# Finding the power of an elevator motor

1. Feb 8, 2012

### petefic

1. The problem statement, all variables and given/known data

A 650kg elevator starts from rest. It moves upward for 3s with constant acceleration until it reaches its cruising speed of 1.75m/s.
A) What is the average power of the elevator motor during this time period?
B) How does this power compare with the motor power when the elevator is at its cruising speed?

2. Relevant equations
I think I did part a correctly, but I would like someone to check it. I don't really understand where to start with part b, how is it different than part a?

3. The attempt at a solution
Pavg = W/dT = (F*d)/t = ((ma+mg)*(v*t))/t = ((m(v/t)+mg)*(v*t))/t = (650kg(((1.75m/s)/3s) + 9.8m/s^2)*(1.75m/s*3s))/3s = 11811.04W

2. Feb 8, 2012

### Delphi51

Welcome to PF, Pete.
All okay except the replacement of d with vt. That would be assuming the motion is at constant speed, which it is not. Use an accelerated motion formula to find d.
The formula P = F*v will also work with the average velocity. Should be half your answer.

3. Feb 8, 2012

### petefic

Thanks. Do you have any idea on how to get started on part b?

4. Feb 8, 2012

### Delphi51

Same thing but leave out the ma term, which is zero.
Velocity is steady on 1.75 now, so you could use d = vt.

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