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Force Equations and Final Speed

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A 7.0-kg rock is subject to a variable force given by the equation
    [tex]F(x) = 6.0N - (2.0N/m)x + (6.0N/m^2)x^2[/tex]
    If the rock initially is at rest at the origin, find its speed when it has moved 9.0m.

    3. The attempt at a solution
    [tex]\sum F_{x} = 6 - 2x + 6x^2[/tex]
    [tex]ma=6 - 2x + 6x^2[/tex]
    [tex]a = \frac{6}{7} - \frac{2}{7}x + \frac{6}{7}x^2[/tex]
    [tex]V_{f}^2 = V_{i}^2 + 2*a*x[/tex]
    [tex]V_{f}^2 = 0 + 2(\frac{6}{7} - \frac{2}{7}x + \frac{6}{7}x^2)x[/tex]
    [tex]V_{f}^2 = 2(\frac{6}{7} - \frac{2}{7}(9) + \frac{6}{7}(9)^2)(9)[/tex]
    [tex]V_{f} = 1219m/s[/tex]

    That seems waaay too fast, and I think I should have used an integral/derivative somewhere since the force changes as x changes. I'm rather lost on this one.
     
  2. jcsd
  3. Oct 18, 2007 #2

    Doc Al

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    Staff: Mentor

    This kinematic equation assumes constant acceleration--not a good assumption for this problem.

    Hint: How much work is done on the rock?
     
  4. Oct 18, 2007 #3
    I think I have it now.
    [tex]W = \int^{9}_{0} 6 - 2x + 6x^2 dx = 1431J[/tex]
    [tex]W = \frac{1}{2}mv_{f}^2 - \frac{1}{2}mv_{0}^2[/tex]
    [tex]1431 = \frac{1}{2}7v_{f}^2[/tex]
    [tex]v_{f} = 20.22m/s[/tex]

    Any improvement?
     
  5. Oct 18, 2007 #4

    Doc Al

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    Staff: Mentor

    Looks good to me!
     
  6. Oct 18, 2007 #5
    Thanks for the help. I've never done a problem like that, all the ones I've completed so far have used the kinematics equations. I guess I'll have to learn how to apply the work equations better.
     
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