Calculate the change in the box's kinetic energy

AI Thread Summary
To calculate the change in the box's kinetic energy, the work done by all forces must be considered, including gravitational potential energy and the work done against friction. The total work is the sum of the gravitational work (mgh) and the work done by the applied force minus the work done against friction. The user initially calculated the gravitational potential energy and frictional work but received incorrect results. After reviewing the calculations, the user identified their mistake, leading to a correct understanding of the energy dynamics involved. The final answer reflects the correct application of the work-energy principle.
lzh
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Homework Statement


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.


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.
 
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lzh said:
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.

The work of all forces equals the change of kinetic energy. That's all you need to know.
 
so what i said was right?
 
lzh said:
so what i said was right?

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.
 
oh, i see. So i would just use W=F*displacemnt= 3369.6J?
 
so essentially, the work of all forces is the sum of the work of mgh and F(diss)*displacement?
 
lzh said:
so essentially, the work of all forces is the sum of the work of mgh and F(diss)*displacement?

There are three forces. You named two of them, and the third one is the force which pulls the crate up.
 
oh i see! ty
 
i tried adding it all up, but my homework 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
ok i figured it out
 
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