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 203: 40



Work Step by Step

The required force can be determined as follows: $\Sigma U_{A\rightarrow B}=F(AC-BC)$ $\Sigma U_{A\rightarrow B}=F(\sqrt{(8)^2+(6)^2})=F(3.675)$ Now according to the principle of work and energy $\frac{1}{2}+\Sigma U_{A\rightarrow B}=\frac{1}{2}mv_B^2$ We plug in the known values to obtain: $0+F(3.675)=\frac{1}{2}(75)(6)^2$ This simplifies to: $F=367.3N$
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