An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms-1. The frictional force opposing the motion is 4000 N. What is minimum power delivered by the motor to the elevator?

Here, m = 1800 kgFrictional force, f = 4000 NUniform speed, v = 2 ms-1Downward force on elevator isF = mg + f     = (1800 kg × 10 ms-2)+4000 N = 22000 NThe motor must supply enough power to balance this force. Hence,P = Fv = (22000 N)(2 ms-1)  = 44000 W = 44×103 W = 44 kW