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
$k = 10^{-2}\, Nm^2$ 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}mv^2 =\int^{x=0.5m}_{x=1.0m} F(dx) = \int^{0.5}_{1.0} \bigg(\frac{-k}{2x^2}\bigg) (dx)$
$ =\frac{-k}{2} \int^{0.5}_{1.0}\frac{dx}{x^2} $
$ =\frac{k}{2}\bigg(\frac{1}{x}\bigg)^{0.5}_{1.0} $
$ =\bigg(\frac{k}{2}\bigg)\bigg(\frac{1}{0.5}-\frac{1}{1.0}\bigg)=\frac{k}{2} $
$ \therefore v = \pm\sqrt{\frac{k}{m}}$
Substituting the values
$ v = \pm\sqrt{\frac{10^{-2}Nm^2 }{10^{-2}kg}} =\pm1\, m/s $
Therefore, velocity of particle at x = 1.0 m is
$ v = - l.0\, m/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}mv^2 =\int^{x}_{1.0} \frac{-k\, dx}{2x^2} =\frac{k}{2}\bigg(\frac{1}{x}\bigg)^2_{1.0}=\frac{k}{2} \bigg(\frac{1}{x}-1\bigg)$
$\therefore v^2 =\frac{k}{m} \bigg(\frac{1}{x}-1\bigg)$
$\therefore v =\sqrt{\frac{1}{x}-1} = \sqrt{\frac{1-x}{x}} $
$ \frac{k}{m}=\frac{10^{-2} \lambda Nm^2}{10^{-2}\, kg}$
but $ v = -\bigg(\frac{dx}{dt} \bigg)= \sqrt{\frac{1-x}{x}} $
$\therefore \int \sqrt{\frac{x}{1-x}}dx= - \int dt $
$ \int^{0.25}_1 \sqrt{\frac{x}{1-x}}dx= - \int^1_0 dt $
Solving this, we get t = 1.48 s.