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 14.6 - Conservation of Energy - Problems - Page 230: 95



Work Step by Step

We can determine the required speed as follow: $T_1+V_1=T_2+V_2$ $\implies \frac{1}{2}mv_1^2+mgh_1+2\frac{1}{2}kx_1^2=\frac{1}{2}mv_2^2+mgh_2+2\frac{1}{2}kx_2^2$ We plug in the known values to obtain: $0+0=\frac{1}{2}(20)v^2+2(\frac{1}{2}(40)\sqrt{(3)^2+(2)^2-2})-(20)(9.81)(3)$ This simplifies to: $v=6.97m/s$
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