Elevator Speed

An elevator operates by cables, pulleys, an 8kW electric motor and controls. Assume the elevator operates with no power lost to friction.

Calculate the maximum rate of ascent of the elevator.
♦ We take our system to be Earth, the elevator frame, cables, pulleys... all of the apparatus out to a cross-section of wires where the electrical power is supplied. We modeled it (in its entirety) to be a Extended BODY. The rate form of the energy equation is applicable:

(1)

The general Energy Equation 
has more terms than we need.

First, the model Extended BODY does not include internal energy. And if we admit the elevator might have gotten warm while functioning, evidence if an internal energy, the internal energy will have become constant or dU/dt = 0. The elevator moves at constant speed hence dKE/dt = 0. dPE/dt for the Earth/elevator is not zero; we return to this term below.

To the right, across the equality, the atmospheric work term, - d(∫dV)dt, is zero because the volume of the elevator (system) does not change. (Bypass ΣW for now). The sum of heats for the event is assumes zero at the beginning of the event and there is no motive for heat change thereafter; ΣQ = 0.

Rewrite Eqn-1 to omit the zero or negligible terms:

(2) 2

To proceed, look more carefully at the term, dPE/dt. Write it explicitly.

(3)

Leave equation terms
in equation form.

Differentiate the term (above right). The mass and gravity are constant. And since we have only one work mechanism, rewrite the work term, subscripted EE for electric work.

(4) 4

A solution is at hand:

(5) 5

Obviously this is the maximum rate of ascent this motor could accomplish. All aspects of friction were assumed equal to zero.