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.
Section P1.6.3 · Power
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
- 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.
Sources for this lesson
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.
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.