2. In $\triangle A B C$, it is known that $A B=4, A C=3, P$ is a point on the perpendicular bisector of side $B C$, then $\overrightarrow{B C} \cdot \overrightarrow{A P}=$
Answer $-\frac{7}{2}$.
Analysis Let the midpoint of $B C$ be $D$, and $D P$ is perpendicular to $B C$,
$$
\begin{aligned}
\overrightarrow{B C} \cdot \overrightarrow{A P} & =(\overrightarrow{A C}-\overrightarrow{A B}) \cdot(\overrightarrow{A D}+\overrightarrow{D P})=(\overrightarrow{A C}-\overrightarrow{C B}) \cdot \frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \\
& =\frac{1}{2}\left(\overrightarrow{A C}^{2}-\overrightarrow{A B}^{2}\right)=\frac{1}{2}\left(3^{2}-4^{2}\right)=-\frac{7}{2}
\end{aligned}
$$

Therefore, the answer is $-\frac{7}{2}$.