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Elastic Potential Energy Question

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data
    PinballMachine.jpg

    A pinball machine launches a 90g ball with a spring driven plunger. The game board is inclined at an angle of 6[tex]\circ[/tex] above the horizontal.

    Assume the plunger's mass and frictional effects are negligible. The acceleration of gravity is 9.8m/s2.

    Find the force constant k of the spring that will give the ball a speed of 101cm/s when the plunger is released from rest with the spring compressed 4 cm from its relaxed position.
    Answer in units of N/m.


    2. Relevant equations
    Not too sure, but...
    Fg=mgsin[tex]\Theta[/tex]
    F=-kx
    [tex]\sum[/tex]F=ma
    And since I see something about speed in there, I assume we'll need a kinematic equation as well? From what I can tell it looks like:
    Vf2=Vi2+2ax

    3. The attempt at a solution

    To start, everything needs converted from cm to m. Next...I'm guessing what needs done is I need to find the acceleration necessary to get the ball from rest to 1.01m/s within 0.04m. Then once that's been found, would I calculate the net force and solve for k in that equation?
     
  2. jcsd
  3. Nov 27, 2009 #2

    kuruman

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    Try using conservation of mechanical energy.
     
  4. Nov 27, 2009 #3
    Ok. So...

    [tex]\frac{1}{2}[/tex]kx2=mgh+[tex]\frac{1}{2}[/tex]mv2?
     
  5. Nov 27, 2009 #4

    kuruman

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    Correct. You need to do some trig to find h.
     
  6. Nov 27, 2009 #5
    Right. Figured. Now, do I need to account for the incline with the force of gravity, or is it good at 9.8?
     
  7. Nov 27, 2009 #6

    kuruman

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    I am not sure what you mean. In mgh, the symbol "h" stands for "change in height". What is the change in height of the ball in this case?
     
  8. Nov 27, 2009 #7
    I get how to do height. My above question was about gravity. The force of gravity on an incline is mgsin[tex]\Theta[/tex]. I was wondering if that was necessary here because height is just straight up and down, like gravity.

    Oh, and height would be x*sin[tex]\Theta[/tex].
     
  9. Nov 27, 2009 #8

    kuruman

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    It is not necessary. The work done by gravity when the ball moves up the incline by distance d is Wg=-mgdsinθ = -mgh which the negative of the potential energy change. So when you put in the potential energy change as mgh, you are taking care of the effects of gravity.
     
  10. Nov 27, 2009 #9
    hmm...ok that makes sense. So I should get 61.9903N/m, right? The equation I got was:

    [tex]\frac{m}{x^2}[/tex](2gxsin[tex]\Theta[/tex]+Vf2)
     
  11. Nov 27, 2009 #10

    kuruman

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    I didn't put in the numbers, but your expression looks right.
     
  12. Nov 27, 2009 #11
    The homework site accepted it as correct. Thank you very much.
     
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