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 systematically:

### Step 1: Understand the System
We have a uniform chain of length \( L \) and mass \( M \) overhanging a horizontal table. Two-thirds of the chain is on the table, and one-third is hanging off. The coefficient of friction between the chain and the table is \( \mu \).

### Step 2: Determine Mass per Unit Length
The mass per unit length \( \lambda \) of the chain can be calculated as:
\[ \lambda = \frac{M}{L} \]

### Step 3: Calculate the Weight of the Hanging Part
The mass of the hanging part of the chain (length \( \frac{L}{3} \)) is:
\[ m_{\text{hanging}} = \lambda \cdot \frac{L}{3} = \frac{M}{L} \cdot \frac{L}{3} = \frac{M}{3} \]
The weight of the hanging part is:
\[ W_{\text{hanging}} = m_{\text{hanging}} \cdot g = \frac{M}{3} g \]

### Step 4: Determine the Normal Force
The normal force \( N \) acting on the part of the chain on the table is equal to the weight of the chain on the table plus the weight of the hanging part:
\[ N = \lambda \cdot \frac{2L}{3} \cdot g = \frac{M}{L} \cdot \frac{2L}{3} \cdot g = \frac{2Mg}{3} \]

### Step 5: Calculate the Frictional Force
The frictional force \( F \) opposing the motion of the chain is given by:
\[ F = \mu N = \mu \cdot \frac{2Mg}{3} \]

### Step 6: Set Up the Work Done by Friction
As the chain slips off the table, we need to calculate the work done by the friction force. The work done \( W_f \) by the friction during the displacement \( dx \) is:
\[ dW_f = -F \cdot dx = -\mu \cdot \frac{2Mg}{3} \cdot dx \]

### Step 7: Integrate to Find Total Work Done
To find the total work done as the chain slips off the table, we integrate from \( x = 0 \) to \( x = \frac{2L}{3} \):
\[ W_f = \int_0^{\frac{2L}{3}} -\mu \cdot \frac{2Mg}{3} \, dx \]
\[ W_f = -\mu \cdot \frac{2Mg}{3} \cdot \left[ x \right]_0^{\frac{2L}{3}} = -\mu \cdot \frac{2Mg}{3} \cdot \frac{2L}{3} \]
\[ W_f = -\frac{4\mu MgL}{9} \]

### Step 8: Relate to Given Work Done Expression
The problem states that the work done by friction is given by:
\[ W_f = -\frac{2}{k} \mu MgL \]
Setting the two expressions for work done equal to each other:
\[ -\frac{4\mu MgL}{9} = -\frac{2}{k} \mu MgL \]

### Step 9: Solve for \( k \)
We can cancel \( -\mu MgL \) from both sides (assuming \( \mu, M, g, L \) are non-zero):
\[ \frac{4}{9} = \frac{2}{k} \]
Cross-multiplying gives:
\[ 4k = 18 \implies k = \frac{18}{4} = \frac{9}{2} \]
Thus, the value of \( k \) is:
\[ k = 9 \]

### Final Answer
The value of \( k \) is \( 9 \).