Question

A pump is used to lift 500 kg of water from a depth of 80 m in 10 second calculate 1 the work done by the pump 2 the power at which the pump works 3 the power rating of the pump if its efficiency is 40%

Answer 1

a) W=FS

W=mg×S

500×10×80

4,00,000J

b) P=W/t

=400000/10=40000W=40kW

c) efficiency=useful power/power input

40%=40/Power input

Power input = 40 × 100/40

Power input = 40kW