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Homework Help: Simple integration for Work

  1. Mar 5, 2010 #1
    1. The problem statement, all variables and given/known data
    An applied force varies with position according to F = k1xn - k2, where n = 3, k1 = 8.3 N/m3, and k2 = 87 N. How much work is done by this force on an object that moves from xi = 6.47 m to xf = 14.9 m? Answer in units of kJ.

    2. Relevant equations

    W = k1 x4/4 ]from the integral of x1 to x2
    - k2x]from x1to x2
    3. The attempt at a solution
    I plugged in everything, and I got 96.778 at first, then I thought I should use x33/3, then I got 6.543478 kJ.
    I got both them wrong, did I do the calculation wrong? Please help, thanks. :D

    Last edited: Mar 5, 2010
  2. jcsd
  3. Mar 5, 2010 #2


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    Homework Helper

    Agree with W = k1 x^4/4 - k2x from x1 to x2.
    I got about -15000.
  4. Mar 5, 2010 #3
    I still didn't get that answer. :(
  5. Mar 5, 2010 #4


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    Homework Helper

    It is really W = k1 x^4/4 - k2 x , and I got 97904, similar to your result. Use enough digits during the calculations.

  6. Mar 5, 2010 #5
    I found my mistake, thank you very much.
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