Answer
$v=\sqrt (\frac{1}{2}gd)$
Work Step by Step
By the last result, we got that
$$\Delta K=\frac{1}{4}mgd$$.
By definition, we have that $\Delta K= \frac{1}{2}mv^2-\frac{1}{2}mv_{0}^2$.
We must remember that the block is initially at rest, so $v_{0}=0$, now we get
$$\Delta K= \frac{1}{2}mv^2-\frac{1}{2}m(0)^2=\frac{1}{2}mv^2$$
In fact,
$$\frac{1}{2}mv^2=\frac{1}{4}mgd$$.
Solving for $v$ we obtain
$$v=\sqrt (\frac{1}{2}gd.$$
