Power Output & Work Done by a Motor

[CPURules]

Homework Statement

A small motor is used to operate a lift that raises a load of bricks weighing 500N to a height of 10m in 40s at constant speed. The lift weighs 300N. What is the work done by the force of the motor? What is the work done by the force of gravity?

Homework Equations

P = \vec{F} \cdot \vec{v}
W = \vec{F} \cdot \vec{l}

The Attempt at a Solution

Since power is work done over a time interval (J/s), I figured that if I found the total power, I could multiply that by the time taken to get the total work done:

P = \vec{F} \cdot \vec{v}<br /> = Fv*cos(\phi)

The angle is 0 since the bricks are being lifted straight up, so that simplifies to P = Fv.

Force is the sum of the weight of the bricks and the lift (500N + 300N = 800N)

Found the average velocity:
v = (10m) / (40s) = 0.25m/s

This gives me an average power output of (800N * 0.25m/s) = 200W. Multiply this by 40 seconds, and I get the total work done, which is 8000 J.

Now I'm stuck. At first I thought that I could then calculate the work done by gravity and then subtract it from total work done to get the work done by the motor, but I'm not sure how to go about doing this or if that's even the right way to go about it.

 

[nasu]

What is the 800N force? The force of what on what?

 

[CPURules]

Well since its the weight of the lift/bricks, I'm guessing it would be the force of gravity on the lift/bricks system

 

[nasu]

If this is so, then what you have calculated is the work done by gravity.

What force must the motor "produce" in order to have the system moving up with constant velocity?

 

[CPURules]

Ah. Okay.

If velocity is constant, then acceleration of the system is 0. So the sum of the net forces should be zero. If gravity is producing a force of -800N (because gravity is going against the direction of displacement) then the motor has to produce a force of 800N. Then you could multiply this by the displacement (10m) to get 8000J / -8000J.

But if velocity is constant then the kinetic energy also does not change, which means that the TOTAL work done is 0; if gravity does -8000J of work then the motor does 8000J. So that works too. Is that right?

 

[nasu]

It looks OK to me.