A force `F=-k/x_2(x!=0)` acts on a particle in x-direction. Find the work done by this force in displacing the particle from. `x = + a` to `x = + 2a`. Where, k is a positive constant.

Explanation

To find the work done by the force \( F = -\frac{k}{x^2} \) when displacing a particle from \( x = +a \) to \( x = +2a \), we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) by a force when moving an object from position \( x_1 \) to \( x_2 \) is given by the integral of the force over the displacement:
\[ W = \int_{x_1}^{x_2} F \, dx \]

### Step 2: Set Up the Integral
In this case, the force \( F \) is given as \( F = -\frac{k}{x^2} \). We need to calculate the work done from \( x = +a \) to \( x = +2a \):
\[ W = \int_{a}^{2a} -\frac{k}{x^2} \, dx \]

### Step 3: Factor Out Constants
Since \( k \) is a constant, we can factor it out of the integral:
\[ W = -k \int_{a}^{2a} \frac{1}{x^2} \, dx \]

### Step 4: Integrate the Function
The integral of \( \frac{1}{x^2} \) can be rewritten as \( x^{-2} \). The integral of \( x^{-2} \) is:
\[ \int x^{-2} \, dx = -\frac{1}{x} \]
Thus, we have:
\[ W = -k \left[-\frac{1}{x}\right]_{a}^{2a} \]

### Step 5: Evaluate the Integral
Now, we evaluate the definite integral from \( a \) to \( 2a \):
\[ W = -k \left[-\frac{1}{2a} + \frac{1}{a}\right] \]
This simplifies to:
\[ W = -k \left(-\frac{1}{2a} + \frac{2}{2a}\right) = -k \left(\frac{1}{2a}\right) \]

### Step 6: Simplify the Result
Combining the terms gives us:
\[ W = -k \left(\frac{1}{2a}\right) = -\frac{k}{2a} \]

### Final Answer
Thus, the work done by the force when displacing the particle from \( x = +a \) to \( x = +2a \) is:
\[ W = -\frac{k}{2a} \]

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