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Pole Vaulting (This should be interesting)

  1. Aug 10, 2009 #1
    1. The problem statement, all variables and given/known data
    At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force that the pole exerts back on the athlete is given by F(x) = (160N/M )x - (180n/m^2 )x^2 acting over a distance of 0.24m.

    How much work is done by the athlete?

    2. Relevant equations
    W=Fdcos(theta)


    3. The attempt at a solution

    I don't understand what x is? is that .24 meters? The distance? It's confusing wording to me. I tried making x = .24 and got 28 and made it negative but that was wrong
     
    Last edited: Aug 10, 2009
  2. jcsd
  3. Aug 10, 2009 #2

    kuruman

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    Is there a question that the problem asks?
     
  4. Aug 10, 2009 #3
    Oh my gosh haha, I'm trying to find how much work is done by the athlete
     
  5. Aug 10, 2009 #4

    kuruman

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    I would assume the work done by the athlete on the pole. To do this you need
    (a) Find the force F(x) the the athlete exerts on the pole. If the problem does not say what x is, I would assume that x is some distance by which the pole bends as the athlete executes the vault.
    (b) Find the work using the expression for a non-constant force
    [tex]W = \int F(x) dx[/tex]

    The limits of integration should be from 0 to 0.24 m. Should the work done by the athlete on the pole be positive or negative? Hint: This is like compressing a spring.
     
  6. Aug 10, 2009 #5
    It should be negative.

    But the problem here is that I'm in a calculus based physics class and have only taken the derivative sections of calculus and not the integral based parts. So is there anyway you could help me through this step by step?
     
  7. Aug 10, 2009 #6

    kuruman

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    Step (a) that I outlined above requires no knowledge of integration. Try that first.
     
  8. Aug 10, 2009 #7
    if x=.24 then F(x) = 28.032
     
  9. Aug 10, 2009 #8

    kuruman

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    We are not looking for a number here. Your answer is not F(x) - it is F(0.24), the force that the pole exerts on the athlete when the vault is bent by 0.24 m. We are looking for the force exerted by the athlete on the pole for any x.

    Prepositions are important here. If the force exerted by the pole on the athlete is F(x) = (160 N/m )x - (180 N/m2 )x2, what is the force exerted by the athlete on the pole?
     
  10. Aug 10, 2009 #9
    Then since they are opposite
    F(x) = -((160 N/m )x - (180 N/m2 )x2)
     
  11. Aug 10, 2009 #10

    kuruman

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    Not quite. Plus signs should change to minus and minus signs should change to plus. As required by Newton's Third Law, the force on the pole exerted by the athlete is

    F(x) = -(160 N/m )x + (180 N/m2 )x2.

    Since you say that you are familiar with the derivative sections of calculus, I will ask you this: Can you find a function W(x) such that, when you take its derivative with respect to x, you get the F(x) that I have above? In other words, can you find W(x) such that

    [tex]\frac{dW}{dx}=-(160 N/m)x + (180 N/m^{2})x^{2}?[/tex]

    Make an educated guess at a function W(x) and then take its derivative and see if it is indeed F(x).
     
  12. Aug 10, 2009 #11


    Honestly I'm so lost and you're obviously much too smart for me so I guess I must find help elsewhere. Sorry for making you do all that work for nothing...
     
  13. Aug 10, 2009 #12

    kuruman

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    I am sorry too. However, I think your course instructor should be made aware of how lost you are. (S)he needs to know.
     
  14. Aug 10, 2009 #13
    The problem is that its online homework and we don't actually learn or need this stuff in class. Any chance for last ditch answer?
     
  15. Aug 10, 2009 #14
    :( :( :(
     
  16. Aug 10, 2009 #15

    diazona

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    Actually that is right. It's just written a little differently than kuruman wrote it.

    The next part of the question requires you to integrate F(x), and the first step in that is just what kuruman said: find some function whose derivative is
    [tex]-160 \frac{\mathrm{N}}{\mathrm{m}}x + 180 \frac{\mathrm{N}}{\mathrm{m}^2}x^2[/tex]
    I think that it won't be as hard as you may think, you just need to try it. If you like, you can make a guess, post it here along with your reasoning (why you guessed it), and we can guide you to the right function.
    No way, we're not too smart for you! There's no such thing. (Well... okay, maybe there is, but that's not a problem here) Just be persistent, we can help you through this; it just may take a while.
     
  17. Aug 11, 2009 #16
    -80(n/m)x^2 + 60(n/m^2)x^3????
     
  18. Aug 11, 2009 #17

    kuruman

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    Excellent!! Persistence pays. :approve:

    What you have found is the "antiderivative" or "indefinite integral". In your case, this is the function

    W(x) = -80 N/m x2 + 60 N/m2 x3

    To complete the problem, you need to find the work as the pole is bent from x = 0 to x = 0.24 m. Another way of saying the same thing is "evaluate the integral from 0 to 0.24 m." The work that you are looking for is summarized with an equation as

    W = W(0.24) - W(0)

    What the above equation says is "First, replace x with 0.24 m in your expression for the antiderivative and calculate the result. Secondly, replace x with 0 m and recalculate (this one requires no calculator). Thirdly, subtract the second number from the first and that's the number you are looking for." The units will multiply out to N.m which in this case become Joules.

    That wasn't too bad, was it?
     
  19. Aug 11, 2009 #18
    God bless
     
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