A spring, placed horizontally on a rough surface is compressed by a block of mass m, placed on the same surface so as to store maximum energy in the spring. If the coefficient of friction between the block and the surface is `mu`, the potential energy stored in the spring is

Options

A. `(mu^(2)m^(2)g^(2))/(2k)`
B. `(2mu m^(2)g^(2))/(k)`
C. `(mu^(2)m^(2)g)/(2k)`
D. `(3mu^(2)mg^(2))/(k)`

Answer

To solve the problem of finding the potential energy stored in a spring that is compressed by a block on a rough surface, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Block:**
- The block of mass \( m \) is resting on a rough surface, and it is compressed against a spring. The forces acting on the block include:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting upward.
- The spring force \( F_s \) acting to the right when the spring is compressed.
- The frictional force \( F_f \) acting to the left opposing the motion.

2. **Establish Vertical Force Balance:**
- Since the block is not moving vertically, we can write the equation for the normal force:
\[ N = mg \]

3. **Determine the Limiting Friction:**
- The limiting frictional force \( F_L \) can be expressed as:
\[ F_L = \mu N = \mu mg \]

4. **Establish Horizontal Force Balance:**
- In the horizontal direction, the spring force must equal the frictional force when the block is on the verge of moving:
\[ F_s = F_f \]
- The spring force can be expressed using Hooke's Law:
\[ F_s = kx_0 \]
- Therefore, we have:
\[ kx_0 = \mu mg \]

5. **Solve for Maximum Compression \( x_0 \):**
- Rearranging the equation gives us the maximum compression of the spring:
\[ x_0 = \frac{\mu mg}{k} \]

6. **Calculate the Potential Energy Stored in the Spring:**
- The potential energy \( PE \) stored in the spring when compressed by \( x_0 \) is given by:
\[ PE = \frac{1}{2} k x_0^2 \]
- Substituting \( x_0 \) into the potential energy formula:
\[ PE = \frac{1}{2} k \left(\frac{\mu mg}{k}\right)^2 \]
- Simplifying this expression:
\[ PE = \frac{1}{2} k \cdot \frac{\mu^2 m^2 g^2}{k^2} = \frac{\mu^2 m^2 g^2}{2k} \]

### Final Answer:
The potential energy stored in the spring is:
\[ PE = \frac{\mu^2 m^2 g^2}{2k} \]