Basic Power Problem: 66 kW Solution

[renob]

Homework Statement

A 1.0 x 10^3 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of
4.0 x 10^3 N retards the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s?

Homework Equations

the final answer turns out to be 66 kW

P=W/time

P=Fv

The Attempt at a Solution

Im really confused here and do not know where to start. all i know is the equations for power

 

[cepheid]

You know the force. You know the velocity. What is the issue?

 

[renob]

You know the force. You know the velocity. What is the issue?

But the force they give is frictional force. I just need some help on how to start out

 

[renob]

I just did this:
The power of the frictional force part is:
(4000 N)(3 m/s)=12000 W

then you find the power of the normal force of the elevator so its:
1000 kg + 800 kg =1800 kg

(1800 kg)(9.8)(3.0 m/s)=52920

then i added both powers:
1800W + 52920W =64920 W

then I converted to kW and got 64.92 W. How did they get 66? sig figs would only bring it up to 65.

 

[cepheid]

Your method is correct! You do indeed have to calculate the power based on the SUM of the two forces, because the motor has to provide enough force to exactly counteract both friction AND gravity (i.e. the elevator's weight), both of which are working against it.

When I first read your problem, I did it in my head, using g = 10 m/s^2 in order to simplify the mental math and get a ballpark answer. However, I got exactly 66 kW. So maybe they were expecting you to use g = 10? I don't know what else to suggest...

 

[renob]

Oh well, 65 is probably acceptable. thanks for the help