ch08-p037 - cm before it stops). 37. (a) The (final)...

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mg ( d + x )sin(30º) = 9.50 J ¡ d = 0.396 m. (b) The block is still accelerating (due to the component of gravity along the incline, mg sin(30º)) for a few moments after coming into contact with the spring (which exerts the Hooke’s law force kx ), until the Hooke’s law force is strong enough to cause the block to being decelerating. This point is reached when kx = mg sin30º which leads to x = 0.0364 m = 3.64 cm; this is long before the block finally stops (36.0
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Unformatted text preview: cm before it stops). 37. (a) The (final) elastic potential energy is U = 1 2 kx 2 = 1 2 (431 N/m)(0.210 m) 2 = 9.50 J. Ultimately this must come from the original (gravitational) energy in the system mgy (where we are measuring y from the lowest “elevation” reached by the block, so y = ( d + x )sin(30º). Thus,...
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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