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An applied force varies with position

  1. Nov 11, 2012 #1
    An applied force varies with position according to F = k1 xn− k2, where n = 3,
    k1 = 2.4 N/m3, and k2 = 56 N.
    How much work is done by this force on an object that moves from xi = 5.13 m to xf = 28.2 m?
    Answer in units of kJ

    i keep using the integration equation: w = ∫f(x) dx
    where k1x^4/4 - 56x is the new equation which i plug 28.2m and 5.13m and subtract
    i get 377736.527
    but its wrong, help please?
     
  2. jcsd
  3. Nov 11, 2012 #2

    Nugatory

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    I haven't checked your arithmetic, but you're attacking the problem in the right way. What's the correct answer that you aren't getting?
     
  4. Nov 12, 2012 #3

    ehild

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    What is the unit of your result? You need to give the result in units of kJ.

    ehild
     
  5. Nov 12, 2012 #4
    About 1515 kJ?
     
  6. Nov 12, 2012 #5

    haruspex

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    No, I also get 377736; but that's J, so the answer should be 377.7kJ.
     
  7. Nov 13, 2012 #6

    ehild

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    That is the correct result. Maybe it should be rounded to three digits: 378 kJ.

    ehild
     
  8. Nov 13, 2012 #7
    Ok. Left the /4 out.
     
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