Two blocks are connected to an ideal spring of stiffness `200 N//m`. At a certain moment, the two block are moving in opposite diretion with speeds `4 ms_(1)` and ' 6 ms_(1) ' the instantaneous elongation of the spring `10 cm`, The rate at which the spring energy `((kx^(2)/(2))` is increasing is: a) 500 J/s b) 400 J/s c) 200 J/s d) 100 J/s

Options

A. `500 J//s`
B. `400 J//s`
C. `200 J//s`
D. `100 J//s`.

Explanation

To solve the problem, we need to find the rate at which the spring energy is increasing when two blocks are connected to an ideal spring. The spring's stiffness is given as \( k = 200 \, \text{N/m} \), the elongation of the spring is \( x = 10 \, \text{cm} = 0.1 \, \text{m} \), and the speeds of the two blocks are \( v_1 = 4 \, \text{m/s} \) and \( v_2 = 6 \, \text{m/s} \).

### Step-by-Step Solution:

1. **Understanding the Energy in the Spring:**
The potential energy stored in the spring is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the elongation of the spring.

2. **Calculating the Potential Energy:**
Plugging in the values:
\[ U = \frac{1}{2} \times 200 \, \text{N/m} \times (0.1 \, \text{m})^2 = \frac{1}{2} \times 200 \times 0.01 = 1 \, \text{J} \]

3. **Finding the Rate of Change of Potential Energy:**
The rate at which the potential energy is increasing can be found using the relationship between the change in potential energy and the change in kinetic energy of the blocks. The increase in potential energy per unit time is equal to the decrease in kinetic energy of both blocks.

4. **Calculating Kinetic Energy of Each Block:**
The kinetic energy of each block is given by:
\[ KE = \frac{1}{2} m v^2 \]
Since we do not have the masses of the blocks, we will denote the mass of the first block as \( m_1 \) and the second block as \( m_2 \).

5. **Finding the Rate of Change of Kinetic Energy:**
The rate of change of kinetic energy for each block is:
\[ \frac{d(KE_1)}{dt} = \frac{1}{2} m_1 \frac{d(v_1^2)}{dt} = m_1 v_1 \frac{dv_1}{dt} \]
\[ \frac{d(KE_2)}{dt} = \frac{1}{2} m_2 \frac{d(v_2^2)}{dt} = m_2 v_2 \frac{dv_2}{dt} \]

6. **Using Newton's Second Law:**
According to Newton's second law, the force exerted by the spring on each block is:
\[ F = kx \]
Therefore, the force on each block is equal to the spring force:
\[ F_1 = -k x \quad \text{and} \quad F_2 = k x \]

7. **Setting Up the Equation:**
The increase in potential energy per unit time is equal to the decrease in kinetic energy:
\[ \frac{dU}{dt} = -\left( \frac{d(KE_1)}{dt} + \frac{d(KE_2)}{dt} \right) \]
We can express this in terms of the velocities and spring force.

8. **Calculating the Rate of Change of Spring Energy:**
The rate at which the spring energy is increasing is given by:
\[ \frac{dU}{dt} = k x \left( v_1 + v_2 \right) \]
Substituting the values:
\[ \frac{dU}{dt} = 200 \, \text{N/m} \times 0.1 \, \text{m} \times (4 \, \text{m/s} + 6 \, \text{m/s}) = 200 \times 0.1 \times 10 = 200 \, \text{J/s} \]

9. **Final Answer:**
The rate at which the spring energy is increasing is \( 200 \, \text{J/s} \).

### Conclusion:
The correct answer is option **c) 200 J/s**.