Calculate signed work by a constant force
Use the component of a force along its point’s displacement, including positive driving work, negative resistance work and zero perpendicular work.
Section 4.5 · Energy, work and power
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.5. Use g = 10 m s⁻² throughout these problems. Masses are positive and constant. Work by a constant force is Fd cos θ, with θ between force and displacement; scalar-product notation is not required. Kinetic energy is ½mv² and gravitational potential-energy change is mgΔh. State the system and count gravitational work or potential-energy change once, not both. Mechanical energy is conserved only under the stated loss-free model; friction and supplied resistance transfer mechanical energy to other forms. Smooth curved-slide examples use overall energy changes and assume contact throughout, rather than requiring a normal-reaction calculation. Average power is work divided by elapsed time; P = Fv applies to a force along the motion. Constant driving power generally gives variable force and acceleration; instantaneous calculations do not justify constant-acceleration formulae over a whole journey. Check string, contact and motion-state changes before continuing a model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Resolve a force using sine or cosine.
- Distinguish a force from the displacement of its point of application.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which part of a constant force contributes to work along a straight displacement?
What you will learn
- Calculate work with the angle between force and displacement.
- Interpret the sign and units of driving, resisting and perpendicular work.
Only the along-displacement component contributes
For a constant force of magnitude F and displacement of magnitude d, the work is W = Fd cos θ, where θ is the angle between their directions. A 25 N pull with cos θ = 4/5 applied while a particle moves 8 m horizontally does work 25 × 8 × 4/5 = 160 J. Its upward component does no work during that horizontal displacement, even though it can change the normal reaction.
Work is measured in joules, with 1 J = 1 N m. It is a scalar and can be positive, zero or negative. Use the displacement of the force’s point of application; force magnitude alone does not specify work. The formula here assumes a constant force and the stated fixed displacement direction. Questions involving a changing force need more information than substituting one instantaneous value into Fd.
The sign follows the force-displacement angle
An opposing resistance Q during forward travel d has θ = 180°, so its work is −Qd. A 12 N resistance during 7 m travel does −84 J. A normal reaction perpendicular to a fixed smooth path does no work in that ideal contact model. A vertical support force also does zero work during a purely horizontal displacement; zero work does not mean that the force is absent.
If θ = 120°, cos θ = −1/2, so a 30 N force during 10 m displacement does −150 J. Use the actual angle between vectors rather than an acute angle copied from an unrelated diagram. When finding net work, calculate each force’s signed work and add them; the result can differ from the work of the drive alone.
Pause and explain
A 12 N resistance opposes forward travel of 7 m. What work does it do?
Worked example
A constant 25 N pull acts at angle θ above a horizontal displacement of 8 m, with cos θ = 4/5. Find its work.
Show the worked solution
- The component along displacement is 25 × 4/5 = 20 N.
- Multiply this component by the point’s displacement: W = 20 × 8.
- Work is +160 J; the force has a component in the displacement direction.
Answer Work +160 J.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A constant 20 N pull acts at angle θ to a 5 m straight displacement, with cos θ = 3/5. Find its signed work.
Give me a hint
Calculate the along-displacement component before multiplying by distance.
Compare my reasoning
- The force component is 20 × 3/5 = 12 N in the displacement direction.
- Work is 12 × 5 = 60 J.
- Its positive sign shows energy transfer by the driving component.
Work +60 J.
Look for these in your work
- Use cosine of the angle between force and displacement.
- Include the sign and joule units.
A particle moves 7 m along a horizontal surface against constant resistance 12 N. A perpendicular normal reaction also acts. Find work by the resistance and by the reaction.
Give me a hint
Use angles 180° and 90° to the horizontal displacement.
Compare my reasoning
- Resistance work is 12 × 7 × cos 180° = −84 J.
- Reaction work is R × 7 × cos 90° = 0, regardless of its magnitude.
- Zero reaction work does not remove the reaction from the force model.
Resistance work −84 J; normal-reaction work 0 J.
Look for these in your work
- Distinguish negative parallel work from zero perpendicular work.
- Do not infer an absent reaction from zero work.
A constant force 30 N makes angle 120° with a 10 m displacement. A separate 80 N force is perpendicular to that displacement. Find each work and their total.
Give me a hint
Use the obtuse angle directly and do not turn its cosine positive.
Compare my reasoning
- The first work is 30 × 10 × (−1/2) = −150 J.
- The perpendicular force contributes 0 J.
- Their signed total is −150 J; both forces can still act physically.
Works −150 J and 0 J; total −150 J.
Look for these in your work
- Retain the negative cosine for the obtuse angle.
- Add signed work values rather than force magnitudes.
Common mistakes
- Use the angle between force and displacement.
- An opposing force does negative work.
- A force can act while doing zero work on the stated displacement.
What can you explain now?
Does a perpendicular normal reaction do work on a particle along a fixed smooth path?
Compare with the explanation
No; its component along the displacement is zero.
The force can constrain the path without transferring energy by work in this ideal model.
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Make it stick
Take the force component along its point’s displacement, multiply by displacement and keep the work sign.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.