Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 230: 56


The mass of the second cheerleader is 53.2 kg

Work Step by Step

When the first cheerleader stands on the spring, the spring's force is equal to the weight of the cheerleader. We can find the spring constant. Let $m_1$ be the mass of the first cheerleader. $F_s = m_1~g$ $kx = m_1~g$ $k = \frac{m_1~g}{x}$ $k = \frac{(65~kg)(9.80~m/s^2)}{0.055~m}$ $k = 11,580~N/m$ We can find the mass $m_2$ of the second cheerleader. Note that the spring compression is $10~cm$ when both cheerleaders stand on the spring. $F_s = (m_1+m_2)~g$ $kx = (m_1+m_2)~g$ $kx-m_1~g = m_2~g$ $m_2 = \frac{kx-m_1~g}{g}$ $m_2 = \frac{(11,580~N/m)(0.10~m)-(65~kg)(9.80~m/s^2)}{9.80~m/s^2}$ $m_2 = 53.2~kg$ The mass of the second cheerleader is 53.2 kg
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