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Homework Help: Maximum speed on conservative forces problem

  1. Oct 12, 2008 #1
    1. A girl swings on a playground swing in such a way that at her highest point she is 3 m from the ground, while at her lowest point she is 0.5m from the ground. The acceleration of gravity is 9.8m/s^2. What is her maximum speed? Answer in units of m/s.

    2. 1/2mVf^2 + mgyf = 1/2mVi^2 + mgyi

    I made initial velocity to equal zero.
    3. [Vf= (mghi - mghf)/ (1/2m)]^(1/2)

    When I do this... I endup square rooting a negative number. What am I doing wrong?
  2. jcsd
  3. Oct 12, 2008 #2


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    What do you use for initial and final height?

    Is the velocity actually zero at the initial height?
  4. Oct 12, 2008 #3
    I'm using .5 as her initial height and 3 as her final height. Is that my problem? And velocity... I just assumed... that may be wrong. (Thanks for the quick reply by the way!)
  5. Oct 12, 2008 #4
    OOh I got it! I did this:


    You sparked my thinking. Thank you.
  6. Oct 12, 2008 #5
    The problem mentions nothing about air resistance so we can assume that our forces our conservative. This meas that the sum of your forces at position 1 (her highest point) is equal to the sum of the forces at position 2 (her lowest point).

    KE1 + PE1 = KE2 + PE2

    You can elimitate your KE1 because at her highest point the girls velocity is 0.

    This leaves us with PE1 = KE2 + PE2 We expand this formula to:

    mgh1 = (1/2)m[v(squared)] + mgh2

    this can be simplified by canceling out all of your masses.

    gh1 = (1/2)[v(squared)] + gh2

    Now just plug in your values and solve for your velocity.

    Edit: Nevermind, you beat me to it.
  7. Oct 12, 2008 #6
    Oh thank you for going ahead and writting up the solution.
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