Work is signed area, even when force changes
You will be able to: Evaluate a force–position integral and interpret positive and negative areas.
How can many tiny work contributions give one total?
A cart is pulled harder as it moves: Fₓ = (2 N/m)x. Between x = 0 and 3 m the force rises from 0 to 6 N. The work is the triangle’s area, 9 J, rather than final force times distance, 18 J.
A useful starting point: Work selects the force along the displacement →
Words and symbols before equations
- Small displacement dx
- A tiny signed change in position, in metres.
- Integral ∫
- A sum of contributions over an interval; here signed work Fₓ dx.
- Integration limits a and b
- Starting and ending positions, in the order traveled.
- Antiderivative
- A function whose derivative is the integrand; evaluate its change between the limits.
What this picture assumes
Prescribed straight path from x = 0 to the chosen endpoint; Fₓ = A + bx. Work is signed area, not absolute area. This investigation calculates work, not whether an unspecified initial speed can reach the endpoint.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- From 0 to 4 m: W = 8 J; endpoint force -2 N. Shading above zero adds work; below zero subtracts it.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Divide the path into short pieces. On each piece, the force is nearly constant and dW = Fₓ(x) dx. Adding and taking the small-piece limit gives W = ∫ₐᵇ Fₓ(x) dx. A graph above the axis contributes positive area for increasing x; below contributes negative area.
If Fₓ = A + bx, the antiderivative is Ax + ½bx². Thus W = A(b_pos − a_pos) + ½b(b_pos² − a_pos²), where b is the force gradient and the position limits are separately labeled. We usually use xᵢ and x_f to avoid that naming clash.
For a curved path the full expression is ∫F · dr. Integrate the component tangent to each displacement, not the force magnitude blindly. The graph model here restricts motion to one straight x-axis segment. Reversing the limits reverses the work of the same position-dependent force.
A worked example, step by step
Fₓ(x) = 6 N − (2 N/m)x. Find work from x = 0 to 4 m, and distinguish the two signed areas.
- Find where Fₓ = 0: 6 − 2x = 0 gives x = 3 m.
- Positive area from 0 to 3 m is ½(3 m)(6 N) = 9 J.
- Negative area from 3 to 4 m is −½(1 m)(2 N) = −1 J.
- W = 8 J. Equivalently, [6x − x²]₀⁴ in SI units gives 24 − 16 = 8 J.
The area is signed. Adding absolute areas gives 10 J in this example, which is not the net work.
What does the slope of accumulated W(x) represent?
Compare with an explanation
dW/dx = Fₓ(x). Accumulated work is largest where force changes from positive to negative.
Predict. Change one thing. Explain.
Keep A = 6 N and gradient b = −2 N/m. Increase the endpoint past 3 m. Predict when accumulated work reaches its largest value and why it then decreases.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
From 0 to 4 m: W = 8 J; endpoint force -2 N. Shading above zero adds work; below zero subtracts it.
Prescribed straight path from x = 0 to the chosen endpoint; Fₓ = A + bx. Work is signed area, not absolute area. This investigation calculates work, not whether an unspecified initial speed can reach the endpoint.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant work, system boundary, energy or calculus 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 questionFₓ = 4 N − (2 N/m)x acts from 0 to 3 m. Sketch the graph, calculate positive and negative work, and state the total.
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Compare with the answer and four-point rubric
- 1 point: Graph is a line from (0 m, 4 N) through (2 m, 0 N) to (3 m, −2 N).
- 1 point: Positive triangle area = ½(2)(4) = 4 J.
- 1 point: Negative triangle area = −½(1)(2) = −1 J.
- 1 point: Total work = 3 J; equivalently [4x − x²]₀³ = 3 J.
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 1Why does an integral replace Fd?
Force can change along the path, so each small displacement has its own force.
RECALL 2What happens when integration limits reverse?
The work changes sign for the same position-dependent force.
RECALL 3How do W(x) and Fₓ(x) connect?
Fₓ is the derivative of accumulated work with respect to x.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Work is signed area, even when force changes
- W = ∫ from xᵢ to x_f Fₓ(x) dx.
- If Fₓ = A + bx, W = A(x_f − xᵢ) + ½b(x_f² − xᵢ²).
- Force–position area has units N·m = J.
Remember: The area is signed. Adding absolute areas gives 10 J in this example, which is not the net work.
Conditions: Prescribed straight path from x = 0 to the chosen endpoint; Fₓ = A + bx. Work is signed area, not absolute area. This investigation calculates work, not whether an unspecified initial speed can reach the endpoint.
Refresh Kid · AP Physics C: Mechanics Unit 3 (official 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 alongside the Fall 2026 clarifications. This is Mechanics Unit 3: Work, Energy, and Power. The unit covers Topics 3.1–3.5. Calculus connects work to force integrals, force to potential-energy derivatives, and power to the rate of energy transfer. Models distinguish object-only and multi-object systems; translational models exclude rotational energy unless explicitly noted. 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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