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Max Length Of Vertical Spring

  1. Nov 13, 2012 #1
    A mass of 0.5 kg hangs motionless from a vertical spring whose length is 0.80 m and whose unstretched length is 0.40 m. Next the mass is pulled down to where the spring has a length of 1.00 m and given an initial speed upwards of 1.5 m/s. What is the maximum length of the spring during the motion that follows?



    2. Uspring = .5kx^2, KE = .5mv^2




    3.First I found the k value of the spring by taking F=-kx, or mg=kx and got 4.9=k(.4) or k = 12.25. Then, I thought the KEinitial + Uinitial = Ufinal. So I solved the equation .5(12.25)(.6)^2+.5(.5)(1.5)^2 = .5(12.25)(x)^2 to find x, or the final chance in the length of the spring. Once I found this, which I got to be.672, I added this to the unstretched length to get 1.07 as the max length of the spring. However, the correct answer is 1.16 and I can't figure out what I'm doing wrong. Any help?
     
  2. jcsd
  3. Nov 13, 2012 #2

    ehild

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    Do not forget the change of gravitational potential energy.

    ehild
     
  4. Nov 13, 2012 #3
    I thought of that, but I am entirely uncertain how to add that in because we won't know the potential of gravity unless we know the final length, right? How would I add this in?
     
  5. Nov 13, 2012 #4

    ehild

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    Express it in terms of x.

    ehild
     
  6. Nov 14, 2012 #5

    ehild

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    You can place the zero of gravitational potential energy at the end of the unstretched spring. Initially, the potential energy is -0.6mg, and the final potential energy is -mgx.

    ehild
     
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