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 202: 34



Work Step by Step

We can determine the required length of the spring as follows: $\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$ $\frac{1}{2}mv_1^2+\frac{1}{2}k(x_1^2-x_2^2)=\frac{1}{2}mv_2^2$ We plug in the known values to obtain: $\frac{1}{2}(\frac{4}{32.2})(9)^2+\frac{1}{2}(50)(s-1.5)^2-(s-1.5-0.2)^2=0$ This simplifies to: $s=1.90ft$
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