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FRQ: Static and Kinetic Friction — Sliding the Tool Chest

Assessments aligned to 2026 AP Physics 1 standards

Translation Between Representations (TBR)  |  MID-LEVEL  |  12 points  |  30 min

▤ Scenario

Two technicians must reposition a steel tool chest of mass m = 10 kg across a level concrete floor. They notice that it takes a hard initial shove to get the chest going, but that once it is sliding a much gentler push keeps it moving.

By experiment they establish two values. The chest is on the verge of sliding when their horizontal push reaches 60 N. Once the chest is sliding, a steady horizontal push of 40 N keeps it moving at constant velocity. Figure 1 shows the chest and the two measured pushes.

Figure 1 — The steel tool chest on the level floor, and the technicians’ two measured pushes.

Throughout, the technicians’ push is horizontal and the chest’s base stays flat on the floor. Treat their combined push as a single horizontal force of magnitude F exerted on the chest. Air resistance is negligible. Take g = 10 m/s². All motion is one-dimensional and horizontal; take the direction of the chest’s motion as positive and the floor as the reference frame.

✎ Free Response Questions

Part A

Consider the chest at the instant the push reaches 60 N, when the chest is still at rest and on the verge of sliding.

On the following dot that represents the chest, draw and label the forces (not components) that are exerted on the chest.

•   Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.

•   Draw the length of each arrow to represent the magnitude of each force, consistent with the scale used in Figure 2.

Figure 2 — The dot representing the chest. The side of each grid square represents 20 N.

Part B

The chest is now sliding, and the technicians hold their push constant at magnitude F. Let μ_k be the coefficient of kinetic friction between the chest and the floor. Starting with Newton’s second law, derive an expression for the acceleration a of the chest. Express your final answer in terms of F, m, μ_k, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

The technicians then raise their steady push to 70 N. Use your expression, together with the observation that a steady 40 N push keeps the sliding chest moving at constant velocity, to predict the chest’s acceleration.

Part C

The chest is brought back to rest. Starting from rest, the technicians then increase their push slowly from 0 N to 70 N. Let f be the magnitude of the friction force exerted on the chest by the floor. Figure 3 shows axes of f against the push F.

On Figure 3, sketch f as a function of F for pushes from 0 N to 70 N.

Figure 3 — Axes for part C: the friction force f against the push F.

Part D

A second, identical tool chest is now stacked on the first, so that the total mass is 20 kg. The lower chest’s base still rests flat on the same concrete floor, the two chests do not slip on each other, and the coefficients of static and kinetic friction, μ_s and μ_k, between the lower chest and the floor are unchanged. A technician makes two statements about the stacked chests.

Statement 1: “Doubling the mass doubles the normal force, so every friction value on the graph in part C doubles at the same push — the whole graph simply stretches upward by a factor of two.”

Statement 2: “The push needed to give the chests any particular acceleration doubles as well.”

Describe how the graph of the friction force f against the push F from part C, extended to pushes as large as 160 N, would change for the stacked chests.

Using that graph and your expression from part B, applied to the stacked chests, indicate which of the technician’s statements, if any, are correct by writing one of the following.

______  Both statements are correct

______  Only Statement 1 is correct

______  Only Statement 2 is correct

______  Neither statement is correct

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 — Model Answer

At the verge of sliding the chest is still at rest, so its acceleration is zero and the forces on it balance in both directions. Exactly four forces are exerted on the chest, and no others: the gravitational force exerted by Earth, the normal force exerted by the floor, the push exerted by the technicians, and the static friction force exerted by the floor.

Vertically, F_g = mg = (10 kg)(10 m/s²) = 100 N downward is balanced by F_N = 100 N upward. Horizontally, the 60 N push, in the direction of the intended motion, is balanced by the static friction force, which has risen to its maximum value, f_s,max = 60 N, in the opposite direction. On the 20 N scale of Figure 2 the two vertical arrows are each five squares long and the two horizontal arrows each three.

Figure 4 — The chest on the verge of sliding: four forces on the Figure 2 scale.

Each pair balances, so the arrows of each pair are equal in length, and the vertical pair is the longer because 100 N is greater than 60 N. Static friction on the resting chest matches the push, up to a maximum of 60 N set by the coefficient of static friction and the 100 N normal force.

Scoring (3 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.

▸ A2 — 1 point: For drawing the push and the static friction force as horizontal arrows in opposite directions, as distinct labelled arrows starting on the dot, with no other horizontal force and no additional arrow in a slanted direction, such as a resultant of the normal and friction forces drawn alongside them.

Scoring Note: Any label that names the force is accepted — F_g, mg, weight or gravitational force; F_N, N or normal force; F, F_push, applied force or push; f, f_s, F_f or friction. The response is not required to name the object exerting each force.

