Question

A water pump pumps 3 kg of water every second from the bottom of a well to the top of the well 50 m above if the pump used 2000 W of electrical power to accomplish the task what is the efficiency of the pump

Answer 1

70% efficiency

Step-by-step explanation:

Power out:

P=W/t

P=mgh/t

P=[(3kg)*(9.8m/s^2)*(50m)]/1sec

P=1470W

Power used: 2000W

efficiency=work out/work in

eff=1470/2000W

eff=0.735=73.5%

When accounting for significant figures:

eff=70%

Answer 2

watts = work per second.

work is mgh = 3x10x50=1500

watts out = 1500

Watts used = 2000

eff=1500/2000=75%