A block is constrained to move along x-axis under a forc `F=4/(x^(2))(xne0)`. Here, F is in newton and x in metre. Find the work done by this force when the block is displaced from `x=4` m to `x=2m`.

Explanation

To find the work done by the force \( F = \frac{4}{x^2} \) when the block moves from \( x = 4 \, \text{m} \) to \( x = 2 \, \text{m} \), we will use the work integral for a variable force.

### Step-by-Step Solution:

1. **Identify the Work Done Formula**:
The work done \( W \) by a variable force is given by the integral:
\[ W = \int_{x_1}^{x_2} F \, dx \]
where \( F \) is the force and \( x_1 \) and \( x_2 \) are the initial and final positions, respectively.

2. **Substitute the Force Function**:
In this case, the force is given as \( F = \frac{4}{x^2} \). So, we substitute this into the work integral:
\[ W = \int_{4}^{2} \frac{4}{x^2} \, dx \]

3. **Factor Out Constants**:
We can factor out the constant \( 4 \) from the integral:
\[ W = 4 \int_{4}^{2} \frac{1}{x^2} \, dx \]

4. **Integrate the Function**:
The integral of \( \frac{1}{x^2} \) is:
\[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} \]
Therefore, we can now evaluate the definite integral:
\[ W = 4 \left[ -\frac{1}{x} \right]_{4}^{2} \]

5. **Evaluate the Limits**:
Now we substitute the limits into the integral:
\[ W = 4 \left( -\frac{1}{2} - \left(-\frac{1}{4}\right) \right) \]
Simplifying this gives:
\[ W = 4 \left( -\frac{1}{2} + \frac{1}{4} \right) = 4 \left( -\frac{2}{4} + \frac{1}{4} \right) = 4 \left( -\frac{1}{4} \right) = -1 \, \text{J} \]

6. **Final Result**:
Thus, the work done by the force when the block is displaced from \( x = 4 \, \text{m} \) to \( x = 2 \, \text{m} \) is:
\[ W = -1 \, \text{J} \]