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Bullet lodged in a block

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A 10.0-g bullet traveling at 200 m/s strikes a fixed wooden block. The bullet comes to rest 22 cm inside the block. The magnitude of the force exerted on the bullet by the block over its 22-cm travel is shown in graph. Find the value of Fmax.

    http://www.scribd.com/doc/135229669/Physics

    2. Relevant equations

    W=Fd
    W=KE=.5mv^2

    3. The attempt at a solution

    I tried to treat is as a work problem and set Fd=.5mv^2.
    I got an answer, but it wasn't one available to me.
    I don't know what I am missing
     
  2. jcsd
  3. Apr 10, 2013 #2

    tiny-tim

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    welcome to pf!

    hi kminkel! welcome to pf! :smile:
    use the more general work done = integral of force "dot" distance :wink:

    (W = ∫ F.ds)

    show us what you get :smile:
     
  4. Apr 10, 2013 #3
    Your idea is a good one, except the equation is only Fd for constant F. The graph shows F varies with distance, so what you're really doing is integral of F*dx.

    Find the area under the curve (triangular + rectangular regions) and try setting that equal to the KE.

    Edit: Looks like I was beaten to it!
     
  5. Apr 10, 2013 #4
    With using the W=.5mv^2, I got the W=200 J.
    But with the integral what would I use as the bounds? Would it be from 0 to .11 m or 0 to .22 m? Then how would you get rid of the integral symbol to solve for F. I'm sort of really confused.
     
  6. Apr 10, 2013 #5
    You are integrating over the whole distance that the force is being applied (0 to 0.22m). If integrals are unfamiliar to you, just think of it like area.

    Set the area under the curve from 0 to 0.22m equal to KE.
     
  7. Apr 10, 2013 #6
    If I set the areas equal to .5mv^2...
    .11x+.5(.22)x=.5(.01)(200)^2
    .11x+.11x= 200
    .22x=200
    x=909.09
    That isn't an option /:
     
  8. Apr 10, 2013 #7

    tiny-tim

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    uhh? :confused: don't make it so complicated! :biggrin:

    the integral of a curve is the area under it :smile:

    (and yes, sometimes physics questions really are that simple! :wink:)
     
  9. Apr 10, 2013 #8

    I see my error. I have the wrong base for the triangle. I fixed it and got 1212.12 which is an option! Thank you Both so much!
     
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