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 201: 31


$R=2.83m$, $v_C=7.67m/s$

Work Step by Step

According to the principle of work and energy $\frac{1}{2}mv_A^2+U_w=\frac{1}{2}mv_B^2$ $\implies \frac{1}{2}mv_A^2+W(3-2)=\frac{1}{2}mv_B^2$ We plug in the known values to obtain: $0-(0.005)(9.81)(1)=\frac{1}{2}(0.005)v_B^2$ $\implies v_B=4.429m/s$ From the equations of kinematics, we get $S_y=S_{\circ}+v_{y_{\circ}}t+\frac{1}{2}gt^2$ $\implies 2=0+0+\frac{1}{2}(9.81)t^2$ $\implies t=0.639s$ Now $S_x=R=S_{x_{\circ}}+v_{x_{\circ}}t$ $\implies R=(4.429)(0.639)$ $\implies R=2.83m$ We know that $\frac{1}{2}mv_A^2+U_w=\frac{1}{2}mv_C^2$ We plug in the known values to obtain: $0+(0.005)(9.81)(3)=\frac{1}{2}(0.005)v_C^2$ This simplifies to: $v_C=7.67m/s$
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