To compare two motors that run for different times, calculate each motor’s input energy using power multiplied by time. Then divide useful energy output by input energy. A lower power rating can still lead to more energy used if the motor runs for longer.
The 2027 SEC G3 Physics K323 syllabus covers work, power and efficiency. This example compares two motors doing the same lifting job at different speeds.
- Identify the useful energy transfer.
- Calculate each motor’s input energy.
- Divide useful output by input for each motor.
- Check that power and energy have the right units.
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Keep the lifting job the same
Two motors each lift the same load through the same vertical distance. Suppose the load’s gravitational potential energy increases by 3,000 J in both trials. Treat that increase as the useful output. The motors have different electrical input powers and take different times to finish.
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Worked example | The lower-power motor uses more energy
Motor A takes 10 s at an input power of 500 W. Its input energy is 500 × 10 = 5,000 J. Its efficiency is 3,000/5,000 × 100% = 60%.
Motor B takes 15 s at 400 W. Its input energy is 400 × 15 = 6,000 J. Its efficiency is 3,000/6,000 × 100% = 50%. Motor B draws less power while running but uses more energy to finish the job.
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Check using power over the same interval
Motor A’s average useful output power is 3,000/10 = 300 W. Dividing 300 W by 500 W gives the same 60%. Motor B’s average useful output power is 200 W, giving 200/400 = 50%. The power method works when input and output refer to the same run.
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Frequently asked questions
What does the rest of the input energy represent?
It is energy transferred in ways that do not contribute to lifting the load. Some can be transferred to the surroundings by heating and sound.
Why can’t I divide 3,000 by 500 to get efficiency?
The first quantity is in joules and the second is in watts. Their ratio is a time. Compare two energies or two powers over the same interval.