Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.2 - Principle of Linear Impulse and Momentum for a System of Particles - Problems - Page 249: 15



Work Step by Step

The required speed can be calculated as $3F=W$ $\implies 3(10t^2+300)=100(9.81)$ $\implies t=1.643s$ Now, $mv_{y1}+\Sigma \int _{t_1}^{t_2}F_ydt=mv_{y2}$ We plug in the known values to obtain: $0+\int_{1.643}^4 3(10t^2+300)dt-100(9.81)(4-1.643)=100v$ $\implies v=4.047m/s$
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