A long slipway!

When the block covers a distance $L$ , it rotates $\dfrac{L}{d}$ rollers to an angular velocity $\omega_{max} = \dfrac{v_{max}}{r}$

Now let us take that $f(t)$ be the kinetic friction acting between the roller and the block. We cannot say whether it has a time dependency or not so that we take it to be dependent. And let's take that it acts for a $\mathrm{d}t$ time.

So,

$\begin{aligned}r\times f(t) \mathrm{d} t &= I \mathrm{d}\omega\\\text{OR }r\times f(t)\Delta t &= I \Delta \omega\\\\\Rightarrow r\int f(t) \mathrm{d}t &= I\omega_{max}\\ \text{And }f(t) &= \dfrac{I}{r}\dfrac{\mathrm{d}\omega}{\mathrm{d}t}\end{aligned}$

Now let us take the loss in energy when the block slips on one of the rollers, since there is kinetic friction. So, loss in Energy is the difference in work done by friction in the block and the roller.

$\begin{aligned}\Rightarrow \mathrm{d}Q &= f(t)\left[v_{max} - r\omega\right]\mathrm{d}t\\ \Rightarrow Q &=r\omega_{max}\int f(t)\mathrm{d}t - r\int f(t)\omega\mathrm{d}t\\ &=r\omega_{max}\int f(t)\mathrm{d}t - r\int\dfrac{I}{r}\omega\mathrm{d}\omega\\ &=I\omega_{max}^2 - I\dfrac{\omega_{max}^2}{2}\\ \therefore Q&=\frac{1}{2}I\omega_{max}^2\end{aligned}$

So heat lost when it covers $L$ distance $=\dfrac{L}{d}Q$

Now we can conserve energy,

$\begin{aligned}MgL\sin\alpha &=\dfrac{L}{d}\left( \frac{1}{2}I\omega_{max}^2 + Q\right)\\ &=\dfrac{L}{d}I\omega_{max}^2\\&=\dfrac{L}{d}\dfrac{m}{2}v_{max}^2 &\left[\because I = \frac{1}{2}mr^2 \text{ and } \omega_{max} = \dfrac{v_{max}}{r}\right] \\\Rightarrow v_{max} &= \sqrt{\dfrac{2dMg\sin\alpha}{m}}\end{aligned}\\\\ \Large \therefore \boxed{v_{max} = \sqrt{\dfrac{2dMg\sin\alpha}{m}}}$

$\begin{aligned}\text{Substituting, } v_{max} &= \sqrt{\dfrac{2\times10\times10\times2\times\sqrt 3}{10\times2}}\\ &=2\sqrt{5\sqrt 3}\\&\approx5.88566...\end{aligned}$

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This solution is not original. It is put to share amazing insight of the problem maker.