Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.4 - Equations of Motion: Rectangular Coordinates - Problems - Page 132: 23



Work Step by Step

We can determine the required velocity as follows: $\Sigma F_y=0$ $\implies F+F-T=0$ $\implies T=2F$ $\implies T=2(150)=300N$ For block A, $\Sigma F_y=ma$ $\implies 2F-W=ma$ $\implies a=\frac{2T}{m}-g$ We plug in the known values to obtain: $a=\frac{2(300)}{50}-9.81$ $\implies a=2.19$ Now, $v^2=v^2_{\circ}+2a(y-0)$ We plug in the known values to obtain: $v=\sqrt{0+0(2.19)(3)}$ This simplifies to: $v=3.62m/s$
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