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: 7

Answer

$v=60.7ft/s$

Work Step by Step

We can determine the required velocity as follows: First, we apply Newton's second law $\Sigma F_y=0$ $\implies N-W=0$ $\implies N=W=10lb$ and $\Sigma F_x=ma_x$ $\implies F-\mu_k N=ma$ $\implies 8t^2-0.2(10)=\frac{10}{32.2}a$ This simplifies to: $a=3.22(8t^2-2)$ We know that $v-v_{\circ}=3.22(\frac{8}{3}t^3-2t)$ $\implies v=4+3.22(\frac{8}{3}t^3-2t)=8.587t^3-6.44t+4$.eq(1) As $v=\frac{ds}{dt}=\int_o^t (8.587t^3-6.44t+4)dt$ $\implies s=2.147t^4-3.22t^2+4t$ This simplifies to: $t=2.009s$ Now from eq(1), we obtain: $v=8.58(2.009)^3-6.44(2.009)+4$ $\implies v=60.7ft/s$
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