# 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

$\begin{array}{1 1}Cx_1^2\\\large\frac{1}{2}\normalsize Cx_1^2\\Cx_1\\\text{Zero}\end{array}$

Answer : $\large\frac{1}{2}$$Cx_1^2 Work done = \int\limits_{x_1}^{x_2} Fdx \qquad\qquad=\int\limits_{0}^{x_1} Cxdx \qquad\qquad=C\big[\large\frac{x^2}{2}\big]_{0}^{x_1} \qquad\qquad=\large\frac{1}{2}$$ Cx_1^2$
edited Sep 2, 2014