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Spring SMH

  1. Nov 19, 2006 #1
    Question: A block of mass .05 kg is pulled .3 m from its equilibrium position and released. The spring constant is 5 N/m

    What is the total energy of the block-spring system?

    My book says E = K + U or E = (1/2)mv^2 + (1/2)kx^2. My book also says E_total = (1/2)kA(amplitude)^2. Which formula is correct? I've just solved for the E_total equation and I found the answer to be .75 J. Can anyone confirm this answer for me? My work is below...

    (1/2)(5)(.3^2) = .75 J

    Thanks
     
  2. jcsd
  3. Nov 19, 2006 #2

    Integral

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    How did you get .75J from that? Try again.
     
  4. Nov 19, 2006 #3
    i lied. .225 J

    Thanks for checking for me. I forgot to square the .3
     
  5. Nov 19, 2006 #4

    Integral

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    With that cleared up.

    The 2 relationships you have posted.

    [tex] E = K + U = \frac 1 2 m v^2 + \frac 1 2 k x^2 [/tex]

    and

    [tex] E = \frac 1 2 k A^2 [/tex]

    are the same. The amplitude is defined as when the displacement is at its maximum so A = x. At the point in time when the displacement is at its maximum, the velocity is zero. So in the first relationship, let x=A and v=0.
     
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