#### Answer

$v_B=4.5m/s$

#### Work Step by Step

We can determine the required speed as follows:
$\Sigma F_y=0$
$N=Wcos\theta=mgcos30=12(9.81)cos 30=101.949N$
and $\Sigma F-x=ma_x Wsin\theta-\mu_k N=ma$
This can be rearranged as:
$a=\frac{mgsin\theta-\mu_kmgcos 30}{m}$
$\implies a=g(sin\theta-\mu_k cos 30)$
We plug in the known values to obtain:
$a=(9.81)(sin30-0.3cos30)=2.356m/s^2$
Now, $v_B=\sqrt{v^2_{A_{\circ}}+2a(s_B-s_A)}$
We plug in the known values to obtain:
$v_B=\sqrt{(2.5)^2+2(2.356)(3-0)}$
This simplifies to:
$v_B=4.5m/s$