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 226: 78



Work Step by Step

The required speed can be determined as follows: We know that $x_A=\sqrt{(1.5)^2+(2)^2}-0.5=2m$ and $x_B=1.5-0.5=1m$ Now according to the equation of conservation of energy $\frac{1}{2}mv_A^2+mgh_A+\frac{1}{2}kx_A^2=\frac{1}{2}mv_B^2+mgh_B+\frac{1}{2}kx_B^2$ We plug in the known values to obtain: $0+(3)(9.81)(2)+\frac{1}{2}(200)(2)^2=\frac{1}{2}(3)v_B^2+(3)(9.81)(0)+\frac{1}{2}(200)(1)^2$ This simplifies to: $v_B=15.5m/s$
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