What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed 3.25 m/s?

Answer in units of kW.

The acceleration of gravity is 9.81 m/s^2.

**A 2.3 x 10^3 kg elevator carries a maximum load of 832.4 kg. A constant frictional force of 3.6 x 10^3 N retards the elevator's motion upward.**

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Given:

Weight of elevator = 2.3 x 10^3 kg = 2300 kg

Load carried by elevator = 832.4 kg

Acceleration of gravity = g = 9.81 m/s^2

Frictional force acting on lift = f2 = 3.6 x 10^3 N = 3600 N

Speed of lift = v = 3.25 m/s

As the lift is moving at constant speed the force required to move it up is force require to overcome gravitational pull on the lift with load plus the frictional force.

Total Load of Elevator plus load carried = m = 2300 + 832.4 = 3132.4 kg

Force required to over come gravitational pull = f1 = m*g

= 3132.4*9.81 = 30728.844

Total force required to move the lift = f = f1 +f2 =

30728.844 + 3600 = 34328.844 N

Power of motor required to operate fully loaded lift = f*v

= 34328.844*3.25 = 111568.743 J

The Power of motor required is 111568.743 J

It can also be converted in units of horse power as 111568.743/746 = 149.55 or approximately 150 hp

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