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 132: 20



Work Step by Step

We can determine the required speed as follows: $\Sigma F_y=0$ $N=Wcos\theta=mgcos30=12(9.81)cos 30=101.949N$ and $\Sigma F-x=ma_x Wsin\theta-\mu_k N=ma$ This can be rearranged as: $a=\frac{mgsin\theta-\mu_kmgcos 30}{m}$ $\implies a=g(sin\theta-\mu_k cos 30)$ We plug in the known values to obtain: $a=(9.81)(sin30-0.3cos30)=2.356m/s^2$ Now, $v_B=\sqrt{v^2_{A_{\circ}}+2a(s_B-s_A)}$ We plug in the known values to obtain: $v_B=\sqrt{(2.5)^2+2(2.356)(3-0)}$ This simplifies to: $v_B=4.5m/s$
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