Power — Question 3

Question 3

A lifting motor rating limits useful speed through efficiency

A motor draws 500 W electrically and converts 80 percent into mechanical power at its cable. It lifts a 20 kg load upward at constant speed; take $g=10$ m/s$^2$ and neglect other losses.

1. Find mechanical output power and lifting speed.

2. Find time and electrical energy needed for a 5 m rise.

3. Why does zero net power on the load not imply zero motor output?

Question 3 - Solution

A lifting motor rating limits useful speed through efficiency

(a) Use cable power at constant speed.

The governing relation is $$\boxed{P_{\mathrm{cable}}=\eta P_{\mathrm{in}}=Tv,\qquad T=mg}$$

$$P_{\mathrm{cable}}=.80(500)=400\ \mathrm W.$$ With $$T=mg=200\ \mathrm N,$$ $P=Tv$ gives $$v=2\ \mathrm{m/s}.$$

(b) Connect distance, time, and input energy.

$$t=5/2=2.5\ \mathrm s,$$ so electrical input is $$500(2.5)=1250\ \mathrm J.$$ Cable work is 1000 J; the difference is conversion loss.

(c) Keep individual powers separate.

Gravity power is $$-mgv=-400\ \mathrm W$$ and cable power is +400 W. Their sum is zero because speed is constant, even though the motor continuously transfers energy to the load.