(5) As shown in the figure, the faces of the squares $A B C D$ and $A B E F$ with side length 2 form a $60^{\circ}$ angle, and points $M$ and $N$ are on segments $A C$ and $B F$ respectively, with $A M=F N$. The range of the length of segment $M N$ is ( ).
(A) $\left[\frac{1}{2}, 2\right]$
(B) $[1,2]$
(C) $[\sqrt{2}, 2]$
(D) $[\sqrt{3}, 2]$
(5. B Hint: Draw $M H / / B C$ intersecting $A B$ at $H$, then $\frac{A M}{A C}=\frac{A H}{A B}$, and since $A M=F N, A C=$ $F B$, we have $\frac{F N}{F B}=\frac{A H}{A B}$, so $N H / / A F$, $N H \perp A B, M H \perp A B$, thus $\angle M H N=$ $60^{\circ}$.

Let $A H=x(0 \leqslant x \leqslant 2)$, then $M H=x$, $N H=2-x$, so
$$
\begin{aligned}
M N & =\sqrt{x^{2}+(2-x)^{2}-2 x(2-x) \cos 60^{\circ}} \\
& =\sqrt{3 x^{2}-6 x+4} \\
& =\sqrt{3(x-1)^{2}+1} .
\end{aligned}
$$

Therefore, $1 \leqslant M N \leqslant 2$.