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FRQ: Work Done by a Collection of Forces — The Rescue Winch Lift

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A rescue helicopter hovers in place above level ground. Its winch cable is attached to a stretcher carrying a patient. The stretcher and patient together have mass m, are treated as a single object referred to below as the stretcher, and start at rest on the ground. The cable raises the stretcher vertically in two phases (Figure 1).

• Phase 1 (0 ≤ y ≤ h): the cable exerts a constant upward tension of magnitude T, where mg < T < 2mg.

• Phase 2 (h ≤ y ≤ 2h): the winch reduces the tension to a smaller constant magnitude T_2, where T_2 < mg. The stretcher continues upward, slowing, and is momentarily at rest at y = 2h.

Figure 1. The two phases of the lift. The helicopter remains stationary throughout.

Air resistance is negligible and the stretcher does not swing. Height y is measured upward from the ground, and the motion is one-dimensional. Use the ground as the reference frame.

✎ Free Response Questions

Part A

On the following dot that represents the stretcher, draw and label the forces (not components) that are exerted on the stretcher during phase 1. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.

• Each arrow must start on, and point away from, the dot.

• Draw the length of each arrow to represent the relative magnitude of each force.

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Part B

Derive an expression for the speed v_h of the stretcher at the instant it reaches y = h. Express your final answer in terms of T, m, and h, 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 kinetic energy K of the stretcher as a function of its height y during phase 1, from y = 0 to y = h. On Figure 2, extend the graph to show K as a function of y during phase 2, from y = h to y = 2h. Your extension should show the shape of the graph during phase 2, the value of K at the instant the tension changes, and the value of K at y = 2h.

Figure 2. Kinetic energy of the stretcher during phase 1. Extend it for phase 2 in part C.

Part D

The lift is repeated with a heavier load of mass m′ > m, using the same constant tension T during phase 1, with T still greater than the new weight m′g. Predict how the graph of K as a function of y for 0 ≤ y ≤ h would change compared with Figure 2. 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

Two forces act on the stretcher during phase 1: the tension of the cable, upward, and the gravitational force exerted by Earth, downward. Because the stretcher is speeding up as it rises, the net force is upward, so the tension arrow is drawn longer than the gravitational-force arrow. There is no separate "upward motion" force — the velocity is not a force and does not appear on the diagram.

Model free-body diagram for part A: T upward and longer than F_g; the velocity is shown separately and is not a force.

Scoring (3 points):

▸ A1 — 1 point: For an upward arrow starting on the dot, labelled as the tension (T, cable force, or equivalent).

▸ A2 — 1 point: For a downward arrow starting on the dot, labelled as the gravitational force (F_g, mg, or weight).

▸ A3 — 1 point: For drawing the tension arrow longer than the gravitational-force arrow, with no additional forces on the diagram.

⚠︎ Common error (partial credit): Correct T and F_g arrows plus an extra upward arrow labelled "motion" or "lift" — earns A1 and A2 (2 of 3 points).

✗ Common error (no credit): Drawing the two arrows with equal lengths — asserts a zero net force on an object that is speeding up; earns 0 for A3.

Part B — Model Answer

Apply the work–energy theorem to phase 1. The stretcher starts from rest. The tension does positive work because it points along the displacement; the gravitational force does negative work because it points opposite the displacement.

W_net = ΔK = K_f − K_i

T·h − mg·h = ½mv_h² − 0

v_h = √( 2(T − mg)h / m )

The speed is set by the net work (T − mg)h — the tension's work minus the magnitude of gravity's work — not by the tension's work alone.

Scoring (4 points):

▸ B1 — 1 point: For writing the work–energy theorem (ΔK = W_net) as the starting principle.

▸ B2 — 1 point: For the work done by the tension, +T·h.

▸ B3 — 1 point: For the work done by the gravitational force, −mg·h (negative, opposite the displacement).

▸ B4 — 1 point: For the correct final expression v_h = √(2(T − mg)h/m).

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

Scoring Note: A derivation from Newton's second law, T − mg = ma, followed by v_h² = 2ah, is a valid path and earns all four points; B1 is then earned for the second-law statement.

⚠︎ Common error (partial credit): Omitting gravity's work, so that v_h = √(2Th/m) — earns B1 and B2 (2 of 4 points).

✗ Common error (no credit): Using the free-fall result v_h = √(2gh) — treats the stretcher as a dropped object and includes no work by the cable; earns 0 for B2, B3 and B4.

Part C — Model Answer

Work through phase 2 one question at a time, using the work–energy theorem: for any small rise Δy, the change in kinetic energy equals the net work, ΔK = F_net·Δy, so the slope of a K-versus-y graph is the net force.

• Where does the extension start? At K_h, the value already reached at y = h. The tension changes at that instant, but a force cannot change the speed instantly — speed changes only as work is done over a distance — so K is continuous and the new line begins where the phase-1 line ends.

• Does K go up or down? Down. During phase 2 the net force is T_2 − mg, which is negative because T_2 < mg. A negative net force means negative net work for every metre of rise, so kinetic energy is being removed and the stretcher slows.

• What shape is the line? Straight. Both forces are constant, so the net force T_2 − mg is the same at every height, and the stretcher loses the same amount of K for every metre it rises: ΔK = (T_2 − mg)·Δy. A constant slope is a straight line, just as in phase 1 — only the sign has changed.

• Where does it end? At K = 0 exactly at y = 2h, because the scenario says the stretcher is momentarily at rest there, and an object at rest has zero kinetic energy.

Putting these together: a straight line from (h, K_h) down to (2h, 0). One check worth noticing: K rises by K_h over the height h in phase 1 and falls by the same K_h over the same height h in phase 2, so the two slopes have equal magnitude — the net force in phase 2 is equal in magnitude and opposite in direction to the net force in phase 1.

Model extension for part C: a straight line from K_h at y = h down to K = 0 at y = 2h.

Scoring (3 points):

▸ C1 — 1 point: For a line that begins at K_h at y = h — continuous with the phase-1 graph.

▸ C2 — 1 point: For a straight (constant-slope) decreasing line between y = h and y = 2h.

▸ C3 — 1 point: For the line reaching K = 0 at exactly y = 2h.

Scoring Note: Equal slope magnitudes are not separately scored.

⚠︎ Common error (partial credit): A decreasing curve (concave up or down) from K_h to zero at 2h — treats the net force as changing during phase 2; earns C1 and C3 (2 of 3 points).

✗ Common error (no credit): A horizontal line at K_h, or a line that keeps rising — ignores that T_2 < mg makes the net work negative; earns 0 for C2 and C3.

Part D — Model Answer

With the same tension T but a larger mass m′, the net force during phase 1, T − m′g, is smaller, so the net work done over the height h, (T − m′g)h, is smaller. By the work–energy theorem the kinetic energy gained by y = h is therefore smaller: the line is still straight and still starts at the origin, but it is less steep and ends at a lower K at y = h.

Part D: the heavier load gives a shallower straight line — less net force, so less net work over h.

Scoring (2 points):

▸ D1 — 1 point: For predicting that the line is less steep (or that K at y = h is smaller), while remaining straight.

▸ D2 — 1 point: For justifying the change with the work–energy theorem or Newton's second law: the net force T − m′g, and hence the net work over h, is smaller.

⚠︎ Common error (partial credit): Predicts a less steep line but justifies it only by "heavier things move slower" with no principle named — earns D1 (1 of 2 points).

✗ Common error (no credit): Predicts a steeper line because the heavier load has "more energy" — confuses mass with kinetic energy gained; earns 0.