Pinball machine problem on energy/power/work

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The discussion revolves around calculating the spring constant of a pinball machine using energy conservation principles. The problem involves a pinball with a mass of 0.05 kg shot up a 15-degree incline, reaching a speed of 3 m/s at the top. Key equations include potential energy (PE = mgh), kinetic energy (KE = 0.5mv^2), and the elastic potential energy of the spring. The final extension of the spring is zero after launching the ball, while the initial extension is 0.08 m. The solution requires applying these energy concepts to find the spring constant.
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Homework Statement


A pinball machine has a spring that loads up, and shoots a small ball up a shallow incline. The mass of the pinball is .05kg. The incline for the pinball is 15 degrees. Assume no friction exists along the distance over which the spring applies its force. The coefficient between the ball and sloped surface is .15. The speed of the ball at the top of the incline is 3m/s. The spring is compressed .08 m. The length of the incline is .75m. What is the value of the spring constant?


Homework Equations



PE=mgh
KE=.5mv^2
KE+PE=w
Spring constant

The Attempt at a Solution



.05(9.8)h=PE

KE=.5(05)(9)
 
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This is another conservation of energy problem: looks like you have both kinetic energy and potential energy being gained, whereas elastic potential energy from the spring is being lost. You can calculate the potential energy of a spring with this equation:

U_{e} = \frac{1}{2}k(x_f^{2} - x_i^{2}), where x_f is the final extension of the spring, and x_i is the initial extension, and k is the spring constant.
 
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How would you go about solving xf and xi?
 
Just think of what they mean: final extension is just that: final extension. After the spring decompresses and launches the ball, it's not compressed anymore, so final extension is zero. Initial extension is also just that. Look at givens in the problem for how much the spring is compressed to start.
 
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