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FRQ: Work by Gravity and Friction on a Round Trip — The Mail-Sorting Ramp

Assessments aligned to 2026 AP Physics 1 standards

Translation Between Representations (TBR)  |  CHALLENGE  |  12 points  |  30 min

▤ Scenario

In a mail-sorting depot, a rigid plastic tote of mass m is given a quick shove at the bottom of a straight ramp inclined at angle φ above the horizontal. The tote leaves the worker's hands at point O with speed v_0, moving up the ramp. The coefficient of kinetic friction between the tote and the ramp is μ_k. The tote slides up the ramp, comes momentarily to rest at its highest point, a distance d from O measured along the ramp, and then slides back down the ramp to O.

Air resistance is negligible, the tote does not tip or rotate, and it may be modelled as a single object. Let s be the tote's distance from O measured along the ramp, and let K_0 = ½mv_0² be its kinetic energy at O at the start. The ramp is at rest in the depot's reference frame, which is inertial.

Figure 1: The tote leaves O with speed v_0 up the ramp, reaches its highest point a distance d along the ramp, and returns.

✎ Free Response Questions

Part A

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

• Draw each force in its correct direction relative to the ramp.

• If any force is zero, do not draw it.

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Part B

Derive an expression for the work W_f done by the friction force on the tote as it moves from O up to its highest point. Express your final answer in terms of m, g, μ_k, φ, and d, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Part C

Figure 2 shows the tote's kinetic energy K as a function of s for the upward trip, from s = 0 to s = d. On Figure 2, sketch the tote's kinetic energy as a function of s for the downward trip, from s = d back to s = 0. Your sketch should show clearly how the kinetic energy at the end of the downward trip compares with K_0, and how the slope of your line compares with the slope of the upward-trip line.

Figure 2: Kinetic energy of the tote against distance s along the ramp for the upward trip (given).

Part D

Justify whether or not your sketch in Part C is consistent with your expression from Part B. Briefly justify your response by referencing a fundamental physics principle.

❖ 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

Three forces are exerted on the tote while it moves up the ramp: the gravitational force exerted by Earth, F_g = mg, vertically downward; the normal force exerted by the ramp, F_N, perpendicular to the ramp surface and pointing away from it; and the kinetic friction force exerted by the ramp, f_k, parallel to the ramp surface and pointing down the ramp, because friction opposes the tote's motion relative to the ramp. The worker's hands are no longer in contact with the tote, so there is no push up the ramp.

A moving object carries no "force of motion" — every force on the tote points down the ramp or perpendicular to it, which is why it slows down.

Part A: Free-body diagram of the tote moving up the ramp — F_g down, F_N perpendicular to the ramp, f_k down the ramp.

Scoring (3 points):

▸ A1 — 1 point: For drawing the gravitational force as an arrow starting on the dot and pointing vertically downward, labelled F_g, mg, W or "gravity".

▸ A2 — 1 point: For drawing the normal force as an arrow starting on the dot, perpendicular to the ramp surface and pointing away from it, labelled F_N, N or "normal".

▸ A3 — 1 point: For drawing the kinetic friction force as an arrow starting on the dot, parallel to the ramp and pointing down the ramp, labelled f, f_k, F_f or "friction", with no extraneous forces drawn.

⚠︎ Common error (partial credit): friction drawn up the ramp, in the direction of motion — earns A1 and A2, 2 of 3 points.

⚠︎ Common error (partial credit): the three correct forces plus an extra arrow up the ramp labelled "push", "F" or "motion" — earns A1 and A2, 2 of 3 points; the extraneous force forfeits the friction point.

⚠︎ Common error (partial credit): normal force drawn vertically upward — earns A1 and A3, 2 of 3 points.

✗ Common error (no credit): a single arrow labelled "F_net" or "motion", or components mg sin φ and mg cos φ drawn in place of the gravitational force with no normal or friction arrows — no criterion is met.

Part B — Model Answer

Perpendicular to the ramp the tote does not accelerate, so Newton's second law in that direction gives F_N − mg cos φ = 0, i.e. F_N = mg cos φ. The kinetic friction force therefore has magnitude f_k = μ_k F_N = μ_k mg cos φ, constant along the ramp. Friction points down the ramp while the tote moves up it, so the angle between the force and the displacement is 180°. From the definition of work, W = Fd cos θ:

W_f = (μ_k mg cos φ)(d)(cos 180°)

W_f = −μ_k mg d cos φ

The work is negative: friction removes kinetic energy on the way up, and its magnitude is friction force × path length along the ramp.

Scoring (4 points):

▸ B1 — 1 point: For beginning the derivation with a fundamental principle: Newton's second law perpendicular to the ramp (F_N − mg cos φ = 0), the definition of work W = Fd cos θ, or the work–energy theorem with gravity's work included.

▸ B2 — 1 point: For the normal force F_N = mg cos φ.

▸ B3 — 1 point: For applying f_k = μ_k F_N with the student's own normal force.

▸ B4 — 1 point: For the final expression W_f = −μ_k mg d cos φ, negative, or W_f = −f_k d evaluated consistently with the student's own expression for f_k.

⚠︎ Common error (partial credit): uses the level-floor normal force F_N = mg and obtains W_f = −μ_k mg d — earns B1, B3 and B4 under the consistency clauses, 3 of 4 points; only B2 is lost.

