Question

a 60hp electric motor lifts an elevator having a max.total load capacity of 2000kg.If the frictional force on the elevator is 4000, the speed of the elevator at full load is close to?

Answer 1

The speed of the elevator at full load is 1.9 m/s.

Explanation:

• We know that:

1 hp = 746 W

60 hp = ( 746 × 60 )W

• Also: P = F. V

Here "F" is the net force acting on the elevator.

By Drawing F.B.D for the elevator we can note that: (Image attached below)

F = Weight of Max Load + Frictional Force.

F = ( 2000 × 10) + 4000

F = 24000 N

Now substituting values of P &  F we get:

V = P / Fn  60 x 746 / 24000 = 1.9 m /s

Hence the speed of the elevator at full load is 1.9 m/s.

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If the power of the motor is 40 kw, at what speed can it raise a load of 20,000 N. ?

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