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 137: 50



Work Step by Step

We can determine the required velocity as follows: $\Sigma F_y=ma_y$ $\implies 3T-W=ma_y$ We plug in the known values to obtain: $3(4000)+(1000)(9.81)=1000a_y$ This simplifies to: $a_y=2.19m/s^2$ We know that $v^2=v_{\circ}^2+2as$ $\implies v=\sqrt{+2(2.19)(6)}$ $\implies v=5.13m/s$
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