# Crate pulled on an incline with constant speed. What is the total work done?

A rope is attached to a 50.0-kg crate to pull it up a frictionless incline at constant speed to a height of 3-meters. Note that the force of gravity has two components (parallel and perpendicular component); the parallel component balances the applied force and the perpendicular component balances the normal force.

Calculate the amount of work done upon the crate.

Answer: Wext = 1470 J

Start with TMEi + Wext = TMEf

KEi + PEi + Wext = KEf + PEf

KEi + 0 J + Wext = KEf + (50 kg) * (9.8 m/s/s) * (3 m)

(KEi = KEf since speed is constant. Thus, both KE terms can be eliminated from the equation.)

Wext = (50 kg) * (9.8 m/s/s) * (3 m) = 1470 J

But why? Isn't that the crate is moving with constant speed, which means net force is zero. So why does the total work done not equal zero?

thank you so much in advance!!!

## 1 Answer

Net force is 0, but you've applied a force to the body which did work to overcome gravity.

You could say gravity also did negative work on the system. That is actually reflected in your energy balance already.

On the left side of your equation you had an external force. Notice what you did when you solved for it. It was equated to the gravitational potential energy after raised on the slope.

The math is telling you that your applied force was exactly equal to the change in potential energy. If we look at the work done by gravity on the block, we can see that it is a negative work (we move opposing gravity). In that sense, the gravity perfectly opposed and the net work is 0. (This is because this system isn't losing any energy)

Generally what the question is concerned about is the applied work. We don't have to apply gravity, it's already there, so the the required work to raise the block is given from your equation.

Total work on the system is 0 though if we aren't losing energy to the environment. That's just often not useful to ask about.