Answer
The minimum speed of the box at the bottom of the incline is $v = \sqrt{2gh\times (1 + \mu_k~cot(\alpha))}$.
Work Step by Step
$KE_1 + W = KE_2$
$\frac{1}{2}mv^2 - mgh - F_f~d = 0$
$\frac{1}{2}mv^2 = mgh + mg~\mu_k~cos(\alpha)~\frac{h}{sin(\alpha)}$
$v^2 = 2gh + 2g~\mu_k~cot(\alpha)~h$
$v = \sqrt{2gh\times (1 + \mu_k~cot(\alpha))}$
The minimum speed of the box at the bottom of the incline is $v = \sqrt{2gh\times (1 + \mu_k~cot(\alpha))}$.