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 198: 14



Work Step by Step

According to the work and energy principle $\frac{1}{2}m_Av_{A_{\circ}}^2+\frac{1}{2}m_Bv_{B_{\circ}}^2+W_A\Delta S_A+W_B\Delta S_B=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2$ This can be rearranged as: $\frac{1}{2}m_Av_{A_{\circ}}^2+\frac{1}{2}m_B2v_{A_{\circ}}^2+W_A\Delta S_A-2W_BS_A=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_B(2v_A)^2$ We plug in the known values to obtain: $_0+0+6(5)-10(10)=\frac{1}{2}(\frac{60}{32.2})v_A^2+\frac{1}{2}(\frac{10}{32.2})(2v_A)^2$ This simplifies to: $v_A=7.18ft/s$
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