An elevator that can carry a maximum load of 1500 kg (elvator+ passengers) is moving up with a constant speed of 2 `ms^(-1).` The frictional force opposing the motion is 3000 N. Find the minmum power delvered by the motor to the elevator in watts as well as in horse power `(g=10ms^(-2))`

Options

A. `36.2 hp`
B. `26.1 hp`
C. `48.3 hp`
D. `12.3 hp`

Answer

`(3)`

Explanation

The downward force on the elevator os
`F = mg + f`
`F = 15000 + 3000 = 18000N` The motor must supply enough power to balance this force.
Hence, `P = FV = 18000 xx 2 = 36000`
`P = 36000/746 = 48.3 hp`