Need help on a work problem -- Pushing a shoe box up a ramp

[BlackJ]

Homework Statement

A constant force (Fa) of magnitude 82.0 N is applied to a 3.00 kg shoe box at an angle of 53.0 degrees, causing the box to move up a frictionless ramp at a constant speed. How much work is done on the box by Fa when the box has moved through a vertical distance h = 0.150 m.

Homework Equations

The Attempt at a Solution

My professor gave a hint that v is constant => a=0, thus, the Wnet= Wg + Wn + Wa =0 (Wn=0)
I feel confused here. If so, work done by Fa will be equal to negative potential energy Wg, which is -mgh and the answer will be in negative?

 

[haruspex]

negative potential energy Wg, which is -mgh

If work done is -Wg and Wg=-mgh, what is the work done?

 

[BlackJ]

Isn't it the work from gravity force?

 

[haruspex]

Isn't it the work from gravity force?

I do not understand your question.
You are asked for the work done by Fa. Call this Wa.
There is an awkwardness in the question. If the box started from rest then it must have accelerated at some stage, so we have to assume it was already moving at speed v when we consider the force first applied. This means there is no change in KE and the net work done on the box is zero.
Gravity does work Wg on the box. Wa+Wg=0.
Wg=-mgh.
So what is Wa?

 

[BlackJ]

Sorry I thought you ask what Wg is.
I understand why the net work done on the box is zero, just still confused how the right answer is in positive but not negative after solving the equation Wa+Wg=0
As I see here
Wa+Wg= 0
=> Wa= -Wg= -mgh
Isn't the answer in negative?

 

[haruspex]

Wa= -Wg= -mgh

No, Wg =-mgh, so what is -Wg?

 

[BlackJ]

does that mean g= -9.8m/s^2 instead of 9.8 because gravity applied a force in opposite direction of force a?

 

[haruspex]

does that mean g= -9.8m/s^2 instead of 9.8 because gravity applied a force in opposite direction of force a?

I was taking g to be +9.8m/s², since that is how most people use 'g'.
(If you are defining up as positive for everything else, it would be more logical to say the weight is mg where g=-9.8m/s², but hardly anyone does that so let's go with the usual style.)
So, g=9.8m/s², weight is -mg (because weight force acts down) and if an object rises by distance h the work done on it by gravity is -mgh.
Since the net work done on the mass in this question is zero, we have Wa+(-mgh)=0.

 

[BlackJ]

Argh I kept thinking -Wg already included the negative sign so I substitute mgh instead of -mgh
It's just clear now, thanks a lot.