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Finding Force from the Potential Energy

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Physics.png

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    2. Relevant equations

    -du/dx = F (in x direction)

    3. The attempt at a solution

    Would I have to use system equations to estimate a cubic equation, and just take the negative derivative of it? Cause it turns out to be a mess and I end up getting a equation not a Force in newtons. Any ideas?
     
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  3. Nov 13, 2012 #2

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    Ummm. :uhh: I'm not sure I'm following you there.
    What's the slope of the line, at any given point on the curve? (Think "rise over run" if it helps.)

    How does the slope of the line relate to -du/dx?
     
    Last edited: Nov 13, 2012
  4. Nov 13, 2012 #3
    The slope of the line at any given point, is the derivative of u(x) at that point. But seeing how the slope is constantly changing how would I estimate the Force.

    And what I was saying is that it looks like the graph can be modelled by a cubic equation. If I take 4 points off the graph, I can use the points to estimate a cubic equation. Then I take the derivative of this, times it by negative 1, and this will give the force equation. However, this is not a number rather an equation, so I can't estimate the amount of Force in Newtons. Unless I am totally off track here.
     
  5. Nov 13, 2012 #4

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    If the problem statement is actually asking for a specific number, I'm guessing that it's asking for the negative of the derivative of u(x) evaluated at x = 4 [m], where particle A happens to placed in the figure.

    But on the other hand, if you wanted to, you could estimate the negative of the derivative of u at all points, from evaluated at x = 0 to 9 [m], and then end up with a plot of F(x).

    (It's not 100% clear to me if the problem is asking for a particular number or a plot.)
    Yes, you could model it by a cubic equation, or even more accurately by a polynomial of a larger order. But that's probably overkill for this problem, I'm guessing. The problem statement did say "estimate" the Force, after all.

    If you could just "eyeball" -du/dx at points x = 0.5, 2.5, 4.0, 6.0, 8.0 and 8.75 [m], it's enough to reproduce a good plot by connecting the points. Or, if the problem is just asking for a number, just eyeball the negative slope around x = 4 [m].
     
  6. Nov 13, 2012 #5
    Ok Thanks so Much :) . I think it asking for it A, or else It would be asking for too much.

    I have to hand this question in. I already found a cubic equation and found the force to be 1.47N.... but when I use a tangent line to find the force I get 1.66N. Which one do you think would be more accurate?
     
  7. Nov 13, 2012 #6

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    Neither, actually.

    When I "eyeball" the particle around x ≈ 4 m (where it's almost linear), it looks to me that when x changes from 3.75 m to 4.25 m, u changes from 4 J to 3 J.

    (I haven't taken the negative of the slope yet. My point above is that I "eyeball" a different, nice, even number.)
    [Edit: of course that's just an approximation.]
     
    Last edited: Nov 13, 2012
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