Answer
$0.496$
Work Step by Step
We can find the coefficient of kinetic friction as follows:
$\Sigma F_x=N-mgcos\theta=0$
$\implies N=mgcos\theta$
Now the vertical component of force is given as
$\Sigma F_y=mgsin\theta-\mu_kN$
$\implies ma=mgsin\theta-\mu_kN$
$\implies mgsin\theta-ma=\mu_kN$
$\implies mgsin\theta-ma=\mu_k(mgcos\theta)$
This simplifies to:
$\mu_k=\frac{gsin\theta-a}{gcos\theta}$
We plug in the known values to obtain:
$\mu_k=\frac{(9.81)sin33.0^{\circ}-1.26}{9.81(cos33.0^{\circ})}$
$\mu_k=0.496$
