Question

A force acting on a particle varies with the displacement x as $F=ax-b{x}^2$. Where $a=1N/m$ and $b=1N/m^2$. The work done by this force for the first one meter ($F$ is in newtons, $x$ is in meters) is :

• A. $\frac{1}{6}J$
• B. $\frac{2}{6}J$
• C. $\frac{3}{6}J$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{1}{6}J$

$W={\int }_0^{x}Fdx=\frac{a{x}^2}{2}-\frac{b{x}^3}{3}$
(since a=1; b=1) $W=\frac{x^2}{2}-\frac{x^3}{3}$
For x= 1 m
$W=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}J$