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Find the potential energy function

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data

    1. Given F = -2 N/m^3 ( (2xyz - z^3)i + x^2z j + (x^2y - 3xz^2)k ).
    a. Find the potential energy function for this force, taking (i + 2 j + k) m as reference point.
    b. Find the work done by this force on a 4-kg block moved from the reference point to the
    origin.
    c. Find the initial acceleration of the block, if at that time this was the only force on it.

    2. Relevant equations

    U(x) = -∫Fdx


    3. The attempt at a solution

    for a. can i integrate i j and k seperately? using the reference points as my limits?
    and then i have no clue where to begin on b. and c.
    any help is greatly appreciated
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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  3. Dec 6, 2012 #2

    Dick

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    You don't integrate Fdx. You integrate the dot product of F.dr where r(t) is a path connecting the reference points. I'd suggest taking a straight line. Once you take the dot product there's no i,j,k left. And you can only play this game if curl(F)=0. Did they want you to check that?
     
  4. Dec 6, 2012 #3
    they didnt specify. but im assuming that the curl is 0.
     
  5. Dec 6, 2012 #4

    Dick

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    It is. You can check that if you need to. But my point is that you need to do a path integral, not just a dx integral. Part c is really much easier than the other two, you really should be able to handle that one even if the other two are giving problems.
     
  6. Dec 6, 2012 #5
    yea i have been working on other problems since i wasnt sure how to start this one. But i did the dot product and for my function i got.
    2*((2xyz-z^3)+(2x^2z)+(x^2y-3xz^2).

    stop me if im wrong here but the rest should be:

    part b looks like i just use (1,2,1) and plug them into my potential function

    part c is just a=f/m
     
  7. Dec 6, 2012 #6

    Dick

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    An alternative to the path integral is basically trying to guess a function U(x,y,z) whose gradient is F. It's not really too hard. Try it.
     
  8. Dec 6, 2012 #7

    Dick

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    Yeah, I'll stop you there. Maybe you aren't really doing path integrals. See my last post. Try to guess a potential function and then see if it works. You know the gradient of U is -F, yes? And sure, once you have that b is easy. c is easy any way you do it and you don't need a and b to do it.
     
  9. Dec 6, 2012 #8
    almost forgot i have to integrate this dot porduct but that shouldnt be too hard.
     
  10. Dec 6, 2012 #9
    ok let me try that then.
     
  11. Dec 6, 2012 #10
    ok so the grad of F is the partial with respect to x,y,z. so if f is the gradient i need to indiviually integrate i j k with respect to x y z.
     
  12. Dec 6, 2012 #11

    Dick

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    Sort of, sure. If you can find a U such that grad(U)=(-F) then you are done. I kind of think of it more like guessing as with that as a hint. Can you find U? Mmm. It doesn't have any i,j,k in it. It's a scalar function.
     
    Last edited: Dec 6, 2012
  13. Dec 7, 2012 #12
    And so now I see where the rest of the class is on the test....:). 11 more to go!
     
  14. Dec 7, 2012 #13
    ok i collected the common terms in i j k so i understand that but when i try to find the work (w = -(change in potential) i get 0.
     
  15. Dec 7, 2012 #14

    Dick

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    What did you get for a potential function?
     
  16. Dec 7, 2012 #15

    HallsofIvy

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    The simplest thing to do is integrate along the straight line from (1, 2, -3) to (x, y, z). However you could also use the "broken" line from (1, 2, -3) to (x, 2, -3), then to (x, y, -3), then to (x, y, z). On the first of those only x varies so only the "dx" part is non-zero, on the second only y varies so only the "dy" part is non-zero, and on the third only z varies so only the "dz" part is non-zero.
     
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