Answer to Question #14597 in Mechanics | Relativity for Jayflord Banac

Question

A 650kg elevator starts from rest. It moves upward for 3s with constant acceleration and it reaches its cruising speed of 1.75m/s.

1. What is the average power of the elevator motor during this period?
2. How does this power compare with the motor when the elevator moves at its cruising speed?

Answer 1

1. What is the average power of the elevator motor during this period?
m =
650 kg, t = 3 s, V0 = 0, V = 1.75 m/s.
V = V0 + a * t
a = V / t
F = m *
(g + a) = 650 * ( 9.81 + 0.58) = 6747N
The distance the elevator travels in 3
seconds is D =a * t^2 / 2 =
0.583 * 3^2 / 2 = 2.62 m
P = F * D / t =
5892.4 W
2. How does this power compare with the motor when the elevator
moves
at its cruising speed:
P = F * V = 6370N x 1.75m/s = 11,147.5 W