Calculate work from force and aligned distance
Use W = Fd for a constant force along the displacement and convert centimetres before multiplying.
Section P1.6.1 · Work
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Convert centimetres to metres.
- Distinguish a named force from a resultant.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many metres is 50 cm?
What you will learn
- Recall and use W = Fd = ΔE.
- State constant-force and direction conditions.
Multiply an aligned force by its displacement
For a constant force acting along a displacement, work done is W = Fd, equal to the energy transferred by that force. Use force in newtons and distance in metres to obtain joules. A constant 20 N pull moving its application point 3 m along the force transfers 60 J. The diagram keeps force and distance arrows separate; they are different quantities, not quantities to add.
A constant 12 N aligned force over 50 cm does 12 × 0.50 = 6.0 J. Do not multiply by 50 while calling the unit metres. Doubling distance doubles this force's work only if its magnitude and alignment remain the same; a longer route with a different force is not that controlled comparison.
Use the force specified, not an invented replacement
For a cart pulled by 20 N over 3 m while a constant 6 N friction force opposes motion, the pull transfers 60 J and friction transfers 18 J to internal stores. The net mechanical contribution to its kinetic store is 42 J if no other stores change. Substituting the 14 N resultant into a question asking for the pull's work answers a different question.
The simple positive Fd calculation assumes the named force acts along the displacement. A perpendicular ideal force transfers no energy through displacement in its direction, and an opposing force removes mechanical energy rather than adding it. These lessons do not require a general angle formula or integration for varying forces: if those conditions are not supplied, do not use total path length with an arbitrary force as if the simple aligned model applied.
Sources for this lesson
Pause and explain
A question asks for work by a 20 N pull. Should 6 N opposing friction be subtracted from the force before answering that question?
Worked example
A constant 20 N pull moves a cart 3 m in the pull direction. Find the work done by that pull.
Show the worked solution
- Use the named constant aligned force, 20 N.
- Displacement along it is 3 m.
- W = 20 × 3 = 60 J.
Answer 60 J by the pull.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Find work by a constant 20 N pull over 3 m aligned displacement.
Give me a hint
Multiply the named force by distance in metres.
Compare my reasoning
- F = 20 N and d = 3 m.
- W = Fd = 20 × 3.
- Work transferred by the pull is 60 J.
60 J.
Look for these in your work
- Use the named force.
- Keep units and alignment explicit.
Find work for 12 N over 50 cm aligned movement and for 8 N over 2.5 m aligned movement.
Give me a hint
Convert the first distance before calculating both products.
Compare my reasoning
- 50 cm = 0.50 m.
- First work = 12 × 0.50 = 6.0 J.
- Second work = 8 × 2.5 = 20 J.
6.0 J; 20 J.
Look for these in your work
- Convert length correctly.
- Use the supplied constant force in each case.
A 20 N pull moves a cart 3 m against 6 N constant friction; no other stores change except kinetic and internal stores. Find pull work, internal increase and kinetic increase.
Give me a hint
Use separate force contributions before the net store balance.
Compare my reasoning
- Pull work is 20 × 3 = 60 J.
- Friction accounts for 6 × 3 = 18 J internal increase.
- The cart's kinetic increase is 60 − 18 = 42 J.
Pull 60 J; internal 18 J; kinetic increase 42 J.
Look for these in your work
- Distinguish specific and net work.
- Balance both destinations.
Common mistakes
- Use metres.
- Name which force does the work.
- The simple Fd model needs an appropriate force and direction.
What can you explain now?
What work does 12 N along a 50 cm displacement do?
Compare with the explanation
6.0 J.
Convert to 0.50 m before multiplying by 12 N.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Recall W = Fd, convert a distance and explain why the pull's work differs from a net kinetic-energy change with friction.
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.