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Energy and Simple Harmonic Motion

  1. Apr 29, 2005 #1
    In preparation for shooting a ball in a pinball machine, a spring (k = 675 N/m) is compressed by 0.0690 m relative to its unstrained length. The ball (m = 0.0540 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 used this equation: v final=square root of k/m(x initial^2 - x final^2)
    but I keep getting the wrong answer. Am I using the wrong equation? Can someone help me?
  2. jcsd
  3. Apr 29, 2005 #2

    Doc Al

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    Staff: Mentor

    Instead of using that equation, think in terms of conservation of energy: Initially, the ball has only spring potential energy ([itex]1/2 k x^2[/itex]), which gets transformed into kinetic energy plus gravitational potential energy as it moves to position B. Set up that equation and solve for the speed.
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