Power: the rate of energy transfer
You will be able to: Calculate average power and distinguish energy transferred from transfer rate.
What is different when the same lift happens faster?
Two motors each raise the same box by the same height. One takes 2 seconds and the other takes 4 seconds. They transfer the same useful energy, but the faster motor supplies twice the average useful power.
A useful starting point: Work as energy transfer →
Words and symbols before equations
- Power P
- Rate of energy transfer or conversion.
- Watt W
- Power unit: 1 W = 1 J/s. The unit W is different from the work variable W.
- Elapsed time Δt
- Duration of the measured transfer, in s.
- Average
- Total transfer divided by the whole chosen interval; not necessarily the rate at each instant.
What this picture assumes
Fixed useful energy transfer 200 J. Graph shows an ideal constant-rate transfer over the selected interval. A real varying transfer with the same total and duration has the same average power.
Connect the picture to the physics
Average power is energy transferred divided by elapsed time: P_avg=ΔE/Δt. For transfer by a force, P_avg=W/Δt. A positive input power adds energy at the stated rate.
The slope of an energy-transferred versus time graph is power. For a straight line, its slope is constant. For a varying curve, use the secant slope across an interval to get the average.
Useful lifting energy is mgh if a load begins and ends with the same kinetic energy. Input electrical energy can exceed useful mechanical energy because of losses; do not silently equate motor input power with useful output power.
| Feature | Energy transfer | Power |
|---|---|---|
| Question | How much? | How quickly? |
| Units | J | W = J/s |
| Same useful lift in less time | Same useful energy | Greater average useful power |
A worked example, step by step
A motor raises a 10 kg load 2 m in 4 s, starting and ending at rest. Use g=10. Find useful energy transfer and average useful power. Compare a 2 s lift.
- The load–Earth system gains mgh=10×10×2=200 J.
- Average useful power for 4 s is 200/4=50 W.
- For 2 s it is 200/2=100 W.
- The useful energy is still 200 J. Only the rate has doubled; these values do not specify electrical input power.
Watts measure joules per second, not an amount of energy.
A 100 W useful output continues for 3 s. How much useful energy is transferred?
Compare with an explanation
300 J: power multiplied by time.
Predict. Change one thing. Explain.
Keep the useful transfer at 200 J. Change the duration from 4 s to 2 s and 8 s. Compare the slopes of the ideal constant-rate energy-transfer lines.
200 J ÷ 4 s = 50 W average useful power. The line shows this rate held constant over the selected duration.
Fixed useful energy transfer 200 J. Graph shows an ideal constant-rate transfer over the selected interval. A real varying transfer with the same total and duration has the same average power.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a work, energy or power relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA student raises a 5 kg mass vertically by 1.2 m in 3 s, with equal initial and final speeds. Use g=10. (a) Find useful work. (b) Find average useful power. (c) Describe how to measure it experimentally. (d) Give one reason the student’s metabolic power is larger.
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Compare with the answer and four-point rubric
- 1 point: 60 J.
- 1 point: 20 W.
- 1 point: Measure mass, vertical height change and elapsed time; repeat trials and report variation.
- 1 point: Body processes and inefficiencies use additional energy that does not become the load’s gravitational potential energy.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What is one watt?
One joule per second.
RECALL 2Does a faster equal-height lift always require more useful energy?
No, for equal initial and final kinetic energy it requires a greater rate.
RECALL 3Average power from a graph?
Energy-transfer change divided by the time interval.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Power: the rate of energy transfer
- P_avg=ΔE/Δt=W/Δt for the relevant work transfer.
- ΔE=P_avgΔt for that interval.
- Energy-transfer versus time: interval slope gives average power.
Remember: Watts measure joules per second, not an amount of energy.
Conditions: Fixed useful energy transfer 200 J. Graph shows an ideal constant-rate transfer over the selected interval. A real varying transfer with the same total and duration has the same average power.
Refresh Kid · Unit 3 · Objectives 3.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.5, objectives 3.5.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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