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: 13



Work Step by Step

The coefficient of kinetic friction can be determined as $v_1=100Km/h=27.78m/s$ and $v_2=40Km/h=11.11m/s$ The momentum in the y-direction is given as $mv_{y1}+\Sigma \int _{t_1}^{t_2} F_y dt=mv_{y2}$ $\implies 0+5N-5(2500)(9.81)=0$ $\implies N=24525N$ The momentum in the x-direction is $mv_{x1}+\Sigma \int _{t_1}^{t_2}F_x dt=mv_{x_2}$ We plug in the known values to obtain: $2500(27.78)-\mu (24525)(5)=2500(11.11)$ $\implies \mu =0.340$
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