Two blocks each of mass m are joined together using an ideal spring of force constant K and natural length `l_(0)` . The blocks are touching each other when the system is released from rest on a rough horizontal surface. Both the blocks come to rest simultaneously when the extension in the spring is `(l_(0))/4`. The coefficient of friction between each block and the surface (assuming it to be same between any of the blocks and the surface) is :

Options

A. `(Kl_0)/(40mg)`
B. `(Kl_0)/(8mg)`
C. `(3Kl_0)/(8mg)`
D. `(17Kl_0)/(20mg)`

Answer

To solve the problem, we will use the work-energy theorem and the properties of springs. Here’s a step-by-step solution:

### Step 1: Understand the Initial and Final Conditions
Initially, the two blocks of mass \( m \) are touching each other, and the spring is at its natural length \( l_0 \). When the system is released, the spring extends by \( \frac{l_0}{4} \).

### Step 2: Calculate the Total Extension of the Spring
The total extension in the spring when the blocks come to rest is given by:
\[ \text{Total extension} = l_0 + \frac{l_0}{4} = \frac{5l_0}{4} \]

### Step 3: Apply the Work-Energy Theorem
According to the work-energy theorem, the work done by friction is equal to the change in potential energy of the spring.

- **Work done by friction**:
\[ W_{\text{friction}} = -\mu mg \left(\frac{5l_0}{4}\right) \]

- **Potential energy of the spring**:
The initial potential energy of the spring when compressed is:
\[ PE_{\text{initial}} = -\frac{1}{2} k l_0^2 \]
The potential energy when extended is:
\[ PE_{\text{final}} = \frac{1}{2} k \left(\frac{l_0}{4}\right)^2 = \frac{1}{2} k \frac{l_0^2}{16} = \frac{k l_0^2}{32} \]

### Step 4: Set Up the Equation
Using the work-energy theorem:
\[ -\mu mg \left(\frac{5l_0}{4}\right) = \left(-\frac{1}{2} k l_0^2 + \frac{k l_0^2}{32}\right) \]

### Step 5: Simplify the Right Side
Combine the potential energies:
\[ -\frac{1}{2} k l_0^2 + \frac{k l_0^2}{32} = -\frac{16k l_0^2}{32} + \frac{k l_0^2}{32} = -\frac{15k l_0^2}{32} \]

### Step 6: Substitute Back into the Equation
Now we can substitute this back into our equation:
\[ -\mu mg \left(\frac{5l_0}{4}\right) = -\frac{15k l_0^2}{32} \]

### Step 7: Solve for the Coefficient of Friction \( \mu \)
Rearranging gives:
\[ \mu mg \left(\frac{5l_0}{4}\right) = \frac{15k l_0^2}{32} \]
\[ \mu = \frac{15k l_0}{32mg} \cdot \frac{4}{5} \]
\[ \mu = \frac{12k l_0}{40mg} = \frac{3k l_0}{10mg} \]

### Final Answer
Thus, the coefficient of friction \( \mu \) is:
\[ \mu = \frac{3k l_0}{10mg} \]

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