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: 71



Work Step by Step

We can determine the required speed as follows: According to the conservation of energy equaiton $\frac{1}{2}m_Bv_{B1}^2+\frac{1}{2}m_Cv_{C_1}^2+V_{B_1}+V_{C_1}=\frac{1}{2}m_Bv_{B_2}^2+\frac{1}{2}m_Cv_{C_2}^2+V_{B_2}+V_{C_2}$ We plug in the known values to obtian: $0+0+0+0=\frac{1}{2}(\frac{200}{32.2})(\frac{-v_C}{2})^2+\frac{1}{2}(\frac{600}{32.2})(v_C)^2+200(15)+600(-30sin20)$ This simplifies to: $v_c=17.7ft/s$
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