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