Force acting on a block moving along x-axis is given by
`F = - ((4)/(x^(2)+2))N`
The block is displaced from `x = - 2m` to `x = + 4m`, the work done will be

Options

A. positive
B. negative
C. zero
D. may be positive or negative

Answer

To find the work done on the block as it moves from \( x = -2 \, \text{m} \) to \( x = 4 \, \text{m} \) under the influence of the force \( F = -\frac{4}{x^2 + 2} \, \text{N} \), we will follow these steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) by a variable force along a straight path is given by the integral of the force over the displacement:
\[ W = \int_{x_1}^{x_2} F \, dx \]
where \( x_1 \) is the initial position and \( x_2 \) is the final position.

### Step 2: Set Up the Integral
In this case, we need to calculate:
\[ W = \int_{-2}^{4} -\frac{4}{x^2 + 2} \, dx \]

### Step 3: Evaluate the Integral
We can simplify the integral:
\[ W = -4 \int_{-2}^{4} \frac{1}{x^2 + 2} \, dx \]
The integral \( \int \frac{1}{x^2 + a^2} \, dx \) is known to be \( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \). Here, \( a^2 = 2 \) implies \( a = \sqrt{2} \).

Thus, we have:
\[ \int \frac{1}{x^2 + 2} \, dx = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) + C \]

### Step 4: Calculate the Definite Integral
Now we need to evaluate:
\[ W = -4 \left[ \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \right]_{-2}^{4} \]

Calculating the limits:
1. For \( x = 4 \):
\[ \tan^{-1} \left( \frac{4}{\sqrt{2}} \right) = \tan^{-1} \left( 2\sqrt{2} \right) \]

2. For \( x = -2 \):
\[ \tan^{-1} \left( \frac{-2}{\sqrt{2}} \right) = \tan^{-1} \left( -\sqrt{2} \right) \]

### Step 5: Substitute the Limits
Now substituting the limits into the integral:
\[ W = -4 \left[ \frac{1}{\sqrt{2}} \left( \tan^{-1} \left( 2\sqrt{2} \right) - \tan^{-1} \left( -\sqrt{2} \right) \right) \right] \]

### Step 6: Simplify the Expression
Using the property of the arctangent function:
\[ \tan^{-1}(-x) = -\tan^{-1}(x) \]
Thus:
\[ W = -4 \left[ \frac{1}{\sqrt{2}} \left( \tan^{-1} \left( 2\sqrt{2} \right) + \tan^{-1} \left( \sqrt{2} \right) \right) \right] \]

### Step 7: Final Calculation
Now we can compute the numerical values of \( \tan^{-1} \left( 2\sqrt{2} \right) \) and \( \tan^{-1} \left( \sqrt{2} \right) \) using a calculator and substitute them back into the equation to find the work done.

### Conclusion
The work done \( W \) will be negative since the force is acting opposite to the direction of displacement.