▸ A3 — 1 point: For drawing the four arrows to the scale of Figure 2: the gravitational force and the normal force each 100 N, five grid squares, and the push and the static friction force each 60 N, three grid squares, each within about half a grid square.

Scoring Note: A3 is scored on the lengths of the arrows drawn for these four forces, whether or not A1 and A2 are earned.

⚠︎ Common error (partial credit): A correct, correctly labelled diagram in which all four arrows are drawn the same length — earns 2 of 3 points, for A1 and A2.

⚠︎ Common error (partial credit): Draws the static friction force two squares long, 40 N, shorter than the 60 N push, with every other arrow correct. The chest is at rest, so the friction balances the push; 40 N is the kinetic value, which applies only once the chest slides — earns 2 of 3 points, for A1 and A2.

⚠︎ Common error (partial credit): Draws the gravitational and normal forces correctly and the push, but no friction force, on the grounds that friction acts only on a moving object. Static friction acts on the resting chest and is what holds it still against a 60 N push — earns 1 of 3 points, for A1.

✗ Common error (no credit): Draws the forces the chest exerts on its surroundings — the chest pressing down on the floor and pushing back on the technicians — rather than the forces exerted on the chest. A free-body diagram carries only the forces exerted on the chosen object, so neither pair is the one A1 and A2 describe, and the arrows drawn are not the four forces A3 measures.

Part B — Model Answer

While the chest slides, the only horizontal forces exerted on it are the push F, in the direction of motion, and the kinetic friction force from the floor, opposite to it. The vertical forces still balance, so the normal force is unchanged: F_N = mg. Starting with Newton’s second law along the direction of motion:

F − f_k = ma,  with f_k = μ_k·F_N = μ_k·mg

a = (F − μ_k·mg) / m = F/m − μ_k·g

No speed appears in the result, so while the push is held constant the acceleration is the same at every speed.

The 40 N observation fixes the friction term without a separate calculation. Setting a = 0 at F = 40 N shows that the friction term μ_k·mg is 40 N for this chest, so while it slides a = (F − 40 N)/(10 kg). At a 70 N push:

a = (70 N − 40 N) / (10 kg) = 3.0 m/s²

Equivalently, μ_k = (40 N)/(mg) = 0.40, and a = 70/10 − (0.40)(10) = 3.0 m/s². The chest accelerates at 3.0 m/s² in the direction of the push. The steady 40 N push only balances the 40 N kinetic friction force, so just the 30 N by which a 70 N push exceeds it accelerates the chest.

Scoring (4 points):

▸ B1 — 1 point: For a multistep derivation that includes Newton’s second law applied to the sliding chest along its direction of motion.

▸ B2 — 1 point: For expressing the friction force on the sliding chest as the kinetic friction force, μ_k·mg, or μ_k·F_N with F_N = mg.

▸ B3 — 1 point: For a correct expression for a: a = F/m − μ_k·g, or any algebraic equivalent such as (F − μ_k·mg)/m. An expression in which the measured 40 N replaces the friction term is not in terms of μ_k and does not earn B3; it can earn B4.

Scoring Note: A correct, isolated, final expression for a earns points B2 and B3.

▸ B4 — 1 point: For predicting the acceleration at a 70 N push by fixing the friction term from the 40 N constant-velocity observation, a = 0 at F = 40 N — for example a = (70 N − 40 N)/(10 kg) = 3.0 m/s², or μ_k = 0.40 substituted into the expression. Award the point when the response reports the value that its own part B expression gives at F = 70 N after the response has fixed that expression’s friction term from a = 0 at F = 40 N, whatever that value is; a response with no symbolic expression may earn it by the first route. An expression that cannot give a = 0 at F = 40 N, such as a = F/m − g or a = F/m, has no friction term the observation can fix and does not earn B4.

Scoring Note: B1 is scored on the starting relation. B2 and B3 are not scored for consistency with an earlier error: B2 is scored on the friction force the response writes, and B3 on its final expression. B4 is scored consistently with the response’s own part B expression.

⚠︎ Common error (partial credit): Substitutes numbers from the outset: finds from the 40 N observation that the kinetic friction force is 40 N, writes 70 N − 40 N = (10 kg)·a and reports 3.0 m/s², but never writes an expression for a in symbols — earns 2 of 4 points, for B1 and B4.

⚠︎ Common error (partial credit): Takes the normal force as F_N = m rather than mg, reaching a = F/m − μ_k. Setting a = 0 at 40 N then gives μ_k = 4.0, and the prediction at 70 N still comes out at 3.0 m/s², because the 40 N observation absorbs the missing g — earns 2 of 4 points, for B1 and B4.

