Quick answer: Average power is work done or energy transferred divided by elapsed time: P_avg = W/t. One watt is one joule per second. Power measures rate, not total energy.
Worked example
A motor does 6,000 J of useful work in 20 s. Find its average useful power.
- P = 6000/20 = 300 W.
- Since 1 kW = 1000 W, this is 0.3 kW.
- If input power is 500 W, efficiency is useful output/input = 300/500 = 60%.
Check the result
300 J/s for 20 s transfers 6,000 J.
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Common mistake to avoid
Do not confuse kilowatts with kilowatt-hours: kW is power, kWh is energy. Convert minutes to seconds when calculating watts.
Try it yourself
1. Find power for 900 J in 30 s. 2. How much energy does a constant 2 kW appliance use in 3 h?
Answers and reasoning
1. 30 W. 2. 6 kWh, equal to 21.6 MJ because 1 kWh = 3.6 MJ.
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