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Spring problem need help

  1. Mar 19, 2006 #1
    A vertical spring (ignore its mass), whose spring stiffness constant is 900 N/m, is attached to a table and is compressed down 0.110 m.
    (a) What upward speed can it give to a 0.300 kg ball when released?
    -----------m/s
    (b) How high above its original position (spring compressed) will the ball fly?
    ------m


    Now if im following my lecture notes correctly for part a)
    I am using the equation is that V^2=KX^2
    -------
    mass of ball
    so when we out it all out it should be v^2=900n/m(-.110m)^2
    -----------------
    .300 kg

    -------------------------->v=6.02 m/s but that answer is wron apparently. I dont know where im going wrong.
     
  2. jcsd
  3. Mar 19, 2006 #2

    Hootenanny

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    I wouldn't mind seeing the derivation for this formula, do you have it?
     
  4. Mar 19, 2006 #3
    the original equation is
    .5mass1v^2+ .5kx^2=1/2mass2v^2+ .5kx^2
    ==>0+1/2mv^2=1/2Kx^2 +0
    ===>v^2=kx^2/mass
     
  5. Mar 19, 2006 #4

    Hootenanny

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    That look's better, however, you are still ignoring gravitational potential. Are you happy assuming the GPE is negligable?
     
  6. Mar 19, 2006 #5
    so how do i intergrate that with this problem?
     
  7. Mar 19, 2006 #6

    Hootenanny

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    What is the equation for GPE?
     
  8. Mar 19, 2006 #7
    the equation that i was taught was that U=MGH
     
  9. Mar 19, 2006 #8

    Hootenanny

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    well there you go, just add this in like so;

    [tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + mgh[/tex]

    However, if you haven't incorperated GPE in this type of porblem before, I would ask you tutor before doing so. :smile:
     
  10. Mar 19, 2006 #9
    ok so since we are ignoring mass the:


    .5(900N/M)(.110^2)=.5(v^2) +(h)(9.8)
     
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