Answer
$v=29.4ft/s$
Work Step by Step
We can determine the required velocity as follows:
According to the impulse momentum principle in the $y$ direction
$m(v_y)1+\Sigma \int_1^2 F_ydt=m(v_y)2$
We plug in the known values to obtain:
$\frac{20}{32.2}(0)+N\Delta t-Wcos30\Delta t=\frac{20}{32.2}(0)$
This simplifies to:
$N=17.32lb$
Now we apply the impulse momentum principle in the $x$ direction
$m(v_x)1+\Sigma \int_{t_1}^{t_2} F_xdt=m(v_x)2$
$\frac{20}{32.2}(2)+Wsin30\Delta t-\mu_k N\Delta t=\frac{20}{32.2}v$
We plug in the known values to obtain:
$\frac{20}{32.2}(2)+20sin30(3-0)-(0.25\times 17.32(3-0))=\frac{20}{32.2}v$
This simplifies to:
$v=29.4ft/s$
