Answer
$v=3.30m/s$, $s=5.04m$
Work Step by Step
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:
$s=0+0+\frac{1}{2}(1.121)(3)^2$
$\implies s=5.04m$
