Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - Questions: 5

Answer

a. Spring 2. b. Spring 1.

Work Step by Step

a. Here, the force is the same. $F = k_{1}x_{1} = k_{2}x_{2}$. The work done on spring 1 is $$W_{1} = \frac{1}{2} k_{1}x_{1}^{2}$$ The work on spring 2 is $$W_{2} = \frac{1}{2} k_{2}x_{2}^{2} = \frac{1}{2} k_{2}\frac{k_{1}^{2} x_{1}^{2}}{ k_{2}^{2}}$$ $$W_{2} = W_{1} \frac{k_{1}}{k_{2}}$$ Because the first spring is stiffer, the work done on spring 2 is greater. b. Now, the distance stretched is the same. As before, the work done on spring 1 is $$W_{1} = \frac{1}{2} k_{1}x^{2}$$ The work on spring 2 is $$W_{2} = \frac{1}{2} k_{2}x^{2}$$ Because the first spring is stiffer, the work done on spring 1 is greater.
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