Answer
$0.042$
Work Step by Step
We can find the coefficient of kinetic friction as follows:
$F=Kx$
$\implies F=(89)(0.22)$
$F=1.96N$
The force in the y direction is given as
$F_y=N+Fsin\theta-W=0$
$\implies N=W-Fsin\theta$.........eq(1)
Similarly, the force in the horizontal x-direction is given as:
$\Sigma F_x=Fcos\theta-\mu_kN=0$
$\implies \mu_k=\frac{Fcos\theta}{N}$
We substitute the value of N from eq(1):
$\implies \mu_k=\frac{Fcos\theta}{W-Fsin\theta}$
We plug in the known values to obtain:
$\mu_k=\frac{(1.96)cos13^{\circ}}{(4.7)(9.81)-(1.96)sin13^{\circ}}$
$\mu_k=0.042$
