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

FRQ: Static and Kinetic Friction — Sliding the Tool Chest

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Translation Between Representations (TBR)  |  MID-LEVEL  |  12 points

▤ Scenario

Two technicians must reposition a 10 kg steel tool chest 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.

Throughout, the technicians' push is horizontal and the chest's base stays flat on the floor. 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

(a) The technicians have increased their push to 60 N, and the chest is at rest on the verge of sliding. Draw a free-body diagram showing the forces exerted on the chest at that instant. Label each force with its name and the object exerting it, and draw the arrows so that their lengths represent the relative magnitudes of the forces.

(b) The chest is now sliding, and the technicians hold their horizontal push constant at a magnitude F. Starting from Newton's second law, derive an expression for the magnitude a of the chest's acceleration in terms of F, the mass m, the coefficient of kinetic friction μ_k and g. Use your expression, together with the two experimental observations, to determine μ_k and μ_s, and then to determine the chest's acceleration when the technicians raise their steady push to 70 N.

(c) The chest is brought back to rest. Starting from rest, the technicians then increase their horizontal push slowly from 0 N to 70 N. On an x–y coordinate system, sketch a graph of the magnitude of the friction force exerted on the chest by the floor as a function of the magnitude of the applied horizontal push, over that range. Label the value of the friction force wherever its behaviour changes.

(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 both coefficients of friction between the lower chest and the floor are unchanged. A technician claims:

“Doubling the mass doubles the normal force, so every friction value on the graph in part (c) doubles — the whole graph simply scales up by a factor of two. And the push needed to give the chests any particular acceleration doubles as well.”

In a clear, paragraph-length response, use your expression from part (b) and your graph from part (c), extended to pushes as large as 160 N, to evaluate this claim. Address both of the technician's statements, and state the condition your reasoning depends on.

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

At the verge of sliding the chest is still at rest, so its acceleration is zero and the forces 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 applied push exerted by the technicians, and the static friction exerted by the floor.

Free-body diagram of the chest at the verge of sliding, with all four arrows drawn to a common scale. Each force carries its name and the object exerting it.

Vertically, F_g = mg = (10)(10) = 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, which has risen to its maximum value f_s,max = 60 N in the opposite direction. Each pair therefore balances, so the arrows of each pair are drawn equal in length, the vertical pair longer because 100 N exceeds 60 N.

Scoring (3 points):

▸ 1 point: draws exactly four forces exerted on the chest — the gravitational force downward, the normal force upward, the applied push horizontally in the direction of the intended motion, and the static friction horizontally opposite to it — with no additional force shown.

▸ 1 point: labels each of the four forces with both its name and the object exerting it (Earth, the floor, the technicians, the floor).

▸ 1 point: draws the two vertical arrows equal in length to each other and the two horizontal arrows equal in length to each other, consistent with zero acceleration in both directions, and draws the vertical pair visibly longer than the horizontal pair, since 100 N is greater than 60 N.

⚠︎ Common error (partial credit): a correct, correctly labelled diagram in which all four arrows are drawn the same length — earns 2 of 3 points, losing the relative-length point.

⚠︎ Common error (partial credit): all four forces drawn with correct directions and correct relative lengths, but labelled by name only with no exerting object given — earns 2 of 3 points.

✗ Common error (no credit): draws the forces exerted by the chest on its surroundings rather than the forces exerted on the chest — for example the chest pressing down on the floor and the chest pushing back on the technicians. A free-body diagram carries only the forces exerted on the chosen body, so the force inventory, the labels and the relative lengths are all wrong and no criterion is earned.

Part (b) — Model Answer

While the chest slides, the only horizontal forces exerted on it are the applied push F, in the direction of motion, and the kinetic friction from the floor, opposite to it. The vertical forces still balance, so the normal force is unchanged.

F_N = mg
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.

Evaluating. A steady 40 N push gives constant velocity, so a = 0 at F = 40 N and μ_k = F/(mg) = 40/100 = 0.40. At the verge of sliding the chest is also in equilibrium, with the static friction at its maximum, so μ_s = f_s,max/(mg) = 60/100 = 0.60. Both coefficients are dimensionless, and μ_s > μ_k as expected for a single pair of surfaces. Substituting F = 70 N:

a = 70/10 − (0.40)(10) = 7.0 − 4.0 = 3.0 m/s²

The chest accelerates at 3.0 m/s² in the direction of the push.

Scoring (4 points):

▸ 1 point: starts from Newton's second law applied to the sliding chest along the direction of motion, with the applied push and the kinetic friction identified as the only horizontal forces exerted on it. This point is earned for the correct starting principle even if the algebra that follows is incorrect.

▸ 1 point: obtains the normal force from the vertical balance, F_N = mg, and substitutes f_k = μ_k·F_N to reach F − μ_k·mg = ma.

▸ 1 point: solves symbolically for a = (F − μ_k·mg)/m = F/m − μ_k·g before substituting any numbers. Consistency credit: award this point for the same symbolic pathway carried through with the student's own expression for the normal force.

▸ 1 point: evaluates μ_k = 0.40 and μ_s = 0.60, both dimensionless, and a = 3.0 m/s² at F = 70 N with units and direction. Consistency credit: award this point for values correctly carried through the student's own expression and the student's own coefficients.

