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

The correct option is (1/2)Cx₁².

Step-by-step Explanation:

1. Identify the given information:
We are given a force $F$ that varies with position $x$ according to the equation:
$F=Cx$
where $C$ is a constant. The particle moves along the x-axis from an initial position $x_i=0$ to a final position $x_f=x_1$.

2. Formulate the work done by a variable force:
Since the force depends on the position $x$, it is a variable force. The work done $W$ by a variable force along a one-dimensional path from $x_i$ to $x_f$ is given by the definite integral of the force with respect to position:
$W={\int }_{x_i}^{x_f}Fdx$

3. Substitute the values and integrate:
Substituting $F=Cx$, $x_i=0$, and $x_f=x_1$ into the work formula:
$W={\int }_0^{x_1}Cxdx$

Since $C$ is a constant, we can pull it outside the integration:
$W=C{\int }_0^{x_1}xdx$

Using the power rule for integration, $\int xdx=\frac{x^2}{2}$, we get:
$W=C{\left[\frac{x^2}{2}\right]}_0^{x_1}$

4. Evaluate the definite integral limits:
Substitute the upper limit $x_1$ and the lower limit $0$:
$W=C(\frac{x_1^2}{2}-\frac{0^2}{2})$
$W=C\left(\frac{x_1^2}{2}\right)$
$W=\frac{1}{2}C{x}_1^2$

Thus, the work done in the process is $\frac{1}{2}C{x}_1^2$, which matches the option (1/2)Cx₁².