Power
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
Power is how fast work is done: P = dW/dt. For a force acting on a moving object, P = F · v. Over time, the work is the area under the power–time graph.
1. The equations
2. Using P = Fv
Constant velocity: the engine’s force equals the resistive force, so P = Fresistv. Lifting at steady speed: P = mgv.
Top speed: a car reaches it when its maximum power can only just balance the drag: vmax = Pmax/Fresist.
A constant net force F on an object starting from rest gives v = (F/m)t, so P = Fv = (F²/m)t grows linearly with time.
For instantaneous power use the instantaneous F and v, not averages.
3. Example, then try it
An 800 kg elevator rises at a steady 2.0 m/s. What power does its motor deliver (ignoring friction)?
Steady speed, so the cable force = mg = 8000 N.
P = Fv = (8000)(2.0) = 16 kW.
A 1200 kg car’s engine delivers a maximum of 30 kW, and the total resistive force is a constant 1500 N. Find its top speed on a level road, and its acceleration at full power when it’s going 10 m/s.
Top speed: engine force = 1500 N, so v = P/F = 30,000 ÷ 1500 = 20 m/s.
At 10 m/s: engine force = P/v = 30,000 ÷ 10 = 3000 N.
Net force: 3000 − 1500 = 1500 N, so a = 1500 ÷ 1200 = 1.25 m/s².
Answer: 20 m/s top speed; 1.25 m/s² at 10 m/s.
4. Common mistakes
Motor A does the same amount of work as motor B in half the time. How does A’s average power compare with B’s?
P = W/Δt. The same work in half the time means twice the power.
Ready to practice?
4 questions on Power, each with a hint and an explanation for every wrong answer.