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 133: 30



Work Step by Step

We can find the required velocity as $2F-W=0$ $\implies 2(625t^2)=mg=(400)(9.81)$ $\implies t=1.772s$ We know that $\Sigma F_y=0$ $\implies 2F-W=0$ $\implies 2(625t^2)-(400)(9.81)=400a$ $\implies a=3.125t^2-9.81$ Now, $v=\int_{1.772}^t(3.125t^2-9.81)dt$ $\implies v=(\frac{3.125 t^3}{3}-9.81t)|_{1.772}^t$ This simplifies to: $v=0.301m/s$
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