A particle of mass m moves on the x-axis under the influence of a force of attraction towards the origin O given by `F=-(k)/(x^(2))hat(i)`. If the particle starts from rest at x = a. The speed of it will attain to reach at distance x from the origin O will be

Options

A. `sqrt((2k)/(m))[(x-a)/(ax)]^(1//2`
B. `sqrt((2k)/(m))[(a+x)/(ax)]^(1//2`
C. `sqrt((k)/(m))[(ax)/(x-a)]`
D. `sqrt((m)/(2k))[(a-x)/(ax)]^(1//2)`

Explanation

To solve the problem of a particle of mass \( m \) moving under the influence of a force given by \( F = -\frac{k}{x^2} \hat{i} \), we will apply the work-energy principle. The particle starts from rest at \( x = a \) and we want to find its speed when it reaches a distance \( x \) from the origin.

### Step-by-Step Solution:

1. **Identify the Force and Potential Energy**:
The force acting on the particle is given by:
\[ F = -\frac{k}{x^2} \hat{i} \]
This is a central force directed towards the origin. The potential energy \( U(x) \) associated with this force can be found by integrating the force:
\[ U(x) = -\int F \, dx = -\int -\frac{k}{x^2} \, dx = \frac{k}{x} + C \]
We can choose the constant \( C = 0 \) for simplicity, so:
\[ U(x) = \frac{k}{x} \]

2. **Calculate the Initial and Final Potential Energy**:
At the initial position \( x = a \):
\[ U(a) = \frac{k}{a} \]
At the position \( x \):
\[ U(x) = \frac{k}{x} \]

3. **Apply the Conservation of Energy**:
The total mechanical energy at the initial position (where the particle starts from rest) is purely potential:
\[ E_{initial} = U(a) = \frac{k}{a} \]
At position \( x \), the total energy is the sum of kinetic energy \( K \) and potential energy \( U(x) \):
\[ E_{final} = K + U(x) = \frac{1}{2} mv^2 + \frac{k}{x} \]
By conservation of energy, we set the initial energy equal to the final energy:
\[ \frac{k}{a} = \frac{1}{2} mv^2 + \frac{k}{x} \]

4. **Rearranging the Equation**:
Rearranging the equation to solve for \( v^2 \):
\[ \frac{1}{2} mv^2 = \frac{k}{a} - \frac{k}{x} \]
\[ \frac{1}{2} mv^2 = k \left( \frac{1}{a} - \frac{1}{x} \right) \]
\[ v^2 = \frac{2k}{m} \left( \frac{1}{a} - \frac{1}{x} \right) \]

5. **Finding the Speed**:
Taking the square root to find the speed \( v \):
\[ v = \sqrt{\frac{2k}{m} \left( \frac{1}{a} - \frac{1}{x} \right)} \]

### Final Answer:
The speed of the particle when it reaches a distance \( x \) from the origin is given by:
\[ v = \sqrt{\frac{2k}{m} \left( \frac{1}{a} - \frac{1}{x} \right)} \]