⚠︎ Common error (partial credit): substitutes numbers from the outset and never writes the symbolic expression, but reaches μ_k = 0.40, μ_s = 0.60 and a = 3.0 m/s² correctly — earns 3 of 4 points, losing the symbolic-solve point.

⚠︎ Common error (partial credit): takes the normal force as F_N = m = 10 N rather than mg, giving μ_s = 6.0 and μ_k = 4.0 — earns 3 of 4 points: the starting principle, the symbolic solve, and, by consistency credit, the evaluation carried through with the student's own coefficients. Only the normal-force substitution point is lost.

✗ Common error (no credit): writes the horizontal equation of motion as F − mg = ma, treating the gravitational force as though it opposed the push — the gravitational force is vertical and is balanced by the normal force, so the starting principle, the substitution, the symbolic result and the evaluation are all lost.

Part (c) — Model Answer

While the chest is at rest, static friction adopts whatever value is required to keep it from slipping, so the friction force is equal in magnitude to the applied push. On the graph that is a straight line through the origin along which friction equals push, rising to the maximum static friction of 60 N.

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, never on how hard the technicians push, so the graph drops abruptly from 60 N to 40 N there and stays at 40 N out to 70 N.

Friction force on the chest against applied push, as the push is raised slowly from rest. The filled point at 60 N belongs to the static branch, where the chest is still in equilibrium; the open point marks a value the friction never takes.

Scoring (2 points):

▸ 1 point: while the chest remains at rest, the graph is a straight line from the origin along which the friction force equals the applied push, rising to a maximum of 60 N, with 60 N labelled.

▸ 1 point: once sliding begins the graph drops to a constant 40 N and remains at 40 N for every larger push up to 70 N, with 40 N labelled.

⚠︎ Common error (partial credit): a correct rising static branch to 60 N, but the friction then shown continuing to rise with the push after sliding begins — earns 1 of 2 points, losing the kinetic-branch point.

⚠︎ Common error (partial credit): a correct drop to a constant 40 N, but the static branch drawn as a horizontal line at 60 N from the origin rather than rising with the push — earns 1 of 2 points, losing the static-branch point.

✗ Common error (no credit): draws 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 static branch is absent and no drop is shown, so neither criterion is earned.

Part (d) — Model Answer

The technician's first statement is wrong and the second is right.

Doubling the mass does double the normal force: with both chests on the floor F_N = (2m)g = 200 N, so the maximum static friction becomes μ_s·F_N = (0.60)(200) = 120 N and the kinetic friction becomes μ_k·F_N = (0.40)(200) = 80 N. The two labelled values on the graph therefore do double.

It does not follow that every friction value doubles. While the chests are at rest, static friction adopts exactly the value needed to prevent slipping, so the friction equals the applied push whatever the mass — at a 30 N push it is 30 N with one chest and 30 N with two. The rising branch is the line “friction = push” and is unchanged. What moves is where that branch ends, 120 N instead of 60 N, and the height of the constant branch that follows, 80 N instead of 40 N. The graph does not scale up by a factor of two; it keeps the same rising line and simply extends it further.

How the graph changes when the mass is doubled. The rising branch is common to both cases; only the breakpoint and the constant kinetic value move.

Over the original 0 N to 70 N window nothing changes at all: 120 N is now needed to start the chests moving, so across that whole window they stay at rest and the friction simply equals the push, exactly as it did with one chest. Every change the technician is asking about appears only further to the right.

The second statement 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 required push is directly proportional to the mass, so doubling the mass doubles it: at a = 3.0 m/s² one chest needs 70 N and the two chests need 140 N.

The reasoning depends on the coefficients being unchanged — the same pair of surfaces, with the base still flat on the same floor — and on the normal force still being the full weight of what is pushed, 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 neither result would hold.

Scoring (3 points):

▸ 1 point: predicts both labelled values correctly for the doubled mass — f_s,max = μ_s(2m)g = 120 N and f_k = μ_k(2m)g = 80 N — on the grounds that the normal force doubles while the coefficients are unchanged.

▸ 1 point: rejects the first statement with its reason: while the chests are at rest the static friction still equals the applied push, so the rising branch of the graph is unchanged and only the breakpoint and the constant kinetic value move. Any of these counts as the evidence: a numerical instance (30 N of friction at a 30 N push in both cases); a correctly described or drawn graph; or the observation that over the original 0 N to 70 N range the chests never slide at all, so nothing on that part of the graph changes.

▸ 1 point: supports the second statement from the part (b) expression, F = m(a + μ_k·g), so at a fixed acceleration the required push is proportional to the mass and therefore doubles; and names the condition the reasoning depends on — the coefficients unchanged and the normal force still equal to the full weight.

⚠︎ Common error (partial credit): correctly doubles both labelled values and correctly rejects “the whole graph scales”, but judges the second statement wrong as well, on the grounds that friction has doubled so the push must more than double — earns 2 of 3 points, losing the third criterion.

⚠︎ Common error (partial credit): agrees with both statements and doubles every value on the graph, but does identify f_s,max = 120 N and f_k = 80 N correctly — earns 1 of 3 points, the first criterion only: the rising branch is misrepresented, and the required-push claim is repeated rather than supported from the derived expression.

✗ Common error (no credit): argues that the friction forces are unchanged because the coefficients are unchanged — this confuses the coefficient with the force. The coefficients are indeed the same, but each friction force is a coefficient multiplied by the normal force, and the normal force has doubled. No criterion is earned.