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Balky Cow (Forces)

  1. Oct 31, 2007 #1
    [SOLVED] Balky Cow (Forces)

    1. The problem statement, all variables and given/known data

    A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component F_x = -[20.0N + (3.0N/m x].

    How much work does the force you apply do on the cow during this displacement?

    3. The attempt at a solution

    Given that: F_x = -[20.0N + (3.0N/m x]

    Displacement of 6.9

    F_x = -[20.0N + (3.0N/m (6.9)] = -40.7N
    F_xo = -[20.0N + (3.0N/m (0)] = -20 N

    w= k2-k1
    k = 1/2mv^2
    W = F*s

    W = (-40.7-20) * (6.9m)
    = 418.83 :(

    Solution is: -209J why?
  2. jcsd
  3. Oct 31, 2007 #2


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    Homework Helper

    [tex]Work = \int_0^{6.9} F_x dx[/tex]

    you need to calculate this integral...

    you can calculate the integral... or you can plot the Fx vs. x graph... and the area under this graph is the work...
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