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Finding velocity when a mass suspended from a spring passes through equilibrium

  1. Nov 11, 2009 #1

    MRM

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    A 1kg mass is suspended from a vertical spring with a spring constant of 100 N/m and the equilibrium position is noted. The spring is then pushed upward (compressed) a distance x=10 cm before the mass is released from rest. How fast will the block be moving when it passes through the equilibrium position?


    PEelastic=1/2kx^2
    KE=1/2mv^2




    I honestly have no idea what I'm doing. I used the equation above and got: 1/2(100)(10)^2 which equals 5000, but I'm honestly not sure what to do after I get the PEelastic (or if I'm even supposed to solve for that). I think if I could figure out KE, I could solve for velocity, but I'm not sure how to solve for kinetic energy...any help in the right direction would be great...
     
  2. jcsd
  3. Oct 17, 2010 #2
    You are going in the right direction by finding PEelastic. By doing this you also found KE since KE=PEelastic so from there you just plug it into the equation.
     
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