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: 12



Work Step by Step

We apply the principle of impulse and momentum as follows: $mv_{x1}+\Sigma \int_{t_1}^{t_2} F_x dt=mv_{x2}$ We plug in the known values to obtain: $(2500)(5.556)+\int_{0}^{5} 600t^2 dt=2500v_2$ This simplifies to: $v_2=15.6m/s$
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