# Homework Help: Calculate the change in the box's kinetic energy

1. Jan 21, 2007

### lzh

1. The problem statement, all variables and given/known data
An 72.4 N box of clothes is pulled 28.8 m up
a 21.8degrees ramp by a force of 117 N that points
along the ramp.
The acceleration of gravity is 9.81 m/s2 :
If the coefficient of kinetic friction between
the box and ramp is 0.27, calculate the change
in the box's kinetic energy. Answer in units
of J.

3. The attempt at a solution
I have already figured out everything, but i'd like to know if the change here is the sum of both the diss. energy from friction and gravitational potential. Since, all the kinetic gets converted to grav. potential and diss in the end.

2. Jan 21, 2007

The work of all forces equals the change of kinetic energy. That's all you need to know.

3. Jan 21, 2007

### lzh

so what i said was right?

4. Jan 21, 2007

Forget about conservative forces, potential changes, etc. The work of all forces equals the change in kinetic energy, as stated, independent of the nature of these forces.

5. Jan 21, 2007

### lzh

oh, i see. So i would just use W=F*displacemnt= 3369.6J?

6. Jan 21, 2007

### lzh

so essentially, the work of all forces is the sum of the work of mgh and F(diss)*displacement?

7. Jan 21, 2007

There are three forces. You named two of them, and the third one is the force which pulls the crate up.

8. Jan 21, 2007

### lzh

oh i see! ty

9. Jan 21, 2007

### lzh

i tried adding it all up, but my hw service keeps saying that i'm wrong.
heres what i found:
mgh+Fdeltax+117:
->72.4*(10.69539)=774.3465J
10.695(height) was founded with:
28.8sin21.8=10.6954
Fdeltax-energy of friction:
first i founded the normal force:
72.4cos21.8=67.22
so force of friction is:
67.22*.27=18.15N
->18.15*(28.8)=522.72J

774.3465J+522.72J+117=1414J
but this isn't correct!
I tried this same step on a friend's version(same quesition w/ different numbers), and it ended up being right.
what did i do wrong?

10. Jan 21, 2007

### lzh

ok i figured it out