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?
1 Answer | Add Yours
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
We’ve answered 319,865 questions. We can answer yours, too.Ask a question