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 253: 33



Work Step by Step

We can determine the required speed as follows: $\Sigma F_y=0$ $\implies mg-N=0$ $\implies N=500\times 9.81=4905N$ and $\Sigma F_x=0$ $\implies F-\mu_s N=0$ $\implies F=\mu_sN=0.5\times 4905=2452.5N$ We apply the principle of impulse and momentum $mv_1+\Sigma \int_{t_1}^{t_2}F_x dt=mv_2$ $500(0)+2\int_{2.476}^3 200t^2dt+2\int_3^5 1800dt-\int_{2.476}^5 \mu_k Ndt=500v$ We plug in the known values to obtain: $(\frac{400}{3}t^3)|^3_{2.476}+(2\times 1800\times (5-3))-[0.4\times 4905\times (5-2.476)]=500v$ This simplifies to: $v=7.65m/s$
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