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Homework Help: Work/force with a magnet problem

  1. Mar 11, 2008 #1
    The strength of magnetic force varies inversely with the square of the distance between the magnets. In other words,

    Force = k/(distance^2)

    Suppose that when two magnets are 4 cm apart, there is a force of 4 newtons. Find the work required to move the magnets from a distance of 3 cm apart to a distance of 11 cm apart.

    4N = k/(0.04[tex]^{2}[/tex])m

    4N = k/(0.0016m)

    k = 2500

    One thing that's thrown me for a loop though, is...

    The distance between the magnets is 4 CM, or 0.04M.

    So do I take the square of 4 CM to get 16 CM, or .16M.

    Or do I square 0.04M to get 0.0016M, or .16CM.

    I think it's the latter, but I'm not totally sure and I don't get second chances on this question.
     
  2. jcsd
  3. Mar 11, 2008 #2

    Hootenanny

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    It is in fact the former, when raising a quantity to a power it's always a good idea to convert it to it's base units.

    [tex]\left(4 cm\right)^2 = \left(4\cdot10^{-2}m\right)^2 = 16\cdot10^{-4}m^2 = 1.6\cdot10^{-3}m^2[/tex]
     
    Last edited: Mar 11, 2008
  4. Mar 11, 2008 #3
    So k is indeed 2500, eh?

    Alright, lemme see...

    As for finding work...

    Work = Integral of ((2500/x^2) * (x)) from 3 to 11.

    Does that sounds about right?
     
  5. Mar 11, 2008 #4

    Hootenanny

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    No, [itex]k\neq2500[/itex] you may want to check you arithmetic.

    For the integral, I'm not sure why your multiplying by x here, perhaps it should be dx? Also take care of your units, your best using meters, i.e. integrate from 0.03 to 0.11.

    Perhaps a minor point, but don't forget to square your units, i.e. m2.
     
  6. Mar 11, 2008 #5
    Wait, does k = .0064?

    I multiply by x because Work is force times distance. I calculate force in the first part. Then I calculate distance with the x. And yes, at the end I forgot to add a (with respect to dx).
     
  7. Mar 11, 2008 #6
    So now I have...

    Integral of (((.0064/x^2) * (x))dx from 0.03 to 0.11.

    Gives 0.0083154 joules.
     
  8. Mar 11, 2008 #7

    Hootenanny

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    Sounds much better :approve:
    Ahh, but the definition of rectilinear [1D] mechanical work is,

    [tex]W = \int^{b}_{a}F(x)dx[/tex]
     
  9. Mar 11, 2008 #8
    Ah, okay. Thanks.

    Now I have...

    Integral of (((.0064/x^2))dx from .03 to .11

    Gives .1551515152 joules.

    Sound accurate?
     
  10. Mar 11, 2008 #9

    Hootenanny

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    Sounds good to me :approve:
     
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