Calculate power for a specified mechanical-work task
Find work by the named aligned force, then divide by its task duration without ignoring other destinations.
Exam scope and assessment notes
Core and Extended · Physics 0972, 2026–2028, section 1.7.4. Power is work done or energy transferred per unit time: P = W/t = ΔE/t. A whole-interval quotient gives average power unless a constant rate is stated. Keep system, units and interval consistent. Written practice is self-checked. Human subject-teacher review, supervised practical experience and cumulative assessment 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
- Calculate work by a constant force along displacement.
- Use a supplied interval and complete energy account.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What work does a constant aligned 20 N force do over 8 m?
What you will learn
- Combine W = Fd with average P = W/t.
- Separate a force's power transfer from net kinetic-energy change.
First identify which force does the work
A constant aligned 20 N pull moves its point of application 8 m in 5 s. Its work is 160 J, and its average power transfer is 160/5 = 32 W. If a constant 5 N friction force acts over the same distance, 40 J transfers to internal stores and the net kinetic increase is 120 J when no other stores change. That net increase per time is 24 W, different from the pull's 32 W.
Use the force and task named in the question. A specified 30 N upward force lifts a load steadily 4 m in 6 s, doing 120 J of work at average 20 W. Its net kinetic change is zero at steady speed, but work by the lifting force is non-zero and increases the gravitational store in the friction-free stated account.
A task comparison holds work fixed
If the same 120 J work task takes 3 s rather than 6 s, average power is 40 W rather than 20 W. Half the duration doubles the average rate for this fixed amount; it does not double the work itself. A different force, distance or destination account changes the amount and cannot be compared on duration alone.
The combined calculation gives interval-average mechanical transfer by the selected force, not necessarily the motor's full electrical input or peak power. Electrical and internal-energy losses may require a larger input. Do not infer those inputs, time profiles or certified performance from the supplied aligned force and distance alone.
Pause and explain
A load rises at steady speed. Does zero net kinetic change prove zero power by its lifting force?
Worked example
A constant aligned 20 N pull moves a cart 8 m in 5 s. Find work and average power by the pull.
Show the worked solution
- Work = 20 × 8 = 160 J.
- Average power = 160/5.
- The pull supplies 32 W on average over the task.
Answer 160 J; average pull power 32 W.
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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 and average pull power for 20 N aligned over 8 m in 5 s.
Give me a hint
Use the named force in Fd, then divide work by time.
Compare my reasoning
- Work = 160 J.
- Average power = 160/5.
- Average pull power = 32 W.
160 J; 32 W.
Look for these in your work
- Use the correct force.
- Distinguish work and rate.
For the same pull, a 5 N friction force acts over 8 m with no other store changes except kinetic and internal. Find internal transfer and net kinetic increase per time over 5 s.
Give me a hint
Account for friction separately before dividing the remaining amount.
Compare my reasoning
- Friction accounts for 5 × 8 = 40 J internal transfer.
- Kinetic increase = 160 − 40 = 120 J.
- Net kinetic increase per time = 120/5 = 24 W.
40 J internal transfer; kinetic increase rate 24 W.
Look for these in your work
- Separate pull transfer from the net account.
- Use the same task interval.
A friction-free steady lift uses a specified 30 N upward force through 4 m. Compare average lifting powers for 6 s and 3 s and explain the unchanged kinetic store.
Give me a hint
Use the same work amount for both duration comparisons.
Compare my reasoning
- Lifting work is 30 × 4 = 120 J.
- Average rates are 20 W and 40 W.
- Steady speed keeps kinetic energy unchanged; the energy increases the gravitational store.
20 W and 40 W; kinetic energy unchanged at steady speed.
Look for these in your work
- Hold force and distance fixed.
- Distinguish individual work from kinetic change.
Common mistakes
- Find the named force's work first.
- Zero net kinetic change does not mean zero lifting-force power.
- Mechanical output need not equal electrical input or peak power.
What can you explain now?
What average power does a 120 J task completed in 3 s require?
Compare with the explanation
40 W.
Divide the fixed task's work by its duration.
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Make it stick
Find mechanical task power, explain a steady lift and distinguish named-force output, net store change and full electrical input.
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