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 129: 6



Work Step by Step

We can determine the velocity of the block as $\Sigma F_y=0$ $\implies N-W=0$ $\implies N=W=10lb$ and $\sigma F_x=ma_x$ $\implies F-\mu_k N=ma$ We plug in the known values to obtain: $8t^2-(0.2)(10)=\frac{10}{32.2}a$ This simplifies to: $a=3.22(8t^2-2)$ We know that $a=\frac{dv}{dt}=\int_0^t 3.22(8t^2-2)dt$ $\implies v-v_{\circ}=3.22(\frac{8t^3}{3}-2t)$ $\implies v=4+3.22[\frac{8}{3}(2)^3-2(2)]$ $\implies v=59.8ft/s$
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