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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} $

1 Answer

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$
answered Aug 28, 2014 by sreemathi.v
edited Sep 2, 2014 by sreemathi.v

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