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

$\theta=22.6^{\circ}$
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: $\theta=22.6^{\circ}$