A particle moves under the effect of a force F = Cx from x = 0 to `x = x_1` The work done in the process is

Options

A. `Cx_1^2`
B. `1/2 Cx_(1)^2`
C. `Cx_1`
D. Zero

Explanation

To solve the problem of finding the work done by a force \( F = Cx \) as a particle moves from \( x = 0 \) to \( x = x_1 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Force**: The force acting on the particle is given by \( F = Cx \), where \( C \) is a constant and \( x \) is the position of the particle. This indicates that the force is variable and depends on the position \( x \).

2. **Work Done Definition**: The work done \( dW \) by a force when moving through a small displacement \( dx \) is given by:
\[ dW = F \cdot dx \]
Since \( F \) is a function of \( x \), we can substitute \( F \) into this equation.

3. **Substitute the Force**: Substitute \( F = Cx \) into the work done equation:
\[ dW = Cx \, dx \]

4. **Integrate to Find Total Work Done**: To find the total work done as the particle moves from \( x = 0 \) to \( x = x_1 \), we need to integrate \( dW \):
\[ W = \int_{0}^{x_1} Cx \, dx \]

5. **Perform the Integration**: The integral of \( Cx \) with respect to \( x \) is:
\[ W = C \int_{0}^{x_1} x \, dx = C \left[ \frac{x^2}{2} \right]_{0}^{x_1} \]

6. **Evaluate the Integral**: Now, evaluate the limits:
\[ W = C \left( \frac{x_1^2}{2} - \frac{0^2}{2} \right) = C \left( \frac{x_1^2}{2} \right) \]

7. **Final Result**: Therefore, the work done by the force as the particle moves from \( x = 0 \) to \( x = x_1 \) is:
\[ W = \frac{C x_1^2}{2} \]