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

Question #14597
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?
1
Expert's answer
2012-09-25T11:13:35-0400
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

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Comments

Assignment Expert
08.10.18, 22:49

Dear Bhavya mohta,
please use panel for submitting new questions

Bhavya mohta
07.10.18, 15:03

By doing it with power =work done/time taken I am getting a different answer .
Please help me with it

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