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: 4



Work Step by Step

We can determine the required distance as follows: $\Sigma F_y=ma_y=0$ $\implies N+500sin45-400sin30-(100\times 9.81)=0$ This simplifies to: $N=827.45N$ We know that $\Sigma U_{1\rightarrow 2}=(500cos45+400cos30-(0.2\times 827.45))s=534.47s$ We also know that $\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$ We plug in the known values to obtain: $\frac{1}{2}(100)(0)^2+534.47(s)=\frac{1}{2}(100)(8)^2$ This simplifies to: $s=5.98m$
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