⚠︎ Common error (partial credit): Starts from Newton’s second law but treats the push as the only horizontal force, so that a = F/m, and predicts 7.0 m/s² at 70 N. The kinetic friction force, and the 40 N observation that measures it, never appear — earns 1 of 4 points, for B1.

⚠︎ Common error (partial credit): Uses the static coefficient for the sliding chest, writing F − μ_s·mg = ma, and predicts (70 N − 60 N)/(10 kg) = 1.0 m/s² from the 60 N verge-of-sliding push. Static friction acts only while the surfaces do not slide over each other; once the chest slides the friction is kinetic — earns 1 of 4 points, for B1.

⚠︎ Common error (partial credit): Writes F − mg = ma for the horizontal motion, treating the gravitational force as though it opposed the push, and evaluates it at 70 N as −3.0 m/s². The gravitational force is vertical and is balanced by the normal force. The expression never uses the 40 N observation, which it cannot satisfy — earns 1 of 4 points, for B1.

✗ Common error (no credit): Writes no equation of motion and reads the 40 N observation as “a steady push gives a steady speed”, concluding that a steady 70 N push keeps the chest moving at a higher constant velocity, with a = 0. A constant 30 N net force gives a constant acceleration; no principle, friction force or expression is written, and no prediction is drawn from one.

Part C — Model Answer

While the chest is at rest, static friction adopts whatever value is required to keep it from sliding, so the friction force is equal in magnitude to the push. On the graph that is a straight line through the origin along which f = F, rising to the maximum static friction force of 60 N. With μ_s the coefficient of static friction, that maximum is f_s,max = μ_s·F_N, so this breakpoint is what fixes μ_s = 60 N / 100 N = 0.60, greater than μ_k = 0.40, as expected for one pair of surfaces.

Static friction cannot exceed that maximum. The instant the push passes 60 N the chest begins to slide, and the force opposing it becomes kinetic friction, f_k = μ_k·F_N = 40 N. Kinetic friction depends only on μ_k and the normal force, not on how hard the technicians push horizontally, so the graph drops abruptly from 60 N to 40 N there and stays at 40 N out to 70 N. Static friction answers the push, up to its maximum, while kinetic friction is set by μ_k and the normal force alone.

Figure 5 — f = F at rest up to 60 N, then a constant 40 N once the chest slides.

Scoring (2 points):

▸ C1 — 1 point: For a straight line from the origin along which the friction force equals the push, f = F, up to F = 60 N, where f reaches 60 N, each value within 5 N, half a grid spacing.

▸ C2 — 1 point: For a drop, where the at-rest branch ends, to a horizontal line at f = 40 N, within 5 N, running from there to F = 70 N. A drop drawn only at F = 70 N does not earn C2.

Scoring Note: C1 and C2 are scored on the sketch itself and do not depend on any other part. Open and filled circles at the breakpoint are not required, and a vertical segment joining the two branches there is accepted.

⚠︎ Common error (partial credit): A correct rising branch to 60 N, but the friction then shown continuing to rise with the push after the chest starts to slide — earns 1 of 2 points, for C1.

⚠︎ Common error (partial credit): A correct rising branch to 60 N, then a flat line at 60 N out to 70 N, as though the friction stayed at its static maximum while the chest slides — earns 1 of 2 points, for C1.

⚠︎ Common error (partial credit): A correct drop to a constant 40 N, but the friction drawn as a flat line at 60 N from F = 0, rather than rising with the push. A resting chest with no push on it feels no friction at all — earns 1 of 2 points, for C2.

✗ Common error (no credit): A single horizontal line at 40 N across the whole range, treating kinetic friction as though it acted while the chest was still at rest — the rising branch is absent and there is no drop, so neither C1 nor C2 is met.

Part D — Model Answer

Only Statement 2 is correct.

Doubling the mass does double the normal force. With the upper chest resting on the lower one, the floor supports both chests, so F_N = (2m)g = 200 N. The maximum static friction force becomes μ_s·F_N = (0.60)(200 N) = 120 N, so the stacked chests stay at rest until the push reaches 120 N, and once they slide the kinetic friction force is μ_k·F_N = (0.40)(200 N) = 80 N.

It does not follow that every friction value doubles at the same push. While the chests are at rest, static friction adopts exactly the value needed to prevent sliding, so the friction force equals the push whatever the mass: at a 30 N push it is 30 N with one chest and 30 N with two. The rising part of the graph is the line f = F and is unchanged. What moves is where that line ends, at a 120 N push instead of 60 N, and the height of the constant part that follows, 80 N instead of 40 N. The graph keeps the same rising line and extends it further; it does not stretch upward by a factor of two.

Figure 6 — One chest and two: the rising line f = F is shared; the breakpoint and plateau move.

