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: 24



Work Step by Step

We can determine the required speed as follows: $\Delta s_A=\frac{v_{A_1}+v_{A_2}}{2}(2)=\frac{6+v_{A_2}}{2}(2)=6+v_{A_2}$ $\implies \Delta s_B=2\Delta s_A=12+2v_{A_2}$ Now according to the principle of work and energy $\frac{1}{2}m_Av_{A_1}^2+\frac{1}{2}m_Bv_{B_2^2}+W_A\Delta s_A-\mu_k N\Delta s_B=\frac{1}{2}m_Av_{A_2}^2+\frac{1}{2}m_Bv_{B}^2$ We plug in the known values to obtain: $\frac{1}{2}(\frac{10}{32.2})+\frac{1}{2}(\frac{4}{32.2})(12)^2+10(6+v_{A_2})-(0.2\times+2v_{A_2})=\frac{1}{2}(\frac{10}{32.2})v_{A_2}^2+\frac{1}{2}(\frac{10}{32.2})4v_{A_2}^2$ This simplifies to: $v_A=26.8ft/s$
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