Calculate the change in the box's kinetic energy

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Homework Help Overview

The problem involves calculating the change in kinetic energy of a box being pulled up a ramp, considering forces such as gravity, friction, and the applied force. The box has a weight of 72.4 N, is pulled 28.8 m up a ramp inclined at 21.8 degrees, and has a coefficient of kinetic friction of 0.27.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss whether the change in kinetic energy should include both the work done against friction and the gravitational potential energy. There is also a focus on the relationship between the work of all forces and the change in kinetic energy.

Discussion Status

Some participants have provided insights into the relationship between work and kinetic energy, emphasizing that the work of all forces equals the change in kinetic energy. Others are exploring the calculations involved and questioning their results, indicating a mix of understanding and confusion regarding the application of these concepts.

Contextual Notes

One participant notes that their homework service indicates their calculations are incorrect, prompting further investigation into their approach. There is mention of differing outcomes based on similar problems, suggesting potential nuances in the problem setup or calculations.

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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