Question

 A hoist operated by an electric motor raises a load of 300kg vertically at a steady speed of 0.2ms^-1. Frictional resistance can be taken to be constant at 1200N. What is the power required?

Answer 1

By the definition of the power, we have:

$$P=Fv=(F_g+F_{fr,r})v=((m_{hoist}+m_{load})g+F_{fr,r})v,$$ $$P=((500\ kg+300\ kg)\cdot9.8\ \dfrac{m}{s^2}+1200\ N)\cdot0.2\ \dfrac{m}{s},$$ $$P=1.81\cdot10^3\ W=1.81\ kW.$$