Answer
$v_{w}=3.48ft/s$
$N=504lb$
$t=0.216s$
Work Step by Step
According to the law of conservation of linear momentum
$\Sigma m_1v_1=\Sigma m_2v_2$
$\implies W_bv_{b_1}(\frac{12}{13})+W_w v_{w_1}=W_b v_{b_2}(\frac{4}{5})+W_w v_{w_2}$
We plug in the known values to obtain:
$(0.03)(1300)(\frac{12}{13})+0=(0.03)(50)(\frac{4}{5})+10v_w$
Solving the above equation, we obtain:
$v_w=3.48ft/s$
Now we determine the normal force on the block as
$mv_1+\Sigma \int Fdt=mv_2$
$(\frac{W_b}{g})v_{b_1}(\frac{5}{13})-W_w t+Nt=(\frac{W_b}{g})v_{b_2}(\frac{5}{3})$
$\implies (\frac{0.03}{32.2})(1300)(\frac{5}{13})-(10)(1\times 10^{-3})+N(1\times 10^{-3})=(\frac{0.03}{32.2})(50)(\frac{5}{3})$
$\implies N=504lb$
To find the required time, we use the principle of impulse and momentum in the x-direction
$\implies mv_1+\Sigma \int Fdt=mv_2$
$\implies (\frac{W_w}{g})v_{w_1}-\mu_k Wt=(\frac{W_w}{g})v_{w_2}$
We plug in the known values to obtain:
$(\frac{10}{32.2})(3.48)-(0.5)(10)t=0$
This simplifies to:
$t=0.216s$
