← All Physics Units

Energy

Physics · Energy

Power

The video lesson for this topic is coming soon.

Essential Question

What makes one machine more powerful than another?

Light bulbs, hair dryers, and car engines all have power ratings. In this lesson, you’ll find out that power is not about how much energy is transferred, but how fast. By the end, you should be able to calculate power with P = W/t and work with watts.

What Is Power?

  • How fast work is done
  • How fast energy is transferred
  • Same work, faster: more power
A staircase with two stick figures. One at the bottom is labeled walker, 10 s. One higher up is labeled runner, 5 s. Caption: same stairs, same work, but the runner has more power.

Power is how fast work is done, or how fast energy is transferred. Two students with the same mass climbing the same stairs do the same amount of work against gravity. But the one who runs up in 5 s does it twice as fast as the one who walks up in 10 s, so the runner has twice the power.

The Power Equation

P = W / t = ΔE / t

  • P = power, in watts (symbol W)
  • W = work, or ΔE = energy transferred, in joules (J)
  • t = time, in seconds (s)
The equation P = W / t = change in E / t in a yellow box. Below it, three cards: P is power in watts, W or change in E is work or energy in joules, and t is time in seconds.

Power is work divided by time. Since work is a transfer of energy, power is also the energy transferred divided by the time it takes. Be careful: the letter W is used for both work (in italics, W) and the unit watt (symbol W). Context tells you which one is meant.

The Watt

  • 1 W = 1 J/s
  • 1 kilowatt (kW) = 1,000 W
  • 1 horsepower ≈ 746 W
Four cards showing typical power ratings: LED bulb about 10 W, laptop about 50 W, microwave about 1,000 W, and car engine about 100,000 W.

One watt (symbol W) is one joule of energy transferred every second. A 10 W LED bulb transfers 10 J of energy every second. Bigger powers use kilowatts (1 kW = 1,000 W), and car engines are often rated in horsepower, where 1 hp is about 746 W. A typical LED bulb uses about 10 W, a laptop about 50 W, a microwave about 1,000 W, and a car engine around 100,000 W.

Same Work, Different Power

  • Same work: 1,500 J
  • Half the time
  • Twice the power
Two cards. The walker does 1,500 J in 10. s, which is 150 W. The runner does 1,500 J in 5.0 s, which is 300 W.

Suppose both students do 1,500 J of work climbing the stairs. The walker takes 10. s, so her power is 1,500 J ÷ 10. s = 150 W. The runner takes 5.0 s, so his power is 1,500 J ÷ 5.0 s = 300 W. Same work, half the time, twice the power. Power and time are inversely related when the work stays the same.

Solving for Any Variable

  • Find power: P = W / t
  • Find work: W = Pt
  • Find time: t = W / P
A triangle split into three parts with W on top and P and t on the bottom. Caption: cover the one you want.

Rearrange P = W/t to find whatever the problem asks for. To find work or energy, multiply power by time: W = Pt. To find time, divide work by power: t = W/P. Always convert times to seconds first, since a watt is a joule per second.

Worked Examples

Try each problem on your own first, then check your work against the solution.

Example 1: Climbing Stairs

A student does 1,800 J of work climbing a flight of stairs in 4.0 s. What is her power?

A staircase with a stick figure partway up, labeled W = 1,800 J and t = 4.0 s.
  1. Knowns: W = 1,800 J, t = 4.0 s, P = ?
  2. Equation: P = W/t
  3. Substitute: P = 1,800 J / 4.0 s
  4. Answer: P = 450 W

Example 2: Motor Lifting a Crate

A motor lifts a 25 kg crate straight up 8.0 m at a constant speed in 14 s. What is the motor’s power?

A motor at the top of a frame pulls a rope attached to a crate. A double arrow shows the crate rises 8.0 m, and a label says t = 14 s.
  1. Knowns: m = 25 kg, h = 8.0 m, t = 14 s, g = 9.8 m/s2
  2. Work: constant speed, so W = mgh = (25 kg)(9.8 m/s2)(8.0 m) = 1,960 J
  3. Power: P = W/t = 1,960 J / 14 s
  4. Answer: P = 140 W

Example 3: Light Bulb

A 60.0 W light bulb transfers 4,500 J of energy. How long is it on?

A glowing light bulb labeled P = 60.0 W and Delta E = 4,500 J, with t = ?
  1. Knowns: P = 60.0 W, ΔE = 4,500 J, t = ?
  2. Rearrange: P = ΔE/t, so t = ΔE/P
  3. Substitute: t = 4,500 J / 60.0 J/s
  4. Answer: t = 75 s

Example 4: Hair Dryer

A 1,500 W hair dryer runs for 2.0 minutes. How much energy does it transfer?

A simple hair dryer shape labeled P = 1,500 W and t = 2.0 min.
  1. Convert time: t = 2.0 min × 60 s/min = 120 s
  2. Rearrange: P = ΔE/t, so ΔE = Pt
  3. Substitute: ΔE = (1,500 J/s)(120 s) = 180,000 J
  4. Answer: ΔE = 1.8 × 105 J

Big Ideas

  • Power: how fast energy is transferred
  • P = W/t = ΔE/t
  • Watt: 1 W = 1 J/s
  • Same work, less time: more power

Back to the essential question: a more powerful machine transfers energy faster. Power is the rate of doing work, P = W/t = ΔE/t, measured in watts, where 1 W = 1 J/s. Doing the same work in less time takes more power. This wraps up our unit on energy.

Take It With You

Printable lesson slides and a practice worksheet for this topic are coming soon.