Question

Along the positive $x-axis$ a particle of mass $2\times 10^{- 2}kg$ is moving under the influence of a force $F\left(\right.x\left.\right)=\frac{- k}{4 x^{2}}$ where, $k=10^{- 2}Nm^{2}$ . Initial conditions at $t=0$ are given as $x=1.5m$ and $v=0$ . If its speed (in $ms^{- 1}$ ) upon reaching $x=0.25m$ is $92\times 10^{- x}$ find $x$ . (Take $\sqrt{30}=5.5$ )

Answer 1

By work-energy thorem,
Change in kinetic energy between $A$ and $B$ $=$ work done by the force between $A$ and $B.$
$\therefore \frac{1}{2}\left(mv\right)^{2}=\displaystyle \int _{x = 1 . 5 m}^{x = 0 . 25 m}F\left(\right.dx\left.\right)$
$=\displaystyle \int _{1 . 5}^{0 . 25}\left(\frac{- k}{4 x^{2}}\right)dx$
$=\frac{- k}{4}\displaystyle \int _{1 . 5}^{0 . 25}\frac{d x}{x^{2}}$
$=\frac{k}{4}\left(\frac{1}{x}\right)_{1 . 5}^{0 . 25}$
$=\frac{k}{4}\left(\frac{1}{0 . 25} - \frac{1}{1 . 5}\right)$
$\therefore \frac{1}{2}mv^{2}=\frac{5 k}{6}$
$\therefore v^{2}=\frac{5 k}{3 m}$
Considering positive root, speed,
$v=\sqrt{\frac{5 k}{3 m}}$
Substituting values,
$v =\sqrt{\frac{5\left(10^{-2}\right)}{3\left(2 \times 10^{-2}\right)}}=\sqrt{\frac{5}{6}}=\frac{\sqrt{30}}{6}=\frac{5.5}{6}$
$\therefore v=0.92m/s=92\times 10^{- 2}m/s$