hw7solution - Homework 7 Solution CHANG 1 Consider the...

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Homework 7 Solution CHANG 1 Consider the energy equation for the collar’s motion up the smooth rod: K A U gA U eA W A B K B U gB U eB Note that even though the rod is frictionless there is an external force applied so W A B W F . The displacement is the distance traveled: s 0.3 m, and the work done by force F is W F F cos s where 30 . For the gravitational potential energy, with the datum at the bottom of rod we have y A 0and y B 0.3 m. The deformations in the spring at the starting and final positions are found from geometry: x A l A l 0 0.4 1.2 0.8 m x B l B l 0 0.4 2 0.3 2 1.2 0.7 m The speed can thus be calculated, with A 0, as 1 2 m A 2 mgy A 1 2 kx A 2 F cos s 1 2 m B 2 mgy B 1 2 kx B 2 0 0 1 2 200 0.8 2 60 cos30 0.3 1 2 5  B 2 5 9.8 0.3 1 2 200 0.7 2 B 2.52 m/s CHANG 2 Take the starting point at the bottom of the hill as position 1 and the highest point the block reaches up on the hill as position 2. There is no elastic potential energy (no spring) so the energy equation for the block’s motion is written as K 1 U g 1 W 1 2 K 2 U g 2 Work produced during the motion has two parts: work done by friction and work done by the pushing force. Work by friction can be computed using W f k Ns where N mg cos F sin from Newton’s 2nd law, and work by the horizontal force is simply W F F cos s where . Using the bottom of the hill as the datum, for the gravitational potential energy calculation we have y 1 0and y 2 h ,where h is the maximum height the block reaches. The displacement
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This note was uploaded on 10/17/2009 for the course PHYS 2305 taught by Professor Tschang during the Spring '08 term at Virginia Tech.

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hw7solution - Homework 7 Solution CHANG 1 Consider the...

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