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Compressed spring, doubling speed of impact

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A 0.211 kg shoe is dropped onto a vertically oriented spring with a spring constant of 102 N/m. The shoe becomes attached to the spring upon contact, and the spring is compressed 0.117 m before coming momentarily to rest.
    If the speed of the shoe just before impact is doubled, what is the maximum compression achieved by the spring?


    3. The attempt at a solution
    Before this question, I figured out the speed the shoe was travelling at just before it hit the spring to be 2.08 m/s, which was correct. I also figured out the work done by the spring force to be -0.698 J and the work done by the weight of the shoe to be 0.242 J.
    I know that when speed doubles, K.E. quadruples, which means the net work done must increase by 4 as well.
    Wnet= Wgrav+Wspring=0.242 -0.698=-0.456 J.
    I found the new K.E. to be 1.826 J (with doubled speed).
    When I equate KE to Wnet, I get a quadratic: m*g*x -0.5*k*x^2= -1.826 J (new K.E.)
    Is there no other way other than solving this quadratic to get to the new compression?
    HELP, please!
     
  2. jcsd
  3. Feb 28, 2012 #2

    SammyS

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    There may be some other way, but what's wrong with solving the quadratic equation?
     
  4. Mar 1, 2012 #3
    I tried solving the quadratic equation, but didn't get the right answer. Can someone tell me where I went wrong with my approach?
    thanks
     
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