A uniform chain of length L and mass M overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is µ. The work done by the friction during the period the chain slips off the table is `[-2/k mu M gL]`. Find the value of k.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the mass per unit length
Let the mass per unit length of the chain be denoted as \( \lambda \). Since the total mass of the chain is \( M \) and its total length is \( L \), we have:
\[ \lambda = \frac{M}{L} \]

### Step 2: Determine the initial mass on the table
Given that two-thirds of the chain is on the table, the length of the chain on the table is:
\[ L_{table} = \frac{2}{3}L \]
The mass of the chain on the table is:
\[ M_{table} = \lambda \cdot L_{table} = \lambda \cdot \frac{2}{3}L = \frac{2M}{3} \]

### Step 3: Calculate the normal force
The normal force \( N \) acting on the chain on the table is equal to the weight of the chain on the table:
\[ N = M_{table} \cdot g = \frac{2M}{3} \cdot g \]

### Step 4: Determine the frictional force
The frictional force \( F \) acting on the chain is given by:
\[ F = \mu \cdot N = \mu \cdot \frac{2M}{3} \cdot g = \frac{2\mu Mg}{3} \]

### Step 5: Set up the work done by friction
As the chain slips off the table, let \( x \) be the length of the chain that has slipped off the table. The length remaining on the table is:
\[ L_{remaining} = \frac{2}{3}L - x \]
The mass remaining on the table is:
\[ M_{remaining} = \lambda \cdot L_{remaining} = \frac{M}{L} \cdot \left( \frac{2}{3}L - x \right) = \frac{M}{L} \cdot \left( \frac{2}{3}L - x \right) \]

### Step 6: Calculate the new normal force
The new normal force \( N' \) as the chain slips is:
\[ N' = M_{remaining} \cdot g = \frac{M}{L} \cdot \left( \frac{2}{3}L - x \right) \cdot g \]

### Step 7: Calculate the work done by friction
The work done by friction \( W \) as the chain slips off the table can be expressed as:
\[ W = -F \cdot dx = -\mu N' \cdot dx \]
Substituting \( N' \):
\[ W = -\mu \cdot \left( \frac{M}{L} \cdot \left( \frac{2}{3}L - x \right) \cdot g \right) \cdot dx \]

### Step 8: Integrate to find total work done
Integrating from \( x = 0 \) to \( x = \frac{2}{3}L \):
\[ W = -\mu g \cdot \frac{M}{L} \int_0^{\frac{2}{3}L} \left( \frac{2}{3}L - x \right) \, dx \]
Calculating the integral:
\[ \int \left( \frac{2}{3}L - x \right) \, dx = \frac{2}{3}Lx - \frac{x^2}{2} \]
Evaluating from \( 0 \) to \( \frac{2}{3}L \):
\[ = \left( \frac{2}{3}L \cdot \frac{2}{3}L - \frac{(\frac{2}{3}L)^2}{2} \right) - 0 = \frac{4}{9}L^2 - \frac{2}{9}L^2 = \frac{2}{9}L^2 \]
Thus,
\[ W = -\mu g \cdot \frac{M}{L} \cdot \frac{2}{9}L^2 = -\frac{2\mu MgL}{9} \]

### Step 9: Relate to the given expression
The work done by friction is given as:
\[ W = -\frac{2}{k} \mu MgL \]
Setting the two expressions for work equal:
\[ -\frac{2\mu MgL}{9} = -\frac{2}{k} \mu MgL \]
Cancelling the common terms:
\[ \frac{2}{9} = \frac{2}{k} \]
Thus, solving for \( k \):
\[ k = 9 \]

### Final Answer
The value of \( k \) is:
\[ \boxed{9} \]