Question

An elevator which 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. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.

Answer 1

Downward force on the elevator is :

F = mg + f = 22000 N

∴ Power supplied by motor to balance this force is

P = Fv = 44000W = \(\frac{44000}{746}\) = 59 hp.