Energy and Simple Harmonic Motion

  • Thread starter shawonna23
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  • #1
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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?
 

Answers and Replies

  • #2
Doc Al
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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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