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 200: 26



Work Step by Step

We can determine the required speed as follows: According to the work-energy principle $T_1+\Sigma U_{1\rightarrow 2}=T_2$ $\implies 0+\frac{1}{2}F(2S_B)=\frac{1}{2}mv_A^2$ We plug in the known values to obtain: $0+\frac{1}{2}(20000)(2)(0.2)=\frac{1}{2}(10)v_A^2$ This simplifies to: $v_A=28.3m/s$
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