Answer
$\theta=22.6^{\circ}$
Work Step by Step
We can determine the smallest incline $\theta$ as follows:
$v^2=v_{A_{\circ}}^2+2a(s_B-s_A)$
We plug in the known values to obtain:
$0=(2.5)^2+2a(3-0)$
$\implies a=1.042m/s^2$
We know that
$\Sigma F_y=0$
$\implies N=Wcos\theta=mgcos\theta$
and $\Sigma F_x=ma_x$
$\implies Wsin\theta-\mu_k N=ma$
$\implies ma=mgsin\theta-\mu_k cos\theta$
This can be rearranged as:
$\frac{a}{g}=\frac{1.042}{9.81}=sin\theta-0.3cos\theta$
This simplifies to:
$\theta=22.6^{\circ}$
