A uniform chain having length L and mass M overhangs a horizontal...

Let $x$ be the length of the chain on the table at any instant.

The mass of the chain on the table is $m = \dfrac{M}{L}x$.

The normal reaction on the chain is $N = mg = \dfrac{M}{L}xg$.

The frictional force acting on the chain is $f = \mu N = \mu \dfrac{M}{L}xg$.

When the chain slips by a small distance $ds$, the length of the chain on the table changes by $dx = -ds$.

The work done by friction during this small displacement is $dW = -f ds = f dx$.

Total work done by friction as the chain slips completely off the table is obtained by integrating from $x = \dfrac{2L}{3}$ to $x = 0$:

$W = \int_{\dfrac{2L}{3}}^{0} \mu \dfrac{M}{L}gx dx$

$W = \mu \dfrac{M}{L}g \left[ \dfrac{x^2}{2} \right]_{\dfrac{2L}{3}}^{0}$

W = \mu \dfrac{M}{L}g \left( 0 -\dfrac{1}{2} \left( \dfrac{2L}{3} \right)^2 \right)$<br><br>$W = $-\dfrac{2\mu MgL}{9}$