Answer
$3.48ft/s$, $0.376ft$
Work Step by Step
We can determine the required speed and distance as follows:
According to conservation of linear momentum
$\Sigma m_1v_1=\Sigma m-2v_2$
$\implies W_bv_{b_1}+W_{w}v_{w_1}=W_bv_{b_2}(\frac{4}{5}+W_wv_{w_2})$
We plug in the known values to obtain:
$0.03(1300)(\frac{12}{13})+0=(0.03)(50)(\frac{4}{5})+10v_w$
This simplifies to:
$v_w=3.48ft/s$
Now, according to the conservation of energy of the box
$T_1+\Sigma U_{1-2}=T_2$
$\implies \frac{1}{2}mv^2-\mu_k Wd=0$
We plug in the known values to obtain:
$\frac{1}{2}(\frac{10}{32.2})(3.48)^2-(0.5)(10)d=0$
This simplifies to:
$d=0.376ft$
