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: 30



Work Step by Step

According to the principle of work and energy $T_A+\Sigma _{A-c}=T_C$ $\implies \frac{1}{2}m_A^2+U_w-U_f=\frac{1}{2}mv_2^2$ $\implies \frac{1}{2}+W(4)-\mu_kWS^{\prime}=\frac{1}{2}mv_2^2$ We plug in the known values to obtain: $0-(30)(4)-(0.6)(30)S^{\prime}=0$ This simplifies to: $S^{\prime}=6.667ft$ Now, $S=10-S^{\prime}$ $\implies S=10-6.667=3.33ft$
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