# A 2.3 × 103 kg elevator carries a maximum load of 880.9 kg. A constant frictional force of 4.6 × 103 N retards the elevator's upward motion. Given that the acceleration due to gravity is -9.81...

A 2.3 × 103 kg elevator carries a maximum load of 880.9 kg. A constant frictional force of 4.6 × 103 N retards the elevator's upward motion. Given that the acceleration due to gravity is -9.81 m/s2, what is the minimum power needed to lift the elevator and its contents at 3.60 m/s?

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### 1 Answer

To determine the power one must know the amount of force the motor must supply. Power will be Force X Velocity.

The Fnet = Fmotor + W + Friction

Because the elevator is moving at a constant velocity, the net force must be equal to zero.

Fmotor+ W + friction = 0

So Fmotor = -(W + friction)

Fmotor = -(mg + friction) = (-3.1809X10^3kg X 9.81 m/s^2 - 4.6X10^3 N)

Fmotor = 3.6X10^4 N

Power = (3.6X10^4 N)(3.60 m/s) = 1.3 X 10^5 W = 130 kW