# 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. 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?

Expert's answer

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

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

## Comments

Assignment Expert08.10.18, 22:49Dear Bhavya mohta,

please use panel for submitting new questions

Bhavya mohta07.10.18, 15:03By doing it with power =work done/time taken I am getting a different answer .

Please help me with it

## Leave a comment