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: 79



Work Step by Step

The required length can be determined as follows: First, we find the velocity: $v_B^2=\frac{\rho Wcos45}{m}=\rho gcos45=(1.5)(32.2)cos45=34.15(m/s)^2$ Now, according to the conservation of energy equation $\frac{1}{2}mv_A^2+\frac{1}{2}kx_A^2=\frac{1}{2}mv_B^2+\frac{1}{2}kx_B^2+Wrsin45$ We plug in the known values to obtain: $0+\frac{1}{2}(2)(1.5\pi-l)^2=\frac{1}{2}(\frac{2}{32.2})(34.15)+(2)(1.5sin45)+\frac{1}{2}(2)(1.5\times 0.75\pi -l)^2$ This simplifies to: $l=2.77ft$
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