Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 18 - Planar Kinetics of a Rigid Body: Work and Energy - Section 18.4 - Principle of Work and Energy - Problems - Page 493: 21

Answer

$$ s=0.304 \mathrm{ft} $$

Work Step by Step

$$ \begin{aligned} & v_G=(0.02) 70=1.40 \mathrm{ft} / \mathrm{s} \\ & T_1+\Sigma U_{1-2}=T_2 \\ & 0+(0.3)(s)=\frac{1}{2}\left(\frac{0.3}{32.2}\right)(1.40)^2+\frac{1}{2}\left[(0.06)^2\left(\frac{0.3}{32.2}\right)\right](70)^2 \\ & s=0.304 \mathrm{ft} \end{aligned} $$
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