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A 120 HP electric motor lifts an elevato...

A 120 HP electric motor lifts an elevator having a maximum total load capacity of 3000 kg. If the frictional force on the elevator is 6000N, the speed of the elevator at full load is close to : (1 HP = 746 `W, g = 10 ms^(-2)`).

A

2m/s

B

2.9 m/s

C

1.7 m/s

D

2.5 m/s

Text Solution

Verified by Experts

The correct Answer is:
D
Power of motor `120 xx 746 w`
Let speed = V
`P_("Gravitational") + P_(tr) + P_("motor") = 0`
` - 30000v - 6000v + P_("motor") = 0`
`rArr" " V = (89520)/(36000) = 2.5 m//s`
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Knowledge Check

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