5. In the pyramid $P-ABCD$, the slant height is $l$. The dihedral angle between a lateral face and the base is $60^{\circ}$, and $E, F$ are the midpoints of $BC, CD$ respectively. The distance from the center $O$ of the base to the section $PEF$ is $\qquad$
$$
\begin{array}{c}
\text { 5. } \frac{\sqrt{21}}{14} \\
PE=1, \angle PEO=60^{\circ} \text { so } OE= \\
OF=\frac{1}{2}, PO=\frac{\sqrt{3}}{2} .
\end{array}
$$

Let $EF$ intersect $OC$ at $\theta$, then $OQ=\frac{\sqrt{2}}{4}$.
Also, $EF=\frac{\sqrt{2}}{2}, PQ=\frac{\sqrt{14}}{4}$. Let the distance from $O$ to the plane $PEF$ be $h, V_{P-OEF}=$ $V_{O-PEF}$, we get:
$$
\frac{1}{3} \cdot PO \cdot S_{\triangle OEF}=\frac{1}{3} \cdot h \cdot S_{\triangle FEF} .
$$
(Notice that $PQ \perp EF$)
So $\frac{1}{2} \cdot OE^{2} \cdot PO=\frac{1}{2} EF \cdot P\theta \cdot h, h=\frac{\sqrt{21}}{14}$.