How Do You Find Work When Force Is at an Angle? FRQ
Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.
FRQ: Work Done by the Gravitational Force on an Incline — The Loading-Ramp Winch
Assessments aligned to 2026 AP Physics 1 standards
Translation Between Representations (TBR) | CHALLENGE | 12 points | 30 min
▤ Scenario
A warehouse uses a straight loading ramp inclined at an angle φ above the horizontal. The ramp surface is fitted with free-turning rollers, so friction between the ramp and a crate on it is negligible. A crate of mass m on the rollers is attached to a light, inextensible cable that runs parallel to the ramp surface up to a winch at the top, as shown in Figure 1. The winch pulls the crate up the ramp at constant speed. During the interval considered, the crate is already moving at that constant speed and travels a distance d along the ramp, from s = 0 at the start of the interval to s = d. Use g for the gravitational field strength. All motion is along the ramp surface; take the up-ramp direction as positive.
Figure 1: The crate on the roller ramp, pulled by a cable parallel to the ramp surface.
✎ Free Response Questions
Part A
On the following dot that represents the crate, draw and label the forces (not components) that are exerted on the crate while it moves up the ramp at constant speed. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
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Part B
Derive an expression for the work W_g done on the crate by the gravitational force as the crate moves the distance d up the ramp. Express your final answer in terms of m, g, φ, 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 a graph of the work W_T done on the crate by the cable tension as a function of the distance s the crate has travelled along the ramp, from s = 0 to s = d. On the same axes, sketch a graph of the work W_g done on the crate by the gravitational force as a function of s over the same range, and label it W_g.
Figure 2: Work done on the crate by the cable tension, W_T, as a function of distance s along the ramp.
Part D
The winch is moved to a second ramp whose angle φ′ is greater than φ. The same crate is pulled up this steeper ramp at constant speed through the same distance d along the ramp. Predict how the line you sketched for W_g in Part C would change for the steeper ramp. 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
Three forces act on the crate: the gravitational force F_g = mg vertically downward (Earth on crate), the normal force F_N perpendicular to the ramp surface (ramp on crate), and the cable tension T directed up the ramp, parallel to the surface (cable on crate). Friction is negligible on the rollers, so there is no friction force and no separate "force of motion". The normal force is perpendicular to the ramp surface, not vertical.
Part A: forces on the crate, with all three arrows drawn to one scale.
Scoring (3 points):
▸ A1 — 1 point: For drawing the gravitational force vertically downward as a labelled arrow (accept mg, weight or F_grav).
▸ A2 — 1 point: For drawing the normal force perpendicular to the ramp surface, directed away from the surface, as a labelled arrow.
▸ A3 — 1 point: For drawing the tension parallel to the ramp surface and directed up the ramp as a labelled arrow, with no additional forces (no friction, no "motion" or "net" force; components drawn in place of a force cost only that force's point).
⚠︎ Common error (partial credit): normal force drawn vertically upward instead of perpendicular to the ramp — earns A1 and A3 (2 of 3).
⚠︎ Common error (partial credit): a friction force added down the ramp although the scenario states friction is negligible — earns A1 and A2 (2 of 3).
⚠︎ Common error (partial credit): the gravitational force replaced by its two components mg sin φ (down the ramp) and mg cos φ (into the ramp), with F_N and T correct — earns A2 and A3 (2 of 3); the prompt asks for forces, not components.
Part B — Model Answer
Start from the definition of work for a constant force in terms of the component of the force parallel to the displacement:
W = F∥ d = F d cos θ, where θ is the angle between the force and the displacement
The gravitational force has magnitude mg and points vertically downward. The displacement d points up the ramp, at angle φ above the horizontal. The angle between a downward vertical and an up-ramp direction is θ = 90° + φ, so
W_g = mg d cos(90° + φ) = −mg d sin φ
Equivalently, the component of the gravitational force along the ramp is mg sin φ, directed down the ramp, opposite to the displacement, so W_g = −(mg sin φ) d. θ is the angle between the force and the displacement — here 90° + φ, not the ramp angle — so the work carries sin φ and is negative.
Scoring (4 points):
▸ B1 — 1 point: For beginning with W = F∥d = Fd cos θ, or with ΔU_g = mg Δy (energy path), from the reference information.
