Question

A particle of mass $10^{-2}kg$ is moving along the positive
x-axis under the influence of a force $F(x)=-k/2x^2$ where
$k=10^{-2}N{m}^2$ At time t = 0, it is at x = 1.0 mand its velocity
v = 0 .
(a) Find its velocity when it reaches x = 0.5 m.
(b) Find the time at which it reaches x = 0.25 m.

Answer 1

$F(x)=\frac{-k}{2x^2},k$ and $x^2$ both are positive. Hence, F(x) is
always negative.
Applying work-energy theorem between points A and B.
Change in kinetic energy between A and B = work done by
the force between A and B
$\therefore \frac{1}{2}m{v}^2={\int }_{x=1.0m}^{x=0.5m}F(dx)={\int }_{1.0}^{0.5}\left(\frac{-k}{2x^2}\right)(dx)$
$=\frac{-k}{2}{\int }_{1.0}^{0.5}\frac{dx}{x^2}$
$=\frac{k}{2}(\frac{1}{x}{)}_{1.0}^{0.5}$
$=\left(\frac{k}{2}\right)(\frac{1}{0.5}-\frac{1}{1.0})=\frac{k}{2}$
$\therefore v=\pm \sqrt{\frac{k}{m}}$
Substituting the values
$v=\pm \sqrt{\frac{10^{-2}N{m}^2}{10^{-2}kg}}=\pm 1m/s$
Therefore, velocity of particle at x = 1.0 m is
$v=-l.0m/s$
Negative sign indicates that velocity is in negative
r-direction.
(b) Applying work-energy theorem between any
intermediate value x = x, we get
$\frac{1}{2}m{v}^2={\int }_{1.0}^x\frac{-kdx}{2x^2}=\frac{k}{2}(\frac{1}{x}{)}_{1.0}^2=\frac{k}{2}(\frac{1}{x}-1)$
$\therefore v^2=\frac{k}{m}(\frac{1}{x}-1)$
$\therefore v=\sqrt{\frac{1}{x}-1}=\sqrt{\frac{1-x}{x}}$
$\frac{k}{m}=\frac{10^{-2}\lambda N{m}^2}{10^{-2}kg}$
but $v=-\left(\frac{dx}{dt}\right)=\sqrt{\frac{1-x}{x}}$
$\therefore \int \sqrt{\frac{x}{1-x}}dx=-\int dt$
${\int }_1^{0.25}\sqrt{\frac{x}{1-x}}dx=-{\int }_0^{1}dt$
Solving this, we get t = 1.48 s.