Power
Power is the rate of energy transfer (or equivalently, the rate of doing work). One watt is one joule per second.
$P = \dfrac{E}{t} = \dfrac{W}{t}$
- $P$ in watts (W)
- $E$ or $W$ in joules (J)
- $t$ in seconds (s)
Why the rate matters
A 60 W lamp and a 600 W toaster might transfer the same total energy in a day if the toaster runs only a few minutes — but the rate at which they do so is different. Power tells you how quickly:
- A high-power kettle boils water faster.
- A high-power car accelerates more quickly because it transfers chemical → kinetic energy at a higher rate.
- A high-power LED is brighter than a low-power one.
✦Worked example— Worked examples
Example 1. A motor lifts a 50 kg crate 2.0 m in 4.0 s. Find the average power.
- GPE gained $= mgh = 50 \times 9.8 \times 2.0 = 980$ J.
- Power $= 980 / 4.0 = 245$ W.
Example 2. A kettle uses 192 kJ of electrical energy to boil water in 2.0 minutes. Find its power rating.
- $E = 192{,}000$ J, $t = 120$ s.
- $P = 192{,}000 / 120 = 1600$ W (1.6 kW).
Example 3. Comparing two motors. Motor A: 200 W, runs 30 s. Motor B: 60 W, runs 100 s. Which transfers more energy in total?
- A: $E_A = 200 \times 30 = 6000$ J.
- B: $E_B = 60 \times 100 = 6000$ J.
- Same total energy — but A does it faster, so A is the more powerful machine.
Useful conversions
- 1 kW = 1000 W
- 1 MW = 1{,}000{,}000 W
- 1 horsepower (hp) ≈ 746 W (not on the spec but useful context)
Common pitfalls
- Time in minutes, not seconds. Convert before substituting.
- Power vs energy. Power is the rate; energy is the total. P × t = E.
- kW vs J. kWh × 3{,}600{,}000 = J. The kilowatt-hour appears in P2 (energy bills) — keep it separate from power in W.
- Confusing input and useful power. A device's stated power is usually the input — only the useful fraction (efficiency × input) does the intended job.
Power in circuits
For an electrical device:
$P = V I = I^2 R = \dfrac{V^2}{R}$
(P2 territory but worth remembering early.) An immersion heater drawing 4 A from a 230 V supply transfers $P = 230 \times 4 = 920$ W.
How exams test power
Common question shapes:
- "An electric drill takes 10 s to drill a hole. The drill transfers 4500 J of energy. What is its power?" — straightforward $P = E/t$.
- "A child of mass 35 kg climbs a 2.5 m rope in 5.0 s. Calculate the useful power developed." — chain GPE with power.
- "Compare the powers of two devices given different times and energies." — calculate both and compare.
➜Try this— Quick check
A weightlifter of mass 80 kg climbs a flight of stairs of height 4.0 m in 6.0 s, with average force per step done against gravity only. Find the useful power.
- GPE = $80 \times 9.8 \times 4.0 = 3136$ J.
- $P = 3136 / 6 = 523$ W (≈ 0.52 kW).
AI-generated · claude-opus-4-7 · v3-deep-physics