The only force acting on a block is along x-axis is given by `P=-((4)/(x^(2)+2))N`, When the block moves from x= -2 m to x = 4 m, the change in kinetic energy of block is-

Options

A. Positive
B. Negative
C. Zero
D. May be positive or negative

Answer

To solve the problem, we need to find the change in kinetic energy of the block as it moves from \( x = -2 \, \text{m} \) to \( x = 4 \, \text{m} \) under the influence of the force \( P = -\frac{4}{x^2 + 2} \, \text{N} \). The change in kinetic energy can be determined by calculating the work done by the force over the specified distance.

### Step-by-Step Solution:

1. **Identify the Force**: The force acting on the block is given by:
\[ P = -\frac{4}{x^2 + 2} \, \text{N} \]

2. **Set Up the Work Done Integral**: The work done \( W \) by the force as the block moves from \( x = -2 \, \text{m} \) to \( x = 4 \, \text{m} \) is given by the integral of the force over the displacement:
\[ W = \int_{-2}^{4} P \, dx = \int_{-2}^{4} -\frac{4}{x^2 + 2} \, dx \]

3. **Calculate the Integral**: We can compute the integral:
\[ W = -4 \int_{-2}^{4} \frac{1}{x^2 + 2} \, dx \]
The integral \( \int \frac{1}{x^2 + 2} \, dx \) can be solved using the formula:
\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
where \( a = \sqrt{2} \). Thus,
\[ \int \frac{1}{x^2 + 2} \, dx = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) + C \]

4. **Evaluate the Definite Integral**:
\[ W = -4 \left[ \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \right]_{-2}^{4} \]
Now, we need to evaluate this at the limits:
\[ W = -4 \left( \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{4}{\sqrt{2}} \right) - \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{-2}{\sqrt{2}} \right) \right) \]

5. **Calculate the Values**:
- For \( x = 4 \):
\[ \tan^{-1} \left( \frac{4}{\sqrt{2}} \right) = \tan^{-1} (2\sqrt{2}) \]
- For \( x = -2 \):
\[ \tan^{-1} \left( \frac{-2}{\sqrt{2}} \right) = \tan^{-1} (-\sqrt{2}) \]

6. **Final Calculation**: Substitute the values back into the equation for work done:
\[ W = -4 \left( \frac{1}{\sqrt{2}} \left( \tan^{-1} (2\sqrt{2}) - \tan^{-1} (-\sqrt{2}) \right) \right) \]

7. **Determine Change in Kinetic Energy**: The change in kinetic energy \( \Delta KE \) is equal to the work done:
\[ \Delta KE = W \]

### Conclusion:
Since the force is negative and the integral evaluates to a negative value, the change in kinetic energy will also be negative. Thus, the final answer is that the change in kinetic energy of the block is negative.