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An elevator weighing 500 kg is to be lif...

An elevator weighing 500 kg is to be lifted up at a constant velociyt of 0.20 m/s. What would be the minimum horsepower of the motor to be used?

A

`10.30 hp`

B

`5.15 hp`

C

2.62 hp

D

1.31 hp

Text Solution

Verified by Experts

The correct Answer is:
D
(d) Given, mass of elevator=500 kg
velocity=0.20 `ms^(-1)`
Weight of elevator`=500xx9.8=F`
Now,power, `P=Fv=500xx9.8xx0.20=980W`
Therefore, hp-rating of motor`=(1)/(746)xx980=1.31hp`
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