Finding the power of an elevator motor

[petefic]

Homework Statement

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?

Homework 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?

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

 

[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.

 

[petefic]

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

 

[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.

 

[asteriatic]

To calculate the average power of the elevator motor during this time period, we can use the formula P = W/dt, where P is power, W is work, and dt is the time interval. In this case, the work done by the elevator motor is equal to the force applied (ma+mg) multiplied by the distance traveled (vt). We can also use the equation v = u + at to find the final velocity (v) of the elevator, which is 1.75m/s.

For part a, the average power of the elevator motor is calculated to be 11811.04W. This means that on average, the elevator motor is using 11811.04 joules of energy per second to lift the elevator.

For part b, we can compare the power of the motor when the elevator is at its cruising speed to the average power calculated in part a. Since the elevator is moving at a constant speed, there is no acceleration and therefore no change in kinetic energy. This means that the power of the motor at cruising speed would be equal to the power needed to overcome the force of gravity, which is just the weight of the elevator (mg).

In this case, the power at cruising speed would be 650kg * 9.8m/s^2 = 6370W. This is significantly lower than the average power calculated in part a, as the elevator motor only needs to provide enough power to counteract the force of gravity to maintain its cruising speed. This also highlights the importance of efficient motor design and using the least amount of energy possible to perform a task.