Box of Mass Work Problem

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  • #1
physicsCU
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Work Problem Need Help Fast

A box of mass m is sliding along a horizontal surface.

The box leaves position x = 0 with speed v_0. The box is slowed by a constant frictional force until it comes to rest at position x = x_1.

Find F_f, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.)
Express the frictional force in terms of m, v_0, and x_1.

I have no idea what to do. I know I need to find the change in kinetic energy, which is -.5*m*v_0

The work done is -.5*m*v_0*x_1 (I think).

But I have no idea what to do to find the force.

After the box comes to rest at position x_1, a person starts pushing the box, giving it a speed v_1.

When the box reaches position x_2 (where x_2 > x_1), how much work W_p has the person done on the box?
Assume that the box reaches x_2 after the person has accelerated it from rest to speed v_1.
Express the work in terms of m, v_0, x_1, x_2, and v_1.

That is the second part of the problem. I have a feeling once I have the first part, the second part won't be terrible.

But other than the first part I have started, any suggestions would be great!!!

I would like to get this problem done tonight, so time is of the essence.
 
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Answers and Replies

  • #2
dextercioby
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The theoreom of variation of KE will do the trick for u...

Daniel.
 
  • #3
physicsCU
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I tried (-.5*m*v_0)/(x_1), but that is not right.

Why not?
 
  • #4
dextercioby
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I can only telly that your answer is incorrect and advise you to follow my first advice...

Daniel.
 
  • #5
physicsCU
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But is my reasoning correct? Because somewhere I am wrong, I am trying to find that step.
 
  • #6
dextercioby
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I didn't look at your reasoning.However,i don't see whay you're reluctant when it comes to mine...The first part offers you a model/a method for handling the second.

Daniel.
 
  • #7
physicsCU
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I am sorry,

I had the right answer, I never read that the answer was supposed to be magnitude, i had a negative sign in there.
 
  • #8
dextercioby
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What about the missing square of the initial velocity?

Daniel.
 
  • #9
physicsCU
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I got that in there too after I realized the magnitude part.
 
  • #10
dextercioby
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I wish there was a way you could assure me that the result (nonetheless correct) was obtained through a "clean" method.

Daniel.
 
  • #11
physicsCU
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I promise that it was. i talked with two people across the hall who were also working on the problem, and we worked on it together. it turns out that when we talked about it, i had that negative problem.
 

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