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

Explanation

The downward force on the elevator is F=mg + `F_(f)=(1800 xx 10)+4000 = 22000` N
The motor must supply enough power to balance this force. Hence,
`P=F.v = 22000 xx 2 = 44000` W = 59 hp