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 199: 20



Work Step by Step

We can determine the required distance as follows: According to the principle of work and energy $\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$ $\implies \frac{1}{2}(\frac{4000}{32.2})(55)^2+U_{1\rightarrow 2}=0$ This simplifies to: $\Sigma U_{1\rightarrow 2}=187.89kip\cdot ft$ We know that the work done can also be calculated as the area under the curve in the given figure: $\implies \Sigma U_{1\rightarrow 2}=(2\times 9)+(5-2)\times 18+(x\times 27)$ $\implies 187.89=(2\times 9)+(5-2)\times 18+(x\times 27)$ This simplifies to: $x=4.29ft$ Thus, $s=x+5=4.29+5=9.29ft$
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