Answer
$2.73\;\rm m/s$
Work Step by Step
First of all, we need to draw a force diagram of the box, as you see below. We analyzed the weight of the box into its $x$ and $y$ components, as you see in the second figure below.
We need to make the nail box moves for 5 meters toward your friend at the edge of the roof. This means that its final speed at the edge of the roof is zero.
$$\overbrace{v_{fx}^2}^{0}=v_{ix}^2+2a_x\Delta x$$
Solving for $v_i$; and plugging the known.
$$v_{ix}^2=-2a_x\Delta x=-2a_x\cdot 5=-10a_x$$
$$v_{ix} =\sqrt{-10a_x }\tag 1$$
Now we need to find the box's acceleration.
We know that the net force exerted on the object in the $y$-direction is zero since it has no motion in this direction.
$$\sum F_y=F_n-mg\cos25^\circ=ma_y=m(0)=0$$
Thus,
$$F_n=mg\cos25^\circ\tag 2$$
The net force exerted on the nail box in the $x$-direction is not zero, since it is slowing down; and is given by
$$\sum F_x=mg\sin25^\circ -f_k=ma_x$$
Recalling that $f_k=\mu_kF_n$, and from (2), we got $f_k=\mu_kmg\cos25^\circ$
Thus,
$$mg\sin25^\circ -\mu_kmg\cos25^\circ=ma_x$$
Thus,
$$ g\sin25^\circ -\mu_k g\cos25^\circ= a_x$$
$$a_x= g\left[\sin25^\circ -\mu_k \cos25^\circ\right] $$
Plugging the known;
$$a_x= 9.8\left[\sin25^\circ -0.55 \cos25^\circ\right] =\bf -0.743\;\rm m/s^2$$
Plugging into (1);
$$v_{ix} =\sqrt{-10 \times-0.743 } =\color{red}{\bf 2.73}\;\rm m/s$$