A sliding rope!

Consider the change in gravitational potential energy of the system. We can think of a length of rope $x$ being taken from the left and transferred to the right (starting with the initial configuration).

The position of the rope segment's center of mass is originally $\large{+\frac{x}{2}}$ , and is $\large{-\frac{x}{2}}$ after being transferred. The overall change in height of the segment's center of mass is therefore $-x$ .

The mass of the segment is $\large{\frac{x}{L} M}$ , where $L$ is the total length and $M$ is the total mass.

The decrease in gravitational potential energy relative to the start position is:

$\large{\Delta U =\frac{x}{L} M g x = \frac{Mg}{L}x^2}$

Equate the change in gravitational potential energy to the kinetic energy:

$\large{\frac{Mg}{L}x^2 = \frac{1}{2}M v^2 \\ v^2 = \frac{2g}{L}x^2 = \frac{2(10)}{5}x^2 = 4x^2 \implies \boxed{v = 2x} }$

To work this question around, you must consider this is a normal Atwood system. That said, we must write the Second Law for each sides of the rope. $\begin{aligned} F_{net1} = T - m_1 g &= m_1 a \\ F_{net2} = m_2 g - T &= m_2 a \end{aligned}$ Where the first equation represents the right side of the rope (going up), and the second one represents the left side. Now to solve it, we need to find a relationship between $m_1$ and $m_2$ . It should be noted that, as the rope slides, $m_1$ and $m_2$ change in time.

Considering the fact that the rope is uniform we can assume that $\mu = \dfrac{M}{2L}$ Where $M$ is the mass of the rope, and $2L$ is the length, in this case, $5m$ .

Now, we have: $\begin{aligned} I) \quad m_1(t) &= \mu \left(L - x \right) \\ II) \quad m_2(t) &= \mu \left(L + x \right) \end{aligned}$

Solving the Newton Laws by Adding up: $\begin{aligned} &I) + II): \\ &\left( m_2 - m_1 \right)g = \left(m_1 + m_2\right) a \rightarrow \\ \rightarrow &\left[ \mu \left(L + x \right) - \mu \left(L - x \right) \right]g = \left[ \mu \left(L + x \right) + \mu \left(L - x \right) \right]a \rightarrow \\ \rightarrow &2\mu x g = 2 \mu L a \rightarrow a= \frac{xg}{L} \rightarrow \\ &\text{Rewriting \$a\$ as } \dfrac{v \mathrm{d} v}{\mathrm{d}x} \\ \rightarrow &\dfrac{v \mathrm{d} v}{\mathrm{d}x} = \frac{xg}{L} \rightarrow v \mathrm{d} v = x \mathrm{d} x \left(\frac{g}{L}\right) \rightarrow \\ \rightarrow &\int_0^v{v \mathrm{d} v} = \left(\frac{g}{L}\right) \int_0^x x \mathrm{d} x \rightarrow v^2 = \frac{gx^2}{L} \rightarrow \\ \rightarrow &\boxed{v = x \sqrt{\frac{g}{L}}} \end{aligned}$ If you didn't understood step 7, be sure to checkout Velocity as a derivative here at Brilliant.