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 14.3 - Principle of Work and Energy for a System of Particles - Problems - Page 197: 12



Work Step by Step

We know that $T_1+\Sigma U_{1-2}=T_2$ $\implies \frac{1}{2}mv^2+\int_0^{0.2} \Sigma U_{1-2}=(0)^2ds$ We plug in the known values to obtain: $\frac{1}{2}(5000)(4)^2-k\frac{(0.2)^3}{3}=0$ This simplifies to: $k=15MN/m^2$
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