6b - equilibrium position of the spring? Assume air...

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PHY2053 – Section 6168 Quiz #6 Friday, February 20, 2004 Name_______________________________ UFID___________________ Please read the problem carefully and be sure to show all of your work. Partial credit will be given. A 0.50 kg block is placed on a light vertical spring (k = 7.0 x 10 2 N/m) and pushed downward, compressing the spring by 5 cm. After the block is released, it leaves the spring and continues to travel upward. What height will the block reach, relative to the
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Unformatted text preview: equilibrium position of the spring? Assume air resistance is negligible. Solution: Conservation of energy: Potential energy in the spring gets converted to kinetic energy as the block leaves the spring and then gets converted to gravitational potential energy as the block reaches its maximum height. Note that the equilibrium position of the spring is where the block leaves the spring (x=0) ! kx 2 = mgh Solve for h: h = kx 2 /2mg = 17.8 cm...
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This note was uploaded on 09/25/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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