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 14.6 - Conservation of Energy - Problems - Page 225: 74


$v_A=1.54m/s$, $v_B=4.62m/s$

Work Step by Step

We can determine the required speed as follows: We know that $3v_A=-v_B~~~~$eq(1) According to the equation of energy conservation $T_{A_{\circ}}+T_{B_{\circ}}+V_{A_{\circ}}+V_{B_{\circ}}=T_A+T_B+V_A+V_B$ $\implies 0+0+0+0=\frac{1}{2}m_Av_A^2+\frac{1}{2}mBv_B^2+W_A\Delta S_A+W_B\Delta S_B$ We plug in the known values to obtain: $0=\frac{1}{2}(20)v_A^2+\frac{1}{2}(30)(-3v_A)^2+(20)(9.81)(\frac{1.5}{3})-(30)(9.81)(1.5)$ This simplifies to: $v_A=1.54m/s$ We plug in this value in eq(1) to obtain: $3(1.54)=-v_B$ This simplifies to: $v_B=4.62m/s$
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