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 197: 11



Work Step by Step

We know that $T+\Sigma U_{1\rightarrow 2}=T_2$ $\implies 0+50=\frac{1}{2}mv^2$ $\implies 50=\frac{1}{2}(6)v^2$ This simplifies to: $v=4.08m/s$
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