Given data

• Length of chain: $L = 3\,\text{m}$
• Mass of chain: $M = 3\,\text{kg}$
• Initially on table: $2\,\text{m}$
• Initially hanging: $1\,\text{m}$
• Table is smooth $\Rightarrow$ no friction
• $g = 10\,\text{m/s}^2$

Mass per unit length

Since the chain is uniform, $\lambda = \frac{M}{L} = \frac{3}{3} = 1\,\text{kg/m}$

Use conservation of mechanical energy

As the chain slips, there is no friction, so loss in gravitational potential energy becomes kinetic energy.

We compare the initial and final positions of the center of mass of the chain.

Initial position of center of mass

Let the tabletop level be the reference level $y=0$, positive downward.

Initially:

• $2\,\text{m}$ of chain is on the table, so its center of mass is at $y=0$.
• $1\,\text{m}$ of chain hangs vertically, so its center of mass is at $y = \frac{1}{2}\,\text{m}$

Masses of these parts:

• On table: $2\,\text{kg}$
• Hanging: $1\,\text{kg}$

Therefore initial center of mass depth is $y_i = \frac{(2)(0) + (1)\left(\frac{1}{2}\right)}{3} = \frac{1}{6}\,\text{m}$

Final position of center of mass

When the chain completely slips off, the whole $3\,\text{m}$ chain hangs vertically.

So the center of mass is at $y_f = \frac{L}{2} = \frac{3}{2}\,\text{m}$

Fall of center of mass

$\Delta y = y_f - y_i = \frac{3}{2} - \frac{1}{6} = \frac{9-1}{6} = \frac{8}{6} = \frac{4}{3}\,\text{m}$

Loss in gravitational potential energy

$\Delta U = Mg\Delta y = 3 \times 10 \times \frac{4}{3} = 40\,\text{J}$

Since there is no friction, this entire loss becomes kinetic energy: $k = 40\,\text{J}$

Final answer

$\boxed{40}$

Comparison with stored correct answer

Stored correct answer = $40$

Derived answer = $40$

Hence, the answers agree.