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: 93



Work Step by Step

We can determine the required speed as follows: According to the conservation of energy equation $T_1+V_1=T_2+V_2$ $\implies \frac{1}{2}mv_1^2+mgh_1+\frac{1}{2}ks_1^2=\frac{1}{2}mv_2^2+mgh_2+\frac{1}{2}ks_2^2$ We plug in the known values to obtain: $0+(10)(9.81)(0.45)+\frac{1}{2}(500)(0.2)^2=\frac{1}{2}v_2^2+\frac{1}{2}(500)(0.3)^2$ This simplifies to: $v_2=1.68m/s$
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