Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.3 - Principle of Work and Energy for a System of Particles - Problems - Page 201: 29



Work Step by Step

According to the principle of work and energy $\frac{1}{2}mv_1^2-U_{s_1}-U_{s_2}=\frac{1}{2}mv_2^2$ $\implies \frac{1}{2}mv_1^2-\frac{1}{2}k_1s^2-\frac{1}{2}k_2s^2=\frac{1}{2}mv_2^2$ We plug in the known values to obtain: $\frac{1}{2}(20)(2)^2-\frac{1}{2}(50)s^2-\frac{1}{2}(100)s^2=0$ This simplifies to: $s=0.730m$
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