qz10sp08sol - April11,2008 PHY2053Discussion...

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April 11, 2008 PHY2053 Discussion Name:                                   UFID: *1. (5 pts) A light spring with spring constant of 200 N/m rests vertically on the bottom  of a large container of water(density = 1000 kg/m ³ ). A 5.00-kg block of wood(density =  600 kg/m ³ ) is attached to the upper end of the spring, and the block-spring system is  allowed to come to static equilibrium. What is the elongation of the spring? Three forces are exerted on the block, buoyancy, gravity and spring force. Since the block   is in equilibrium, the net force acting on it is zero. Thus we have B-mg-kΔL = 0   ΔL = (B-mg)/k The buoyant force is equal to the weight of displaced water and the volume of the wood   is given by the mass divided by the density. Therefore, ΔL = ( ρ Vg-mg)/k = [( ρ
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This note was uploaded on 12/13/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

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qz10sp08sol - April11,2008 PHY2053Discussion...

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