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

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

## Homework Statement

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?

## Homework 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!

ehild
Homework Helper
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!

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)

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)

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

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

It is a pleasure if a student feels so.

ehild