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 203: 36



Work Step by Step

We can determine the required speed as follows: $\Delta U_{A\rightarrow B}=W(3+0.25)(\frac{3}{5})-\mu_k N_B(3+0.25)+\frac{1}{2}k(x_1^2-x_2^2)$ We plug in the known values to obtain: $\Sigma U_{A\rightarrow B}=4(3+0.25)(\frac{3}{5})-0.2\times 3.2(3+0.25)+\frac{1}{2}(50)(2-1.5)^2-(2-1.25)^2=-2.0925$ Now, according to the principle of work and energy $\frac{1}{2}mv_A^2+\Sigma U_{A\rightarrow B}=\frac{1}{2}mv_B^2$ We plug in the known values to obtain: $\frac{1}{2}(\frac{4}{32.2})v_A^2-2.0925=0$ This simplifies to: $v_A=5.8ft/s$
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