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

$v=3.30m/s$, $s=5.04m$
We can determine the required distance and velocity as follows: We know that $\Sigma F_y=0$ $\implies N-W+Psin\theta=0$ $\implies N=W-Psin\theta$ $\implies N=(50)(9.81)-200sin30=390.5N$ and $\Sigma F_x=ma_x$ $\implies -\mu_k N+Pcos\theta=ma$ We plug in the known values to obtain: $200 cos 30-(0.3)(390.5)=50a$ $\implies a=1.212m/s^2$ As $v=v_{\circ}+at$ $\implies v=0+(1.121)(3)=3.30m/s$ and $s=s_{\circ}+v_{\circ}t+\frac{1}{2}at^2$ We plug in the known values to obtain: $s=0+0+\frac{1}{2}(1.121)(3)^2$ $\implies s=5.04m$