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An elevator has full load capacity 800 k...

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`)

A

`-10.72 HP`

B

`-32.17 HP`

C

`+10.72 HP`

D

`+ 32.17 HP`

Text Solution

Verified by Experts

The correct Answer is:
A
`P = vecF.vecv = F v cos 180^@`
`= -Fv`
`= -(8000 - 4000) xx 2`
`= - 8000 W`
on `P = -(8000)/(746) HP`.
`= -10.72 HP`.
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