## Engineering Mechanics: Statics & Dynamics (14th Edition)

$0.340$
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: $2500(27.78)-\mu (24525)(5)=2500(11.11)$ $\implies \mu =0.340$