Work: force along a displacement
You will be able to: Calculate signed work from force, displacement and the angle between them.
Why can a large force do zero work?
You pull a small wagon 3 m forward with a horizontal 10 N force. That force transfers 30 J. If the same force acted straight upward while its point of application moved only horizontally, its work would be zero.
A useful starting point: Vector components →
Words and symbols before equations
- Work W
- Energy transferred by a force through displacement; measured in J.
- Displacement d
- Here the magnitude of the point of application’s displacement, in m.
- Angle θ
- Angle between the force and displacement vectors.
- Parallel component
- The part of the force pointing along the displacement: F cosθ.
What this picture assumes
Constant force 10 N; application point moves 4 m right. The diagram is a vector construction, not a claim that only this force controls the path. F and its projection are one force, not two interactions.
Connect the picture to the physics
For a constant force and straight displacement, W=Fd cosθ. First identify the direction of displacement. Only the force component along that direction contributes to work.
A force with a forward component does positive work; a backward component does negative work. A perpendicular force does zero work. Work is scalar: its sign describes energy transfer, not east or west.
The displacement belongs to the point where the force acts. A person can feel tired holding a stationary bag while doing zero mechanical work on that bag in this ideal description. For a deforming system, do not automatically replace application-point displacement with center-of-mass displacement.
| Quantity | Force | Work |
|---|---|---|
| Meaning | Interaction between objects | Energy transferred by a force through displacement |
| Units | N | J = N·m |
| Can be nonzero at rest? | Yes | Mechanical work over zero displacement is zero |
A worked example, step by step
A rope pulls a cart 4 m horizontally with constant force 10 N at 60° above horizontal. Find the rope’s work. Compare a 5 N friction force acting opposite the displacement.
- Rope’s parallel component is 10 cos60°=5 N forward.
- Rope work is 5×4=+20 J.
- Friction work is 5×4 cos180°=−20 J.
- If these are the only forces doing work, net work is zero. Weight and normal do zero work for horizontal motion on this level surface.
Use the angle between force and displacement, not an angle chosen from an unrelated axis.
Does zero work by one force prove the object is not moving?
Compare with an explanation
No. A force can be perpendicular to a nonzero displacement.
Predict. Change one thing. Explain.
Keep force 10 N and displacement 4 m. Change the angle through 0°, 60°, 90° and 180°. Explain how the parallel component determines the sign of work.
Work = 20 J. Parallel force 5 N times displacement 4 m.
Constant force 10 N; application point moves 4 m right. The diagram is a vector construction, not a claim that only this force controls the path. F and its projection are one force, not two interactions.
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 box moves 5 m right. A constant 8 N force acts at 60° to the rightward displacement. (a) Find its parallel component. (b) Calculate its work. (c) Calculate work by 2 N friction left. (d) Give the sum of these two works.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: 4 N right.
- 1 point: +20 J.
- 1 point: −10 J.
- 1 point: +10 J; include any other working forces before calling it total net work.
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 1Does a perpendicular force do work here?
No; its parallel component is zero.
RECALL 2Can work be negative?
Yes, when the force component opposes displacement.
RECALL 3Which point’s displacement matters?
The point of application of the force.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Work: force along a displacement
- W=Fd cosθ for constant force and straight displacement.
- Positive / zero / negative work at θ=0° / 90° / 180°.
Remember: Use the angle between force and displacement, not an angle chosen from an unrelated axis.
Conditions: Constant force 10 N; application point moves 4 m right. The diagram is a vector construction, not a claim that only this force controls the path. F and its projection are one force, not two interactions.
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.
Want to work through this with a tutor?
Bring your question about work: force along a displacement. Your explanation and answers remain free to access.