Question

A water pump is run by an electric motor with a power rating of 750W. It is used to pump water from a reservoir up to a height of 37 m and into a water tower at a rate of 1.48 kg per second. What is the useful energy output

Answer 1

power of motor =750W=750joules/s

potential energy generated by 1.48kg of water at 37 m =mgh=1.48*9.8*37=536.648joules

536.648joules of PE is generated in 1 second by the pump .
useful energy output=(536.648/750)*100=71.55%

sorry if I went wrong.