# Finding the change in K.E. G.P.E ignored?

1. Sep 23, 2011

### Icetray

Hi guys! Just a little confused on the solution provided.

1. The problem statement, all variables and given/known data

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The
pulling force is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal.
The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m.

(d) What is the change in kinetic energy of the crate?

2. Relevant equations

Change in word done = Fs - mgsin$\Theta$s - friction

I understand this part but I want to know why the solution key takes this change as the change in kinetic energy itself. Isn't there a gain in G.P.E. as well?

Thanks for the help!

2. Sep 24, 2011

### ehild

The work done by all forces is

Fs - mgsin$\Theta$s - (work of friction)=ΔKE

The term -mgsin$\Theta$s is the work of gravity, and the opposite is the change of the gravitational potential energy, as
s*sin$\Theta$=h, the height reached. You can rewrite the equation as ΔKE+ΔPE=Fs+(work of friction)

3. Sep 24, 2011

### Icetray

Thanks! (: Haha I can't believe that I didn't realize that. :-X

4. Sep 24, 2011

### ehild

It is a pleasure if a student feels so.

ehild