(5) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with the left vertex $A_{1}$ and the right focus $F_{2}$. Point $P$ is a moving point on the ellipse. When $\overrightarrow{P A_{1}} \cdot \overrightarrow{P F_{2}}$ takes the minimum value, the value of $\left|\overrightarrow{P A_{1}}+\overrightarrow{P F_{2}}\right|$ is ( ).
(A) $2 \sqrt{2}$
(B) 3
(C) $2 \sqrt{3}$
(D) $\sqrt{13}$
(5) From the conditions, we know $a=2, b=\sqrt{3}$, so $c=1$, hence $A_{1}(-2,0) 、 F_{2}(1,0)$. Let $P(x, y)$, then
$$
\overrightarrow{P A_{1}}=(-2-x,-y), \overrightarrow{P F_{2}}=(1-x,-y) .
$$

Thus,
$$
\begin{aligned}
\overrightarrow{P A_{1}} \cdot \overrightarrow{P F_{2}} & =(x+2)(x-1)+y^{2} \\
& =x^{2}+x-2+3\left(1-\frac{x^{2}}{4}\right) \\
& =\frac{1}{4} x^{2}+x+1 \\
& =\frac{1}{4}(x+2)^{2} \\
& \geqslant 0
\end{aligned}
$$

When $x=-2$, the equality holds. That is, when $P$ coincides with $A_{1}$, $\overrightarrow{P A_{1}} \cdot \overrightarrow{P F_{2}}$ reaches its minimum value, at which time $\left|\overrightarrow{P A_{1}}+\overrightarrow{P F_{2}}\right|=\left|\overrightarrow{A_{1} F_{2}}\right|=3$. Therefore, the answer is B.