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

[Icetray]

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]

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)

 

[Icetray]

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]

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

It is a pleasure if a student feels so.

ehild