# Total work on a system

You are a member of an Alpine Rescue Team. You must project a box of supplies up an incline of constant slope angle a so that it reaches a stranded skier who is a vertical distance $$h$$ above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient $$μ_k$$. Use the work–energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of $$g$$, $$h$$, $$μ_k$$, and $$α$$.

I have seen the answer to this problem on many sites, it is this answer:

I understand everything except one thing: Why is it that the total work done on the box must be equal only to the work done by kinetic friction and gravity? what about the force that pushes the box up the incline?

The answer to this problem is the same everywhere. Can you please explain me why the force that pushes the box is not being included in the total work?

The pushing force that acts on the box does so up to the instant before the box begins to travel up the incline. That is, at $$t=0$$. After this instant, $$t \gt 0$$ the pushing force no longer acts. This initial pushing force does not act on the box throughout its transit up the hill.
However, once the box is pushed, for all $$t\gt 0$$ there are only two forces that act upon it:
So you have a box that begins with an initial velocity, say $$v$$, at the bottom of the slope, and so the work done by these two forces are needed to bring the box to the point where it stops, or where $$v=0$$ at the needed distance. The work energy theorem means that $$F_N \times \text{distance} = \Delta \text{Kinetic Energy}=\frac{1}{2}mv^2$$ where $$F_N$$ is the net force, and as stated, is the sum of those two force acting aboe and no other forces for $$t>0$$