⚠︎ Common error (partial credit): correct magnitude but positive sign, W_f = +μ_k mg d cos φ — earns B1, B2 and B3, 3 of 4 points.

⚠︎ Common error (partial credit): begins with the work–energy theorem but sets the friction work equal to the whole loss of kinetic energy, W_f = −½mv_0², ignoring the work done by gravity — earns B1 only, 1 of 4 points.

Scoring Note: A correct, isolated, final expression for W_f earns points B2, B3 and B4.

Part C — Model Answer

On the upward trip both the gravitational force (component mg sin φ down the ramp) and friction (μ_k mg cos φ down the ramp) oppose the motion, so by the work–energy theorem K falls linearly with s at the rate mg(sin φ + μ_k cos φ), reaching zero at s = d. On the downward trip gravity's component helps the motion while friction, now pointing up the ramp, still opposes it, so K rises linearly as s decreases at the smaller rate mg(sin φ − μ_k cos φ). The downward line therefore starts at (d, 0), is straight, is less steep than the upward line, and reaches s = 0 at K_f, greater than zero but less than K_0.

The two straight lines do not retrace each other: friction takes energy on both legs, so the tote returns to O with less kinetic energy than it started with.

Part C: Downward-trip line added (teal) — straight, less steep than the upward line, ending at K_f < K_0.

Scoring (3 points):

▸ C1 — 1 point: For a downward-trip line that begins at (d, 0), continuous with the end of the given upward line, and reaches s = 0 with a kinetic energy greater than zero.

▸ C2 — 1 point: For the kinetic energy at s = 0 on the downward trip drawn clearly less than K_0.

▸ C3 — 1 point: For a straight line whose slope magnitude is clearly smaller than that of the upward line.

⚠︎ Common error (partial credit): a straight line from (d, 0) back up to K_0 at s = 0, retracing the upward line — a round trip treated as lossless; earns C1 only, 1 of 3 points.

⚠︎ Common error (partial credit): a curved line with the correct end points (K_f between 0 and K_0) — earns C1 and C2, 2 of 3 points.

⚠︎ Common error (partial credit): a straight line from (d, 0) to a value above K_0 at s = 0 — earns C1 only, 1 of 3 points.

✗ Common error (no credit): a horizontal line at K = 0 from s = d to s = 0, or no downward-trip line drawn — no criterion is met.

Part D — Model Answer

The sketch is consistent with the expression from Part B. By the work–energy theorem, K_f − K_0 = W_gravity + W_friction over the round trip. The gravitational force is conservative, so its round-trip work is zero: −mgd sin φ on the way up, +mgd sin φ on the way down. Friction opposes the motion on both legs and does the same negative work on each, so its total is 2W_f = −2μ_k mg d cos φ. Hence K_0 − K_f = 2μ_k mg d cos φ: the gap between K_0 and the end of the downward line is twice the magnitude of the Part B expression, positive for any nonzero μ_k, which is why the line must end below K_0.

Scoring (2 points):

▸ D1 — 1 point: For stating, from the work–energy theorem, that the gravitational force does zero total work over the round trip (equal and opposite work on the two legs, or "conservative"), so that only friction changes the tote's kinetic energy over the round trip.

▸ D2 — 1 point: For identifying the gap K_0 − K_f as twice the magnitude of the Part B expression, 2μ_k mg d cos φ (the same negative work on both legs), and concluding that the end point below K_0 is consistent — or a conclusion that correctly follows from the student's own Parts B and C.

⚠︎ Common error (partial credit): states that gravity's round-trip work is zero and that friction removes energy on both legs, but never links the gap to the Part B expression — earns D1 only, 1 of 2 points.

⚠︎ Common error (partial credit): states that gravity's round-trip work is zero but equates the gap to the Part B expression alone (friction's work on one leg) — earns D1 only, 1 of 2 points.

✗ Common error (no credit): argues that gravity's work on the way down "gives back" what friction took, so the tote returns with K_0 — gravity cancels its own work, not friction's; no criterion is met.

Why Does a Tote Come Back Down with Less Kinetic Energy?

In short: Gravity is conservative, so its work over the round trip is zero: −mgd sin φ on the way up and +mgd sin φ on the way down. Kinetic friction opposes the motion on both legs, so it takes energy on both. Each leg costs μₖmgd cos φ, so the round trip loses 2μₖmgd cos φ. That is why the tote returns to O with less than K₀, but still moving.

Common doubts

Does gravity give back the energy that friction took? No. Gravity only repays its own work: what it removed on the way up, it returns on the way down. Friction's loss on the way up is never repaid, and friction takes a fresh amount on the way back.

Why is friction's work negative on the way down too? Friction always points against the motion. Going up it points down the ramp, and going down it points up the ramp, so the angle with the displacement is 180° on both legs. Each leg gives −μₖmgd cos φ, never a positive value.

Why is the downward line on the K–s graph less steep? Going up, gravity and friction both oppose the motion, so K falls at mg(sin φ + μₖ cos φ). Coming down, gravity helps and friction still opposes, so K rises at the smaller rate mg(sin φ − μₖ cos φ). The line is gentler and ends below K₀.

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