Power Equations

Question

A 2.0 × 103 kg elevator carries a maximum load of 796.7 kg. A constant frictional force of 2.9 × 103 N retards the elevator’s motion upward. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed 2.82 m/s? The acceleration due to gravity is 9.81 m/s 2 . Answer in units of kW.

Answer 1

To find the minimum power required should be an application of the equation P = W/t

Where W is the work needed to lift the elevator and its maximum load.

To find the work needed you can use the equation W= F*d where F is the force you must overcome to lift the elevator and d is the distance travelled (which you don't need to know since you're told the elevator moves at a constant speed).

In this instance that force should be the sum of the frictional force and the force of gravity (m*g) that acts on the elevator and its load, since both work against the upward motion of the elevator.

That gives you the equation P = F*d/t where d/t is the constant speed that the elevator moves at.

Final form of the equation before substitution should be

P = (friction force + force of gravity)*constant speed

P = (2.9*10^3 + (2.0 *10^3 + 796.7)*9.81)*2.82

P = 85546.47 W = 85.55 kW