Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 6 - Exercises and Problems - Page 110: 55

Answer

Please see the work below.

Work Step by Step

We know that $W=(\frac{1}{2})kx^2$ Now we can find the required ratio as $\frac{k_B}{k_A}=(\frac{W_B}{W_A})(\frac{x_A}{x_B})^2=(\frac{2}{1})(\frac{2}{1})^2=8$
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