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Physics question - object released at equillibrium position of spring

  1. Nov 11, 2007 #1
    Hello everybody,

    I am having a lot of trouble with the following question and would appreciate it if someone showed me the correct way to do it. Thanks in advance for your help! :)

    1. The problem statement, all variables and given/known data
    A 0.20 kg ball attached to a vertical spring (k = 28 N/M) is released from rest from the unstretched equilibrium position of the spring. Determine how far the ball falls, under negligible air resitance, before being brought to a momentary stop by the spring.

    So,
    m = 0.20 kg
    k = 28 N/M
    x = ?

    2. Relevant equations
    Fspring = kx
    Fg = mg
    EE = 1/2 kx^2
    EG = mg (delta h)

    3. The attempt at a solution
    My friend and I both attempted to solve this problem. We used different methods and obtained different answers, but neither of us know which answer is right and why the other is wrong.

    Here is the 1st way we did it:
    When the ball is brought to a momentary stop, Fnet = 0, so
    Fg = Fspring
    mg = kx
    x = mg/k

    And here is the 2nd way:
    Since energy is conserved,
    Ebefore = Eafter
    EG lost = EE gained
    mg(delta h) = 1/2 kx^2

    delta h = x (since the ball falls the same distance that the spring stretches), therefore
    mgx = 1/2 kx^2
    x = 2mg/k

    But this is twice as large as the solution obtained using the first method. Am I overlooking something in one of the methods? Which method is correct?

    Again, thanks in advance!
     
  2. jcsd
  3. Nov 11, 2007 #2

    Doc Al

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    Staff: Mentor

    Fnet = 0 means that the acceleration is zero, not the speed.
     
  4. Nov 11, 2007 #3

    rl.bhat

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    Homework Helper

    x = mg/k is true only during stretched equilibrium position of the spring. Therefore 2nd method is correct.
     
  5. Nov 11, 2007 #4
    Omg. I can't believe I made such a dumb mistake >_<. Thank you so much for your help!
     
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