Topic 4 of 6
Power and measurements
Power is the rate of energy transfer. The same energy transfer can happen quickly or slowly.
Energy and work are measured in J. Time is measured in s. A power calculation needs both the amount transferred and the time taken for that transfer.
P = E / tP = average power in watts, W; E = energy transferred in J; t = elapsed time in s. 1 W = 1 J/s.
Same energy transfer, different times
Each motor transfers 240 J. The energy bars use one common scale; the time bars use a separate common scale.
Motor A: 240 J in 8.0 s
Average power = 240 J / 8.0 s = 30 W.
Motor B: 240 J in 4.0 s
Average power = 240 J / 4.0 s = 60 W.
Transferring the same energy in half the time gives twice the average power.
Worked comparison
The same transfer in half the time
- Motor A: P = 240 / 8.0 = 30 W.
- Motor B: P = 240 / 4.0 = 60 W.
B transfers the energy twice as quickly. It does not transfer twice as much energy in this comparison: both amounts are 240 J.
Rearrange to E = Pt when power and time are given, or t = E / P when the transfer amount and power are given. For example, transferring 240 J at a constant 60 W takes 240 / 60 = 4.0 s. Convert minutes to seconds and kilowatts to watts before combining with joules.
If the rate changes during the interval, total energy divided by total elapsed time gives the average power. It does not show how the rate varied at each instant.
Determine useful lifting power from measurements
A motor steadily raises a load. Its intended output is the increase in the load-Earth gravitational store. Measure that store change and the time over the same part of the lift.
- Measure total lifted mass: include the load and its hanger. Record mass in kg; for example, 500 g = 0.500 kg.
- Mark two vertical levels: use a fixed point on the load to identify its start and end positions. Measure their vertical separation h, not the length of a sloping string.
- Time the marked interval: start as the chosen point passes the lower mark and stop at the upper mark. Use the steady part of the lift so the load's kinetic energy does not change over the interval.
- Calculate the output: use the supplied g to find mgh, then divide by the corresponding time.
| Quantity | Value |
|---|---|
| Total lifted mass / kg | 0.500 |
| Vertical rise / m | 0.80 |
| Time for that rise / s | 2.0 |
| Supplied g / N/kg | 10 |
The intended energy increase is mgh = 0.500 x 10 x 0.80 = 4.0 J. Average useful lifting power = 4.0 / 2.0 = 2.0 W.
This method determines the rate at which the gravitational store increases. It does not determine the motor's total electrical input power: energy can also increase internal stores. The input requires a separate measurement.
Link improvements to a specific measurement
- Repeat timings for the same mass, rise and operating conditions to estimate random variation. A longer suitable steady interval can make reaction time a smaller fraction of the measured time.
- Keep the ruler vertical and use the same point on the load at both marks. Overestimating h makes the calculated energy and output power too large.
- Keep the timing interval matched to those height marks. Timing extra motion outside the measured rise makes t too large and the calculated power too small.
- Including the hanger matters: omitting part of the lifted mass makes the calculated gravitational change too small.
Repeated readings do not repair an incorrect height reference or a systematically mismatched timing interval. Correct the method causing that error.
Keep the unit attached to the claim. J describes an energy amount. W describes an amount per second. A high power alone does not establish a large total transfer or a high efficiency.