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`. Here, k is a positive constant.

Options

A. `-k/2a`
B. `-k/a`
C. `-2k/a`
D. `-a/k`

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 will follow these steps:

### Step 1: Understand the Work Done by a Variable Force
The work done \( W \) by a variable force when moving from position \( x_1 \) to \( x_2 \) is given by the integral:
\[ W = \int_{x_1}^{x_2} F \, dx \]
In our case, \( F = -\frac{k}{x^2} \), \( x_1 = a \), and \( x_2 = 2a \).

### Step 2: Set Up the Integral
Substituting the values into the work integral, we have:
\[ W = \int_{a}^{2a} -\frac{k}{x^2} \, dx \]

### Step 3: Evaluate the Integral
To evaluate the integral, we can rewrite it as:
\[ W = -k \int_{a}^{2a} \frac{1}{x^2} \, dx \]
The integral of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \). Thus, we can evaluate it as follows:
\[ W = -k \left[ -\frac{1}{x} \right]_{a}^{2a} = -k \left( -\frac{1}{2a} + \frac{1}{a} \right) \]

### Step 4: Simplify the Expression
Now, simplifying the expression:
\[ W = -k \left( \frac{1}{a} - \frac{1}{2a} \right) = -k \left( \frac{2}{2a} - \frac{1}{2a} \right) = -k \left( \frac{1}{2a} \right) \]
Thus, we have:
\[ W = -\frac{k}{2a} \]

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

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