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 195: 3



Work Step by Step

The required distance can be determined as follows: $\Sigma F_y=ma_y=0$ $\implies N+1000(\frac{3}{5})-800sin30-(100)(9.81)=0$ $\implies N=781N$ We know that $\Sigma U_{1\rightarrow 2}=\Sigma F_x(s_2-s_1)$ $\Sigma U_{1\rightarrow 2}=1000(\frac{4}{5})+800cos 30-(0.2\times 781)(s)$ $\implies \Sigma U_{1\rightarrow 2}=1336.62(s)$ As $\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$ We plug in the known values to obtain: $\frac{1}{2}(100)(0)^2+1336.62(s)=\frac{1}{2}(100)(6)^2$ This simplifies to: $s=1.35m$
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