Calculate work from force and aligned distance
Use W = Fd for a constant force along the displacement and convert centimetres before multiplying.
Exam scope and assessment notes
Core and Extended · Physics 0972, 2026–2028, section 1.7.2. Mechanical or electrical work equals energy transferred; W = Fd = ΔE for a constant force along the stated displacement. Name which force does the work and distinguish its transfer from a resultant or a store change with other contributions. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
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.
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.
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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.
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.