Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.3 - Conservation of Linear Momentum for a System of Particles - Problems - Page 262: 36



Work Step by Step

We can determine the required distance as follows: $m_bv_b+m_{sb}v_{sb}=(mb+m_{sb})v$ $\implies (50)(5)+5(0)=(50+5)$ $\implies v=4.545m/s$ According to the law of conservation of energy $T_A+V_A=T_B+V_B$.eq(1) $T_A=\frac{1}{2}(m_b+m{sb})v_A^2=\frac{1}{2}(50+5)(4.545)^2=568.068J$ and $V_A=(m_b+m_{sb})gh_A=0$ $T_B=\frac{1}{2}(m_b+m_{sb})v_B^2=0$ $V_B=(m_b+m_{sb})gh_B$ $\implies V_B=(m_b+m_{sb}) gs \space sin 30^{\circ}=55\times 9.81 sin30^{\circ} s=269.775s$ We plug in the known values in eq(1) to obtain: $568.068+0=0+269.775s$ This simplifies to: $s=2.1m$
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