Calculate work done and energy transfer
Calculate work from force and displacement in the force direction, then connect the work to an energy transfer.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 5. These original explanations and calculations support theory study; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use force, displacement, speed, mass and gravitational field strength with consistent SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
When does a constant force do no work through a displacement?
What you will learn
- Use work done = force × displacement in the force direction
- Relate work done to an energy transfer
- Distinguish positive, zero and negative work from the force-displacement alignment
Work requires displacement along a force
Work done by a force is the force multiplied by the displacement in the direction of that force. If force and displacement are parallel, W = Fs. If only part of the displacement is along the force, use that component.
When a force acts opposite to a displacement, it transfers energy out of the object's motion and its work on that motion is negative. A force perpendicular to the displacement does no work through that displacement. Work is measured in joules (J), equivalent to N m.
Work tracks energy transferred by a force
When a force does work, energy is transferred between stores or into the surroundings. For example, lifting an object transfers energy to its gravitational potential store, while friction can transfer mechanical energy into internal energy of surfaces and surroundings.
Keep the chosen object or system clear: work done by an external force and work done against that force can have opposite signs. The scalar amount of energy transferred is non-negative, while signed work records the direction of transfer relative to the motion.
Pause and explain
A force acts opposite to an object's displacement. What is the sign of work done by that force on the motion?
Worked example
A student pushes a box with a constant horizontal force of 50 N through 4.0 m in the direction of the force. Find the work done by the push.
Show the worked solution
- Force and displacement are parallel, so all 4.0 m is in the force direction.
- W = Fs = 50 × 4.0.
- Give the result in joules.
Answer 200 J transferred by the pushing force.
Try a topic test
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Test this topic →From guided practice to a new situation
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 30 N horizontal force moves a crate 6.0 m horizontally in the same direction. Find the work done.
Give me a hint
Use the displacement component along the force.
Compare my reasoning
- W = Fs because force and displacement are parallel.
- W = 30 × 6.0.
- Write the answer in joules.
180 J.
Look for these in your work
- Alignment identified
- Work equation applied
- Joule unit included
A 12 N friction force acts opposite to a sled's 5.0 m displacement. Find the work done by friction on the sled's motion.
Give me a hint
Use a negative sign because force and displacement are opposite.
Compare my reasoning
- Take the direction of displacement as positive.
- W = −Fs = −12 × 5.0.
- Interpret the sign as energy transferred out of the sled's motion.
−60 J; friction removes 60 J from the sled's mechanical motion in this interval.
Look for these in your work
- Opposite direction represented
- Work calculated with sign
- Energy-transfer meaning stated
A 40 N force pulls a trolley 3.0 m at 60° to the direction of travel. Find the work done by this force.
Give me a hint
Only the force component along the displacement contributes: F cos θ.
Compare my reasoning
- Parallel component = 40 cos 60° = 20 N.
- W = (20 N)(3.0 m).
- Give the transferred energy in joules.
60 J.
Look for these in your work
- Force resolved along displacement
- Work calculated from parallel component
- Unit correct
Common mistakes
- Multiplying by time instead of displacement
- Using the full displacement when only its component along the force matters
- Treating a perpendicular force as doing work through that displacement
What can you explain now?
When does a constant force do no work through a displacement?
Compare with the explanation
When the force is perpendicular to the displacement
There is no displacement component along the force.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Which part of a force contributes to work done through a displacement?
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.