Inside the part C window the difference appears as the push passes 60 N. There the single chest breaks free and its friction drops to 40 N, while the stacked chests are still at rest, so their friction keeps rising with the push, reaching 70 N at a 70 N push — neither the single chest’s 40 N nor the 80 N that doubling at the same push would give. The stacked graph is exactly the one-chest graph enlarged by a factor of two along both axes — a doubling of the push and the friction together, not of the friction at the same push.

Statement 2 is correct, and the expression from part B shows why. Rearranging a = F/m − μ_k·g gives F = m(a + μ_k·g). At a fixed acceleration the push needed is proportional to the mass, because both the part that accelerates the chests, m·a, and the part that balances kinetic friction, μ_k·mg, double: at a = 3.0 m/s² one chest needs 70 N and the two chests need 140 N.

The doubled values, 120 N and 80 N, depend on the coefficients being unchanged and on the normal force still being the full weight of both chests, which requires a level floor and a horizontal push. Were the technicians to push downward at an angle, the normal force would exceed the total weight and grow with the push, so the friction would no longer be 120 N at breakaway or a constant 80 N while sliding. Statement 2 needs less: the push needed, M·a + μ_k·Mg for a load of mass M, doubles because both of its terms are proportional to M. Doubling the mass doubles the maximum static and the kinetic friction forces, not the friction on a resting chest at a given push.

Scoring (3 points):

▸ D1 — 1 point: For predicting that the stacked chests stay at rest until the push reaches 120 N and then slide against a constant kinetic friction force of 80 N, because the normal force doubles to 200 N while the coefficients are unchanged — stated directly, or as a maximum static friction force of 120 N reached along the line f = F. The reason may appear anywhere in part D.

▸ D2 — 1 point: For rejecting Statement 1, in the selection and in the description or justification, because while the chests are at rest the static friction force equals the push whatever the mass, so the rising part of the graph, f = F, is unchanged. Any of the following counts as the evidence: a numerical instance at a push below 60 N, such as 30 N of friction at a 30 N push for one chest and for two; a numerical instance at a push between 60 N and 120 N, such as the stacked chests still at rest at a 70 N push with 70 N of friction, where doubling the single chest’s 40 N would give 80 N; or a correctly described or drawn extended graph.

▸ D3 — 1 point: For a verdict on Statement 2, in both the selection and the justification, drawn from the response’s part B expression for a in terms of F, m, μ_k and g — for example by rearranging it as F = m(a + μ_k·g), or by substituting 2m for every m, including inside the friction term, and holding a fixed. For the correct expression, the part of the push that accelerates the chests and the part that balances kinetic friction are each proportional to the mass, so the push needed doubles. Award the point wherever the verdict is correctly drawn from the response’s own part B expression in those terms, whatever verdict it leads to. A form in which the single chest’s 40 N stands in for the friction term does not count, because that term does not stay 40 N when the mass doubles. A response whose part B contains no symbolic expression may earn the point by writing the push needed for the stacked chests, (2m)·a + μ_k·(2m)g or an equivalent, in part D.

Scoring Note: D1 and D2 are scored on the physics of the stacked chests and may be earned whatever graph the response drew in part C. D2 turns on the reason: static friction matching the push while the chests are at rest.

⚠︎ Common error (partial credit): Places the breakaway at a 120 N push and the plateau at 80 N and rejects Statement 1 correctly, but judges Statement 2 wrong as well, on the grounds that the friction has doubled so the push must more than double — earns 2 of 3 points, for D1 and D2.

⚠︎ Common error (partial credit): Selects “Only Statement 2 is correct” with D1 and D2 correct but, although its own part B expression contains the friction term, supports Statement 2 from F = ma alone. The push needed is m·a plus the kinetic friction force, and it doubles only because both terms double — earns 2 of 3 points, for D1 and D2.

✗ Common error (no credit): Accepts both statements, drawing the rising part as f = 2F up to 120 N at a 60 N push. That line puts a static friction force twice the push on a resting chest, which could not then stay at rest; it places the breakaway at a 60 N push, rejects neither statement, and repeats Statement 2 without drawing it from part B.

✗ Common error (no credit): Argues that the friction forces on the stacked chests are unchanged because the coefficients are unchanged, and so that the push needed for a given acceleration less than doubles. This confuses the coefficient with the force: each friction force — the static maximum and the kinetic value — is a coefficient multiplied by the normal force, and the normal force has doubled. It predicts neither 120 N nor 80 N; its reason for rejecting Statement 1 — unchanged coefficients — is not that static friction matches the push, so it does not earn D2 even where it quotes the correct 30 N instance; and its verdict on Statement 2 contradicts its own part B expression for a in terms of F, m, μ_k and g.