Aim for FivePhysics 1

Unit 3 · Topic 3.5

3.5 Power

Power is the rate at which energy is transferred or converted. Average power is the energy (or work) divided by the time it took, measured in watts, and the instantaneous power a force delivers is its component along the velocity times the speed. Two people can do the same work at very different powers.

Key terms

  • power
  • watt (W)
  • average power
  • instantaneous power
  • rate of energy transfer

Power as a rate

Power measures how quickly energy is moved or converted. The energy might be flowing into or out of a system, or switching from one form to another inside it. Average power is P_avg = ΔE/Δt = W/Δt. The unit is the watt: 1 W = 1 J/s. A 60 W light bulb converts 60 joules of electrical energy every second.

Walking up a flight of stairs and running up it take the same work against gravity (mgh), because you end at the same height. Running takes less time, so it takes more power.

Instantaneous power

The power a force delivers at a given moment is P = F∥v = Fv cos θ, where v is the object's speed and θ is the angle between the force and the velocity. This follows from P = W/Δt with W = F∥d and d/Δt = v.

A force perpendicular to the velocity delivers no power, just as it does no work. A force opposing the motion, like friction, has negative power: it removes energy at a rate F_f v.

Constant speed still takes power

A car cruising at constant speed on a level road has zero net work done on it, so its kinetic energy stays the same. But the engine still has to deliver power, because air resistance and rolling friction remove energy at the rate F_resistive × v. The engine's power replaces that energy as fast as it's removed.

The same idea explains why going faster is so costly: the resistive force grows with speed, and the power needed is that force times the speed.

Power on graphs

On a graph of energy against time, the slope is the power. On a graph of power against time, the area under the curve is the energy transferred.

If an engine delivers a constant power, P = Fv means the force it can supply shrinks as the speed grows. That's why a car accelerates quickly at low speeds and more slowly at high speeds.

Units to recognize

A kilowatt (kW) is 1000 W. A kilowatt-hour (kWh) is a unit of energy, not power: 1 kW delivered for 1 hour, which is 1000 J/s × 3600 s = 3.6 × 10⁶ J. Electric bills charge for kilowatt-hours because they charge for energy used.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Running up the stairs

    A 60 kg student runs up stairs that rise 4.0 m vertically in 5.0 s. Use g = 9.8 m/s². What average power does she deliver to raise her body?

    Show the solution
    1. Step 1: Energy needed: the increase in gravitational potential energy, mgh = (60)(9.8)(4.0) = 2352 J.
    2. Step 2: P_avg = ΔE/Δt = 2352 J ÷ 5.0 s ≈ 470 W.
    3. Step 3: Walking up in 10 s would take the same energy but only half the power, about 235 W.

    Answer: About 470 W

  2. Example 2Calculator allowed

    Cruising car (classic trap)

    A car drives at a constant 25 m/s on a level highway. The total resistive force from air and the road is 600 N. What power must the engine deliver to the wheels?

    Show the solution
    1. Step 1: Constant velocity means the forward force equals the resistive force: 600 N.
    2. Step 2: P = Fv = (600 N)(25 m/s) = 15,000 W = 15 kW.
    3. Step 3: The trap is saying no power is needed because the kinetic energy isn't changing. The engine must replace the energy that resistance removes every second.

    Answer: 15 kW

  3. Example 3Calculator allowed

    Energy from a power–time graph

    A motor's power output is 300 W for the first 10 s, then decreases steadily to zero at t = 20 s. How much energy does it deliver over the 20 s?

    Show the solution
    1. Step 1: Energy is the area under the power–time graph.
    2. Step 2: Rectangle from 0 to 10 s: (300 W)(10 s) = 3000 J.
    3. Step 3: Triangle from 10 to 20 s: ½(10 s)(300 W) = 1500 J.
    4. Step 4: Total = 4500 J.

    Answer: 4500 J (4.5 kJ)

Common mistakes

  • Mixing up energy and power. Joules measure an amount of energy; watts measure how fast it moves.
  • Thinking a kilowatt-hour is a unit of power. It's an amount of energy.
  • Assuming no power is needed at constant speed. Power is needed whenever a force does work against resistance.
  • Using the total force instead of the component along the velocity in P = Fv cos θ.

On the exam

  • Power often appears as a ratio question: same work in half the time means twice the power.
  • In lab questions, you might measure power by timing how long a motor or person takes to lift a known mass a known height. List the measurements (mass, height, time) and the equation P = mgh/Δt.

Connected topics

Videos

Check yourself

4 questions on 3.5 Power. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A motor lifts a 50 kg crate 10 m straight up at constant speed in 20 s. What is the motor's average power output? Use g = 10 m/s².

Question 2 of 4Calculator allowed

A car travels at a constant 25 m/s while air resistance and road friction exert a total backward force of 600 N on it. How much power must the engine deliver to the wheels?

Question 3 of 4Calculator allowed

Two students of equal mass climb the same flight of stairs. Student 1 walks up in 20 s and student 2 runs up in 10 s. Which statement is correct?

Question 4 of 4Calculator allowed

An electric company charges customers for kilowatt-hours. What kind of quantity is a kilowatt-hour?

0 of 4 answered