Let $ABCD$ be a square with side length $5$, and let $E$ be the midpoint of $CD$. Let $F$ be the point on $AE$ such that $CF=5$. Compute $AF$.
1. **Define the coordinates:**
   - Let the coordinates of the square $ABCD$ be $A(0, 0)$, $B(5, 0)$, $C(5, 5)$, and $D(0, 5)$.
   - The midpoint $E$ of $CD$ has coordinates $E\left(\frac{5+0}{2}, \frac{5+5}{2}\right) = E(2.5, 5)$.

2. **Equation of line $AE$:**
   - The line $AE$ passes through points $A(0, 0)$ and $E(2.5, 5)$. The slope of $AE$ is:
     \[
     \text{slope of } AE = \frac{5 - 0}{2.5 - 0} = 2
     \]
   - The equation of the line $AE$ in slope-intercept form is:
     \[
     y = 2x
     \]

3. **Find point $F$ on $AE$ such that $CF = 5$:**
   - Let $F$ have coordinates $(x, 2x)$ since it lies on $AE$.
   - The distance $CF$ is given by:
     \[
     CF = \sqrt{(x - 5)^2 + (2x - 5)^2}
     \]
   - Set $CF = 5$:
     \[
     \sqrt{(x - 5)^2 + (2x - 5)^2} = 5
     \]
   - Square both sides to eliminate the square root:
     \[
     (x - 5)^2 + (2x - 5)^2 = 25
     \]
   - Expand and simplify:
     \[
     (x - 5)^2 = x^2 - 10x + 25
     \]
     \[
     (2x - 5)^2 = 4x^2 - 20x + 25
     \]
     \[
     x^2 - 10x + 25 + 4x^2 - 20x + 25 = 25
     \]
     \[
     5x^2 - 30x + 50 = 25
     \]
     \[
     5x^2 - 30x + 25 = 0
     \]
     \[
     x^2 - 6x + 5 = 0
     \]

4. **Solve the quadratic equation:**
   - Factorize the quadratic equation:
     \[
     (x - 1)(x - 5) = 0
     \]
   - Thus, $x = 1$ or $x = 5$.

5. **Determine the correct $x$ value:**
   - If $x = 5$, then $F$ would be at $(5, 10)$, which is not on the line segment $AE$.
   - Therefore, $x = 1$.

6. **Find $AF$:**
   - The coordinates of $F$ are $(1, 2)$.
   - The distance $AF$ is:
     \[
     AF = \sqrt{(1 - 0)^2 + (2 - 0)^2} = \sqrt{1 + 4} = \sqrt{5}
     \]

The final answer is $\boxed{\sqrt{5}}$.