Work from a force–displacement graph
You will be able to: Find signed work using rectangle, triangle and trapezoid areas.
How do you find work when the force changes?
You stretch a spring gently. The farther you pull, the harder you must pull. Multiplying the final force by the whole distance would pretend that the force was large from the beginning.
A useful starting point: Work and its sign →
Words and symbols before equations
- F_parallel
- Signed component of force along the displacement, in N.
- Graph area
- Height in N multiplied by width in m gives work in J.
- Signed area
- Area above the zero-force axis is positive; area below is negative for rightward travel.
What this picture assumes
Force varies linearly during monotonic rightward motion from x=0 m to x=4 m. Orange dot marks final force. Signed area gives work; no calculus is required.
Connect the picture to the physics
For monotonic rightward motion, graph the signed parallel force against position. The signed area between the curve and the position axis is work. For constant force use a rectangle; for a line from zero use a triangle.
A straight line from F₀ to F₁ across distance d gives a trapezoid: W=½(F₀+F₁)d. The average-endpoint shortcut works because the force varies linearly; it is not general for every curve.
The graph’s height is force, not work. Its slope is change in force per meter, not work. If motion reverses, split the motion into legs and account for the direction of displacement.
A worked example, step by step
During rightward motion, force rises linearly from 0 N to 8 N over the first 3 m. It is then −2 N for the next 2 m. Find each contribution and the total work.
- First region is a triangle: W₁=½(3 m)(8 N)=12 J.
- Second region is below zero: W₂=(2 m)(−2 N)=−4 J.
- Net work from the graphed force is 12−4=8 J.
- The total unsigned area would be 16 J, which would incorrectly discard the negative contribution.
Area gives work; slope does not.
A line rises from −4 N to +4 N over 4 m. What is total work for rightward travel?
Compare with an explanation
Zero: the equal triangular areas below and above the axis cancel.
Predict. Change one thing. Explain.
A force changes linearly over 4 m. Hold its initial value at 0 N and vary its final value. Then choose opposite-signed endpoints and compare positive and negative areas.
Work = ½(0+8)×4 = 16 J. Shaded area below zero subtracts from area above zero.
Force varies linearly during monotonic rightward motion from x=0 m to x=4 m. Orange dot marks final force. Signed area gives work; no calculus is required.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a work, energy or power relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA force decreases linearly from +6 N to −2 N as a cart moves right through 4 m. (a) Find average force for this line. (b) Calculate work. (c) State the graph feature used. (d) Explain why final force times distance fails.
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Compare with the answer and four-point rubric
- 1 point: (6−2)/2=2 N.
- 1 point: 8 J.
- 1 point: Signed area of the force–position graph.
- 1 point: Force was not −2 N throughout; that shortcut ignores its changing values.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Work units from the axes?
N times m gives J.
RECALL 2What happens below zero force?
It contributes negative work for positive displacement.
RECALL 3When can endpoints be averaged?
When force changes linearly over the displacement.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Work from a force–displacement graph
- Rectangle: W=Fd. Triangle: W=½(base)(height).
- Linear force: W=½(F₀+F₁)d, with signed forces.
- Use areas geometrically; calculus is not needed for these graphs.
Remember: Area gives work; slope does not.
Conditions: Force varies linearly during monotonic rightward motion from x=0 m to x=4 m. Orange dot marks final force. Signed area gives work; no calculus is required.
Refresh Kid · Unit 3 · Objectives 3.2.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.2, objectives 3.2.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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