14. (15 points) A rectangular container without a lid, with a length of 40, width of 25, and height of 60 (neglecting the thickness), is filled with water to a depth of $a$, where $0<a \leqslant 60$. Now, a cubic iron block with an edge length of 10 is placed on the bottom of the container. What is the water depth after the iron block is placed in the container?
【Analysis】From the problem, we know that the container is a lidless rectangular prism with a length of 40, width of 25, and height of 60 (neglecting the thickness). It is filled with water to a depth of $a$, where $0<a \leqslant 60$. This means the water depth is unknown and uncertain, so we need to discuss different scenarios based on the water depth after the iron block is placed inside.

【Solution】Given:
The base area of the container $S=40 \times 25=1000$,
The volume of the container $V=1000 \times 60=60000$,
The base area of the iron block $S_{\text {iron }}=10 \times 10=100$,
The volume of the iron block $V_{\text {iron }}=10 \times 10 \times 10=1000$,
(1) If the water depth in the container is exactly 60 after the iron block is placed,
$1000 a+1000=60000$, we get $a=59$.
Therefore, when $59 \leqslant a \leqslant 60$, the water depth is 60 (excess water overflows).
(2) If the water depth in the container is exactly 10 after the iron block is placed, $1000 a+1000=10000$, we get $a=9$.
Therefore, when $9 \leqslant a<59$, the water depth is $\frac{a \times 40 \times 25+10 \times 10 \times 10}{40 \times 25}=a+1$;
(3) From (2), when $0<a<9$, let the water depth be $x$, then $1000 x=1000 a+100 x$, we get $x=\frac{10}{9} a$;
Answer: When $0<a<9$, the water depth is $\frac{10}{9} a$; when $9 \leqslant a<59$, the water depth is $a+1$; when $59 \leqslant a \leqslant 60$, the water depth is 60.