Question

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

Answer 1

Answer: 2

By work-energy thorem,
Change in kinetic energy between $A$ and $B$ $=$ work done by the force between $A$ and $B.$
$\therefore \frac{1}{2}{\left(mv\right)}^2={\int }_{x=1.5m}^{x=0.25m}F(dx)$
$={\int }_{1.5}^{0.25}\left(\frac{-k}{4x^2}\right)dx$
$=\frac{-k}{4}{\int }_{1.5}^{0.25}\frac{dx}{x^2}$
$=\frac{k}{4}{\left(\frac{1}{x}\right)}_{1.5}^{0.25}$
$=\frac{k}{4}\left(\frac{1}{0.25}-\frac{1}{1.5}\right)$
$\therefore \frac{1}{2}m{v}^2=\frac{5k}{6}$
$\therefore v^2=\frac{5k}{3m}$
Considering positive root, speed,
$v=\sqrt{\frac{5k}{3m}}$
Substituting values,
$v=\sqrt{\frac{5\left(10^{-2}\right)}{3\left(2\times 10^{-2}\right)}}=\sqrt{\frac{5}{6}}=\frac{\sqrt{30}}{6}=\frac{5.5}{6}$
$\therefore v=0.92m/s=92\times 10^{-2}m/s$