▸ B2 — 1 point: For identifying the component of the gravitational force along the ramp as mg sin φ, or the angle between the gravitational force and the displacement as 90° + φ (or Δy = d sin φ on the energy path).
▸ B3 — 1 point: For recognising that the along-ramp component of the gravitational force is directed opposite to the displacement, so that the work done by the gravitational force is negative.
▸ B4 — 1 point: For a correct final expression, W_g = −mg d sin φ.
Scoring Note: A correct, isolated, final expression for W_g earns points B2, B3 and B4.
Scoring Note: A derivation that begins from the change in gravitational potential energy, W_g = −ΔU_g = −mg Δy with Δy = d sin φ, is a correct alternative path and earns all four points.
⚠︎ Common error (partial credit): W_g = −mg d cos φ (taking φ itself as the angle between the force and the displacement) — earns B1 and B3 (2 of 4).
⚠︎ Common error (partial credit): W_g = +mg d sin φ (correct magnitude, wrong sign) — earns B1 and B2 (2 of 4).
⚠︎ Common error (partial credit): W_g = −mg d (the whole gravitational force treated as acting along the ramp) — earns B1 and B3 (2 of 4).
Part C — Model Answer
The work done by the gravitational force grows in proportion to the distance travelled, W_g(s) = −mg s sin φ, so the graph is a straight line through the origin with negative slope. Because the crate moves at constant speed, the cable tension is T = mg sin φ (the along-ramp component of the gravitational force), so W_T(s) = +mg s sin φ: the two lines are mirror images of each other about the s-axis, and the W_g line reaches −W_0 at s = d. The normal force is perpendicular to the displacement and does no work. At constant speed the net work is zero over every interval, so the work done by the gravitational force is the exact negative of the work done by the tension.
Part C: W_g is a straight line through the origin, the mirror image of W_T, reaching −W_0 at s = d.
Scoring (3 points):
▸ C1 — 1 point: For a straight line through the origin, labelled W_g.
▸ C2 — 1 point: For a line that lies below the s-axis for all s > 0 (negative work).
▸ C3 — 1 point: For a line whose slope has the same magnitude as the W_T line, reaching −W_0 at s = d (the mirror image of W_T), consistent with the net work being zero at constant speed or with T = mg sin φ.
⚠︎ Common error (partial credit): a straight line from the origin with negative slope but reaching a value of visibly different magnitude from W_0 at s = d — earns C1 and C2 (2 of 3).
⚠︎ Common error (partial credit): a horizontal line at −W_0 (the work treated as a constant, like the force) — earns C2 only (1 of 3).
⚠︎ Common error (partial credit): a straight line from the origin with positive slope lying on top of W_T ("the crate gains height, so gravity does positive work") — earns C1 only (1 of 3).
Part D — Model Answer
The line for W_g would be steeper — still straight and through the origin, but with a more negative slope, so that at s = d it reaches a value of greater magnitude than W_0. From Part B, |W_g| = mg d sin φ: m, g and d are unchanged and sin φ′ > sin φ, so the gravitational force does more (negative) work over the same distance on the steeper ramp. Physically, more of the gravitational force lies along the ramp — its along-ramp component mg sin φ is larger — so it takes more energy out of the crate per metre travelled. The work done by the gravitational force depends on the component of the force along the displacement, mg sin φ, which grows as the ramp steepens.
Part D: on the steeper ramp the W_g line is steeper, reaching a larger magnitude −W_0′ at s = d.
Scoring (2 points):
▸ D1 — 1 point: For predicting that the W_g line becomes steeper (more negative slope), reaching a value of greater magnitude than W_0 at s = d — or a prediction consistent with the student's own expression from Part B and sketch from Part C.
▸ D2 — 1 point: For justifying the prediction from the angle dependence of the student's own derived expression (sin φ for a correct derivation: the along-ramp component of the gravitational force, mg sin φ, is larger for the steeper ramp).
⚠︎ Common error (partial credit): predicts a steeper line but justifies only with "the ramp is steeper, so gravity does more work", without identifying the along-ramp component or the sin φ dependence — earns D1 (1 of 2).
✗ Common error (no credit): predicting no change because m, g and d are all unchanged, from a Part B expression that does contain the ramp angle — the work depends on the component of the gravitational force along the displacement, mg sin φ, which increases with φ; neither criterion is earned. (A student whose own Part B expression has no angle dependence is scored by the carry-through clauses in D1 and D2.)
