Physics: Principles with Applications (7th Edition)

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

Chapter 11 - Oscillations and Waves - Problems - Page 323: 19

Answer

(a) $k = 430~N/m$ (b) The mass is 4.7 kg

Work Step by Step

(a) When the spring is compressed, the potential energy in the spring is equal to the work $W$ done to compress the spring. Therefore, $U_s = W$ $\frac{1}{2}kx^2 = W$ $k = \frac{2W}{x^2}$ $k = \frac{(2)(3.6~J)}{(0.13~m)^2}$ $k = 430~N/m$ (b) The mass experiences maximum acceleration when the spring is at maximum compression, because then the spring exerts the strongest force on the mass. We can find the mass $M$ as: $Ma = kx$ $M = \frac{kx}{a}$ $M = \frac{(430~N/m)(0.13~m)}{12~m/s^2}$ $M = 4.7~kg$ The mass is 4.7 kg.
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