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Conservation of Forces

  1. Jan 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A 70-kg, starting from rest, is pulled across a floor with a constant horizontal force of 200 N. For the first 10 m the floor is frictionless, and for the next 10 m the coefficient of friction is 0.30. What is the final speed of the crate?

    2. Relevant equations

    W= -Delta PE
    Wnc= DELTA PE+DELTA KE
    F=ma
    Ff=MuFn
    W=Fd
    3. The attempt at a solution

    What I tried is to calculate the total work. So I calculate the work being done frictionless.
    W=FD
    W=200(10)
    W=2000
    Now I had to calculate work on friction
    W=FD
    W=(686)(.30)(10)
    W=2058
    This is how far I have gotten.
     
  2. jcsd
  3. Jan 5, 2010 #2

    ideasrule

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    OK, so total work =?
     
  4. Jan 5, 2010 #3
    I dont know if I am doing it right or not
     
  5. Jan 5, 2010 #4

    ideasrule

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    Yes, you're doing it right so far.
     
  6. Jan 5, 2010 #5
    total work=-58?
     
  7. Jan 5, 2010 #6

    vela

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    Check your signs on the two values of work you calculated, otherwise you're off to a good start. Actually, the work you calculated for the applied force is wrong. Over what distance did the 200-N force act?

    Both the applied force and friction are non-conservative forces, so the work they do contribute to the lefthand side of the equation

    [tex]W_{nc} = \Delta{PE}+\Delta{KE}[/tex]

    Now you just have to figure out the righthand side of the equation.
     
  8. Jan 5, 2010 #7
    I don't understand how to calculate work at all. What kind of PE is calculated in this problem or how to calculate it
     
  9. Jan 5, 2010 #8

    vela

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    Oh, okay. First, determine what forces are acting on the block, and then identify which of those are conservative. Each conservative force will have a corresponding PE (though the change could very well be zero). At this point, you probably only know of two conservative forces: gravity and springs. The potential energy function for gravity is [tex]U=mgh[/tex], and for springs, it's [tex]U=1/2 kx^2[/tex].
     
  10. Jan 5, 2010 #9
    Can the PE for this equation be U=MGD or is it zero since there is no height?
     
  11. Jan 5, 2010 #10

    vela

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    Both! You can be pedantic and include the gravitational potential energy terms. When you plug everything in, you will find they cancel. Or you can just neglect them right from the start since you know the change in gravitational PE will be zero because the block didn't move up or down.
     
  12. Jan 5, 2010 #11
    Oh ok, i got 7.4 for final velocity

    So I did what you told me and calculate the the force applied again i got 4k. I subtract that from 2058. I got 1942, with that I just set it equal to 1/2mv^2 and i solve for v and got 7.4?
     
  13. Jan 5, 2010 #12

    vela

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    Sounds good. Don't forget the units!
     
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