Work, Energy and Power
Power
Use P = W/t = F·v to solve power problems.
Two cranes can lift the same load through the same height, so they do the same work. The one that does it faster is more powerful. Power is the rate of doing work:
- Average power = total work ÷ total time, W/t.
- Instantaneous power P = dW/dt. Since dW = F·dr, this is P = F·v, force × velocity.
The unit of power is the watt: 1 W = 1 J s⁻¹. Bigger units are the kilowatt (10³ W) and the megawatt (10⁶ W). Motors are often rated in horsepower: 1 hp = 746 W.
Example: a motor lifts 200 kg through 15 m in 20 s (g = 10 m s⁻²). The work is mgh = 200 × 10 × 15 = 30 000 J, so the power is 30 000/20 = 1500 W, about 2 hp.
No machine turns all its input into useful work; some is always lost to friction and heat. Efficiency = useful power out ÷ power in. If the motor above is 75% efficient, it must draw 1500/0.75 = 2000 W from the mains.
Careful with one unit: the kilowatt hour (kWh) is a unit of energy, not power. It is 1 kW used for 1 hour: 1000 W × 3600 s = 3.6 × 10⁶ J. This is the 'unit' on an electricity bill: a 1.5 kW heater left on for 2 hours uses 3 units.
Other energy units you will meet: the electron volt, 1 eV = 1.6 × 10⁻¹⁹ J, for atoms; and the calorie, 1 cal ≈ 4.2 J, for heat.
P = Fv also explains why a body under a constant force from rest needs more and more power as it speeds up (P = ma × at grows with t), while a motor of constant power gives smaller and smaller pushes as the speed rises.
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