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Finding speed using conservation of mechanical energy

  1. Jul 28, 2005 #1
    In preparation for shooting a ball in a pinball machine, a spring (k = 675 N/m) is compressed by 0.0650 m relative to its unstrained length. The ball (m = 0.0585 kg) is at rest against the spring at point A. When the spring is released, the ball slides (without rolling) to point B, which is 0.300 m higher than point A. How fast is the ball moving at B?

    I drew three springs vertically each having a ball on the top. The first spring on my left, is unstrained. The second spring is compressed by 0.065 m. The third spring is the tallest, it is .3 m above the second spring. The second spring having a ball at the top is called point A. The third spring having the ball at the top is called point B. I made my h=0 at point B.
    From the way I understand the problem, speed at A is zero and mgh at B is also zero. This leads me to the following equation:
    (mgh at A) + (1/2)(k)(x^2 at A) = (1/2)(m)(V^2 at B) + (1/2)(k)(x^2 at B)

    I don't know if this is right, but I think that h at A is -0.3. Also, x^2 at A is 0^2 - (0.065^2). My x^2 at B is (-(.065^2)) - (.235^2). Plugging everything in, I get my speed at B to be 23.105 m/s. Is this right? Am I using the right approach. Are my values right? Unfortunately, this question is an even question from my textbook, so I do not have an answer for it. If someone is certain, please help me.
     
    Last edited: Jul 28, 2005
  2. jcsd
  3. Jul 28, 2005 #2

    LeonhardEuler

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    Gold Member

    I don't think the spring expands all .3 meters with the ball. I think the ball leaves the spring when it expands. I think you can neglect the potential energy of the spring after it has expanded.
     
  4. Jul 30, 2005 #3
    thanks, your right, the answer works that way
     
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