Chapter_8_solved_problems_Physics_1301

# Chapter_8_solved_problems_Physics_1301 - Chapter 8 solved...

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Chapter 8 solved problems 4. Picture the Problem : The physical situation is depicted at right. Strategy: Use equation 2 1 2 W kx = (equation 7-8) to find the work done by the spring, but caution is in order: This work is positive when the force exerted by the spring is in the same direction that the block is traveling, but it is negative when they point in opposite directions. One way to keep track of that sign convention is to say that ( 29 2 2 1 i f 2 W k x x = - . That way the work will always be negative if you start out at i 0 x = because the spring force will always be in the opposite direction from the stretch or compression. Solution: 1. (a) Sum the work done by the spring for each segment of path 1: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 { } ( 29 ( 29 2 2 2 2 1 1 1 2 2 3 2 2 2 2 2 1 2 1 550 N/m 0 0.040 m 0.040 m 0.020 m 0.44 J 0.33 J 0.11 J W k x x x x W = - + - = - + - = - + = - 2. Sum the work done by the spring for each segment of path 2: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 { } ( 29 ( 29 2 2 2 2 1 2 1 4 4 3 2 2 2 2 2 1 2 2 550 N/m 0 0.020 m 0.020 m 0.020 m 0.1 J 0 J 0.11 J W k x x x x W = - + - = - - + - - = - + = - 3. (b) The work done by the spring will stay the same if you increase the mass because the results have no dependence on the mass of the block. Insight: The work done by the spring is negative whenever you displace the block away from x = 0, but it is positive when the displacement vector points towards x = 0. 34.

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## This note was uploaded on 01/05/2012 for the course CHEM 3321 taught by Professor Bean during the Spring '11 term at American Jewish University.

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Chapter_8_solved_problems_Physics_1301 - Chapter 8 solved...

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