About this experiment
Lifting a load higher requires more work. Lifting it in less time requires more power, even if the work is unchanged. Compare loads and heights with an ideal hoist, then change the duration to separate energy transfer from the rate of energy transfer.
Learning goals
- Investigate how lift height affects work while controlling the other variables.
- Use the graph and numerical readings to test a prediction.
- Explain the physical assumptions and where this model stops applying.
How to use it
- 1Lift 20 kg through 3 m in 5 s. Check work is 588.6 J and power is 117.72 W.
- 2Save the curve, then double mass. Compare work at each height.
- 3Halve duration at fixed mass and height. Which readout doubles?
- 4Set height to zero. Explain why the supporting force remains nonzero while work is zero.
The model
Assumptions
- The load moves vertically at constant speed during the lift. Start and stop transients are ignored.
- There are no mechanical losses. Work and power refer to the useful transfer to the load, not electrical input.
Try this in class
- Q1Two motors lift the same load the same height in different times. Which quantities match?
- Q2How much useful power is needed to lift 50 kg at 0.4 m/s?
Frequently asked questions
Why does the force not change with duration?
At constant speed the lifting force balances weight. Shorter duration increases speed, not the required steady lifting force.
Is average power the same as instantaneous power?
During the ideal constant-speed part of this lift, yes. Real motors also need power to accelerate the load and overcome losses.