An elevator has full load capacity 800 kg. Frictional force on the elevator is 4000 N. If elevator is going down with constant speed 2.0 m/s, then the power applied by electric motor at full load is lose to : (1 HP = 746 W, g = 10 `m//s^2`)

Options

A. -10.72 HP
B. -32.17 HP
C. +10.72 HP
D. +32.17 HP

Answer

A

Explanation

`P=vecF.vecv=Fvcos180^(@)`
`=-Fv`
`=-(8000xx4000)xx2`
`=-8000W`
on `P=-(8000)/(746)HP`.
`=-10.72HP`.