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 200: 25



Work Step by Step

According to the work-energy principle $T_1+\Sigma U_{1\rightarrow}=T_2$ $\implies \frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$ $\implies \frac{1}{2}mv_1^2+W(3+s)+\frac{1}{2}k(x_1^2-x_2^2)=\frac{1}{2}mv_2^2$ We plug in the known values to obtain: $\implies \frac{1}{2}(\frac{5}{100})(10)^2+5(3+s)+\frac{1}{2}(4)(0.75)^2-(0.75+s)^2=0$ This simplifies to: $s=0.0735ft$
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