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. How much is the total force

Answer 1

$$F_{tot}=W+F_{fr},$$ $$F_{tot}=(m_{hoist}+m_{load})g+F_{fr},$$ $$F_{tot}=(500\ kg+300\ kg)\times9.8\ \dfrac{m}{s^2}+1200\ N=9040\ N.$$