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A 60 HP elect...

Question

A 60 HP electric motor lifts an elevator
having a max. total load capacity of 2000 kg.
If the frictional force as the elevator is
4000N, the speed of the elevator at full load
is close to:
(1 HP = 746W, g = 10ms?2
)

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Solution

Dear Student

Acceleration due to gravity g=10m/s²
So total force need to apply =(2000 x10+4000) =24000N

Power delivered by the motor P =force⋅velocity
Velocity = P/Force = 60 x 746 /24000 = 1.865 m/s

Regards

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