From a power graph to transferred energy
You will be able to: Integrate P(t), compare average and instantaneous power, and use the energy to predict speed.
How much energy is transferred when power changes with time?
A device delivers 10 W for 3 s. It transfers 30 J. If its power instead rises steadily from 0 to 10 W over those 3 s, it transfers only 15 J: the triangular area under the power graph.
A useful starting point: Power measures how quickly a force does work →
Words and symbols before equations
- Average power P_avg
- Total work or energy transferred divided by elapsed time.
- Instantaneous power P(t)
- The local rate of transfer at time t, in J/s.
- Accumulated work W(t)
- The signed integral of power from the initial time to t.
- Net power
- Sum of the powers of all forces on the particle, which determines dK/dt.
What this picture assumes
Prescribed transfer P(t) = A + bt, integrated from t = 0. This is an energy-transfer graph, not a complete motor or kinetic-energy model. Negative work means net energy leaves through this transfer; average power at zero elapsed time is undefined.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- At t = 3 s: instantaneous P = 12 W; transferred work = 18 J. Average power from 0 to t = 6 W.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
The definitions are P_avg = ΔW/Δt and P(t) = dW/dt. Integrating gives ΔW = ∫P(t) dt. Positive and negative areas count with opposite signs, just as signed force–position areas do for work.
If P(t) = A + bt, integration from 0 to t gives W(t) = At + ½bt². Divide by t, for t > 0, to get P_avg = A + ½bt. For a straight-line power graph, this also equals the mean of its endpoint values.
When the stated P is the net power delivered to a particle, K(t) = K₀ + ∫P_net dt and v(t) = √[2K(t)/m] wherever K ≥ 0. If it is only input power to a machine, first account for losses and other stored energy. The graph here represents a prescribed transfer, not a complete motor model.
| Quantity | Work / transferred energy | Power |
|---|---|---|
| Meaning | Amount transferred | Transfer rate |
| Unit | Joule (J) | Watt (W = J/s) |
| Graph connection | Signed area under P(t) | Slope of W(t) |
A worked example, step by step
Net power to a 2 kg cart is P(t) = (4 W/s)t for 0 ≤ t ≤ 3 s. Its initial speed is 1 m/s. Find transferred energy, final speed and average power.
- Initial K is ½(2)(1²) = 1 J.
- Integrate net power: W = ∫₀³4t dt = [2t²]₀³ = 18 J.
- Final K = 19 J, so speed = √(2×19/2) = √19 = 4.36 m/s.
- Average power is 18 J / 3 s = 6 W. Final instantaneous power is 4×3 = 12 W; these are different quantities.
Use the whole signed area under P(t), not final power times duration unless power is constant.
Can zero net work hide nonzero power earlier in an interval?
Compare with an explanation
Yes. Equal positive and negative power areas cancel, while transfers occurred during the interval.
Predict. Change one thing. Explain.
Set A = 0 W and b = 4 W/s, then increase elapsed time. Compare the triangle area with accumulated work and average power. Try a negative gradient to see why net transfer can decrease.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At t = 3 s: instantaneous P = 12 W; transferred work = 18 J. Average power from 0 to t = 6 W.
Prescribed transfer P(t) = A + bt, integrated from t = 0. This is an energy-transfer graph, not a complete motor or kinetic-energy model. Negative work means net energy leaves through this transfer; average power at zero elapsed time is undefined.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant work, system boundary, energy or calculus 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 questionNet power is P(t) = 6 W − (2 W/s)t from t = 0 to 4 s. Calculate the signed transfer, average power, and the time at which accumulated work is greatest.
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Compare with the answer and four-point rubric
- 1 point: W = ∫₀⁴(6 − 2t)dt = [6t − t²]₀⁴ = 8 J.
- 1 point: Average power = 8/4 = 2 W.
- 1 point: P becomes zero at t = 3 s and then negative, so W is greatest at 3 s.
- 1 point: W(3 s) = 9 J; the final negative area removes 1 J, leaving 8 J net.
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 does the area under P versus t measure?
Signed transferred energy or work, in joules.
RECALL 2When does P_final × duration give the correct work?
When power is constant throughout the interval, or happens to equal its average.
RECALL 3What is needed to infer speed from power?
Net particle power, initial K and mass, plus a check that the predicted K remains nonnegative.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
From a power graph to transferred energy
- ΔW = ∫P(t) dt; P_avg = ΔW/Δt.
- If P = A + bt, W from 0 to t is At + ½bt².
- For net particle power: K(t) = K₀ + ∫P_net dt.
Remember: Use the whole signed area under P(t), not final power times duration unless power is constant.
Conditions: Prescribed transfer P(t) = A + bt, integrated from t = 0. This is an energy-transfer graph, not a complete motor or kinetic-energy model. Negative work means net energy leaves through this transfer; average power at zero elapsed time is undefined.
Refresh Kid · AP Physics C: Mechanics Unit 3 (official 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 alongside the Fall 2026 clarifications. This is Mechanics Unit 3: Work, Energy, and Power. The unit covers Topics 3.1–3.5. Calculus connects work to force integrals, force to potential-energy derivatives, and power to the rate of energy transfer. Models distinguish object-only and multi-object systems; translational models exclude rotational energy unless explicitly noted. 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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