Aim for FivePhysics C: Mechanics

Unit 3 · Topic 3.5

3.5 Power

Power is the rate at which energy is transferred or converted, measured in watts (joules per second). Average power is total energy over total time, instantaneous power is P=dWdtP = \frac{dW}{dt}, and for a force on a moving object P=F⃗⋅v⃗P = \vec{F}\cdot\vec{v}.

Key terms

  • power
  • average power
  • instantaneous power
  • watt
  • rate of energy transfer

Average and instantaneous power

Power tells you how fast work is done or energy changes. Average power over a time interval is:

Pavg=ΔEΔt=WΔtP_{avg} = \frac{\Delta E}{\Delta t} = \frac{W}{\Delta t}

Instantaneous power is the rate at one moment: P=dWdtP = \frac{dW}{dt} (or dEdt\frac{dE}{dt}). The unit is the watt: 1 W = 1 J/s. Two climbers of the same weight who reach the same summit do the same work against gravity; the one who gets there faster has more power.

Going backward, the energy transferred is the integral of power over time, W=∫P dtW = \int P\,dt, which is the area under a power–time graph.

A kilowatt-hour (kWh), the unit on an electric bill, is energy, not power: 1 kWh = (1000 W)(3600 s) = 3.6 × 10⁶ J.

Power from force and velocity

A force F⃗\vec{F} acting on an object moving with velocity v⃗\vec{v} delivers power:

P=F⃗⋅v⃗=Fvcos⁡θP = \vec{F}\cdot\vec{v} = Fv\cos\theta

This follows from dW=F⃗⋅dr⃗dW = \vec{F}\cdot d\vec{r}: divide by dt and dr⃗dt\frac{d\vec{r}}{dt} is v⃗\vec{v}. Only the force component along the velocity delivers power. A force perpendicular to the motion delivers none, and a force against the motion (like drag) delivers negative power, removing energy.

For an engine moving something at constant speed against a resisting force, the engine's force equals the resistance, so P = (resisting force)(speed). That's why a car's power needs rise quickly at high speed: drag grows with speed and is multiplied by the speed again.

A sense of scale

A 60 kg student running up a 3.0 m flight of stairs in 4.0 s gains (60)(9.8)(3.0)≈1760(60)(9.8)(3.0) \approx 1760 J of gravitational energy, an average power of about 440 W. A car engine might deliver tens of kilowatts, and a bright LED bulb uses about 10 W.

Power and energy answer different questions. A small motor and a large motor can lift the same load to the same height, transferring the same energy. The large motor just does it faster.

Power and the work–energy theorem

The net power on an object is the rate its kinetic energy changes: Pnet=dKdtP_{net} = \frac{dK}{dt}. If the net force does positive work at 30 W, the object gains kinetic energy at 30 J/s.

With constant power from rest (and no losses), K=PtK = Pt, so speed grows like t\sqrt{t}, not linearly. The acceleration shrinks as the speed grows, because the same power at higher speed means less force (F = P/v).

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Lifting at constant speed

    An elevator motor lifts a 1000 kg car at a constant 2.0 m/s. What power does the motor deliver, ignoring friction? How much energy does it use in 15 s? Use g = 9.8 m/s².

    Show the solution
    1. Step 1: Constant speed means the cable tension equals the weight: T=(1000)(9.8)=9800T = (1000)(9.8) = 9800 N.
    2. Step 2: P=Fv=(9800)(2.0)=19,600P = Fv = (9800)(2.0) = 19{,}600 W, about 19.6 kW.
    3. Step 3: Energy in 15 s: W=Pt=(19,600)(15)=2.94×105W = Pt = (19{,}600)(15) = 2.94 \times 10^5 J. Check: the car rises 30 m, and mgh = (1000)(9.8)(30) = 2.94 × 10⁵ J. ✓

    Answer: About 19.6 kW; about 2.9 × 10⁵ J in 15 s.

  2. Example 2

    Power from a time-dependent force

    A force F(t)=4tF(t) = 4t N (t in seconds) pushes a 2.0 kg block from rest along a frictionless floor. Find the power delivered at t = 2.0 s and the total work done from 0 to 2.0 s.

    Show the solution
    1. Step 1: a=4t2.0=2ta = \frac{4t}{2.0} = 2t m/s², so v=∫0t2t dt=t2v = \int_0^t 2t\,dt = t^2 m/s.
    2. Step 2: P=Fv=(4t)(t2)=4t3P = Fv = (4t)(t^2) = 4t^3 W. At t = 2.0 s, P = 32 W.
    3. Step 3: W=∫024t3 dt=[t4]02=16W = \int_0^2 4t^3\,dt = \left[t^4\right]_0^2 = 16 J.
    4. Step 4: Check with the work–energy theorem: at t = 2.0 s, v = 4.0 m/s and K=12(2.0)(4.0)2=16K = \frac{1}{2}(2.0)(4.0)^2 = 16 J. ✓
    5. Step 5: The trap is computing W = Pt = (32)(2.0) = 64 J, which treats the final power as if it lasted the whole time.

    Answer: 32 W at t = 2.0 s; 16 J total.

Common mistakes

  • Treating power and energy as the same thing. Power is a rate; energy is an amount (and a kWh is energy).
  • Using W = Pt when the power changes. Integrate P over time, or use the area under the P–t graph.
  • Using the whole force in P=FvP = Fv when the force is at an angle to the velocity. Use Fvcos⁡θFv\cos\theta.

On the exam

  • Expect questions comparing how fast two systems transfer the same energy, or asking for the power needed to move something at constant speed against friction or gravity.
  • On graphs, the slope of an energy–time graph is power, and the area under a power–time graph is energy transferred.

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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 worker lifts a 50 kg sack 10 m up at constant speed in 20 s. What average power does the worker supply? Use g = 10 m/s².

Question 2 of 4Calculator allowed

A car cruises at a constant 25 m/s while air resistance and friction exert a total of 2000 N against it. What power must the engine deliver to the wheels?

Question 3 of 4Calculator allowed

A rope pulls a sled with a 100 N force directed 60° from the sled's velocity. The sled moves at 3.0 m/s. At what rate does the rope do work on the sled?

Question 4 of 4Calculator allowed

The power delivered to a machine is P(t)=6t2P(t) = 6t^2, with P in watts and t in seconds. How much energy is delivered from t = 0 to t = 2.0 s?

0 of 4 answered