·g − f_k, and evaluates the resulting a = g(1 − μ_k) as +8.0 m/s², forward — earns 2 of 3 points, A3 and A5. A4 is lost: the gravitational force is vertical, so this is not the horizontal equation Newton’s second law gives for this crate, whatever cancels inside it.

✗ Common error (no credit): States that the acceleration is −g = −10 m/s² because the crate is slowing down under gravity, writing no horizontal force equation and no friction relationship. A3 is not earned, neither Newton’s second law in the horizontal direction nor the friction relationship having been written; nothing is derived for A4; and A5 has no derivation or force analysis of the response’s own to evaluate.

Part A (iii) — Model Answer

Both crates leave the worker’s hands at the same speed, and by part A (ii) both slide with the same acceleration, of magnitude μ_k·g = 2.0 m/s². Using the constant-acceleration relationship with a final velocity of zero:

v = v_0 + a·t   with v = 0   so   t = v_0 / |a|
t = (2.0 m/s) / (2.0 m/s²) = 1.0 s

Each crate slides for 1.0 s, so the results do not support the worker’s claim.

The worker is right that crate A has four times the inertia of crate B, and right that a given force would therefore change crate A’s velocity four times as slowly. What the reasoning does not take into account is the size of the retarding force, which is not the same for the two crates. Kinetic friction is μ_k·m·g, so crate A, pressing on the floor four times as hard, also experiences four times the friction — 24 N against 6.0 N — and the two factors of four cancel exactly. The worker counts the inertia and forgets the friction, and both of them grow with the crate’s mass in exactly the same proportion.

Scoring (2 points):

▸ A6 — 1 point: For obtaining a stopping time for each crate from the response’s own analysis and the given initial speed, by any correct route — a constant-acceleration relationship with a final velocity of zero, or impulse and momentum, or work and energy followed by the one further step at the average speed v_0/2 that turns the stopping distance into a time. The expected value is t = 1.0 s for each crate. Award this point for times consistent with the response’s own acceleration or friction force from part A (ii).

▸ A7 — 1 point: For stating the verdict on the worker’s claim that follows from the response’s own two times, and identifying the quantity the worker’s reasoning does not take into account — the size of the retarding force on each crate, which depends on that crate’s mass just as its inertia does. For two equal times, the verdict is that the claim is not supported.

Scoring Note: A6 and A7 are independent of each other and of the response in part A (ii). A7 does not require the factor of four to be named, or either friction force to be evaluated; a response stating that the friction force grows with mass in the same way the inertia does has identified the omission in full. A response whose part A (ii) acceleration depends on mass obtains two different times, and A7 then asks for the verdict those times support together with the same omitted quantity, whether or not its own analysis makes that force mass-proportional.

⚠︎ Common error (partial credit): Computes 1.0 s correctly for both crates and concludes only that the times are equal, so the worker is wrong, with no account of what the worker’s reasoning does not take into account — earns 1 of 2 points, A6. The numerical agreement is the evidence, not the explanation.

⚠︎ Common error (partial credit): Identifies the omission in full, and correctly obtains the stopping distance Δx = v_0²/(2|a|) = 1.0 m, but then takes the stopping time as t = Δx/v_0 = (1.0 m)/(2.0 m/s) = 0.50 s for each crate — earns 1 of 2 points, A7. Dividing by the initial speed gives the time the crate would take at that speed throughout; a uniformly decelerating crate covers the distance at an average speed of v_0/2, which returns the stopping time to 1.0 s.

✗ Common error (no credit): Endorses the worker’s claim, predicting a longer sliding time for crate A on the strength of its inertia alone and computing no stopping time for either crate. Neither a time nor the omitted quantity is obtained.

Part B — Model Answer

v_A = v_B

On the frictionless surface the friction force is absent, and the worker exerts no force on either crate after release. The gravitational force and the normal force are vertical, equal in magnitude and opposite in direction, so on each crate they cancel. The net force on each crate is therefore zero:

ΣF = 0   so   a = 0   and   v = v_0 = 2.0 m/s for all t > 0

By Newton’s first law, a system on which the net force is zero moves with constant velocity, so neither crate’s velocity can change. Each still has speed 2.0 m/s at t = 3.0 s, the two speeds are equal, and neither crate ever comes to rest. On the concrete floor both had stopped 2.0 s earlier.

Figure 4 — On the frictionless surface only the balanced vertical pair remains.

Inertial mass is the property that determines how strongly an object’s motion resists being changed when something acts on it, and it enters the physics only through a = ΣF/m. The condition under which the two masses would make a difference is therefore a non-zero net force, and that condition is not met here: with ΣF = 0 the expression gives a = 0 whatever the value of m. Crate A’s greater inertia would show itself only if a net force were acting — under a 6.0 N net force crate A would accelerate at 0.50 m/s² while crate B accelerated at 2.0 m/s². Inertia is a property of an object rather than a force exerted on it, so it can change how a net force acts but can do nothing when there is none.

Scoring (3 points):

▸ B1 — 1 point: For indicating v_A = v_B.

▸ B2 — 1 point: For establishing that the net force on each crate is zero — the vertical forces balancing and no horizontal force being exerted — and concluding that each crate’s velocity therefore cannot change.

▸ B3 — 1 point: For stating the condition under which the crates’ inertial masses would affect their motion, and that it is not met here: mass bears on the motion only through a = ΣF/m, so it matters only when ΣF ≠ 0, and here ΣF = 0.

Scoring Note: B1, B2 and B3 are independent of one another and of the response in part A. Restating the value 2.0 m/s is not required for B2, and the conclusion that the velocity cannot change may be stated directly or attributed to Newton’s first law. B3 is a content criterion, not a route criterion: it requires the response to establish that mass bears on the motion only through the net force, whether by writing a = ΣF/m, by contrasting a case in which a net force does act, or by any equivalent route. Supporting it with a contrasting non-zero-force case is one way to earn B3 and is not required.

⚠︎ Common error (partial credit): Indicates v_A = v_B and gives correct first-law reasoning, but the discussion of inertial mass stops at the observation that mass does not matter here, without naming the condition that would make it matter — earns 2 of 3 points, B1 and B2. The missing step is what stops “mass never matters” being carried away as a general rule.

⚠︎ Common error (partial credit): Indicates v_A = v_B and justifies it only by asserting that momentum is conserved, with no appeal to Newton’s first law and no account of the role of inertial mass — earns 1 of 3 points, B1. The conservation argument rests on the same zero-net-force condition, but the response neither establishes that condition nor names the law the justification asks for.

✗ Common error (no credit): Indicates v_A < v_B, on the grounds that crate A has more inertia and so will come to rest first, or treats inertia as something that acts on the crate to slow it down. Inertia never appears on a free-body diagram, and with no horizontal force exerted on either crate neither slows at all.

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