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))`

Explanation

The downward force on the elevator is
`F=mg+F_(f)=(1500xx10)+3000=18000N`
The major must supply enough power to balance this force.
Hence, `P=vecF.vecv=18000xx2=36000W=(36000)/(746)=48.3hp`