Question

A 645-kg elevator starts from rest. It moves upward for t = 3.13 s with a constant acceleration until it reaches its constant cruising speed of v = 1.26 m/s. It moved 2.75m during that time.
a) What is the average power (W)is delivered by the motor during the initial acceleration of the eleator during the first 3.13s?
b) How much power by the elevator motor while the elevator moves upward now at cruising speed?

Answer 1

Fe = mg = 645 kg * 9.8 N/kg = 6321 N. =
Force of elevator.

a. P = F * d/t = 6321 * 2.75/3.13 = 5554 J/s. = 5554 W.

b. P = F*V = 6321 * 1.26 = 7965 J/s. =
7965 W.

Answer 2

Fe = mg = 645 kg * 9.8 N/kg = 6321 N. =
Force of elevator.

a. P = F * d/t = 6321 * 2.75 = 17382.75J
That is the increase in potential, now we must add the increase in kinetic energy, which is:
Ws = 1/2 m(Vf)^2 - 1/2 m(Vi)^2
Ws = 1/2 645kg(1.26m/s)^2 - 0
Ws = 512.001J
adding the two together we get:
512.001J + 17382.75J = 17894.751J

17894.751/3.13 = 5717.173 J/s. = 5717 W.

b. P = F*V = 6321 * 1.26 = 7965 J/s. =
7965 W.