Question

The potential energy of a particle of mass $0.1$ kg, moving along the x-axis, is given by $V = 5x(x - 4)J$ where x is in meters. It can be calculated that
(A) The particle is acted upon by a variable force
(B) The minimum potential energy during motion is $- 20J$
(C) The speed of the particle is maximum at $x = 2m$
(D) The period of oscillation of the particle is $\left( {\dfrac{\pi }{5}} \right)$sec

Answer 1

In order to solve this type of problems we can use the relation between potential energy and force
$\vec F = - \dfrac{{dV}}{{dx}}$
And we have used the minima and maxima concept in which $\dfrac{{dV}}{{dx}}$ should be equal to zero.

Complete step by step answer:
First we are calculating force
$\therefore F = - \dfrac{{dV}}{{dx}}$
Given that $V = 5x(x - 4)$
$F = - \dfrac{{d[5x(x - 4)]}}{{dx}}$
$F = - 5\left[ {x\dfrac{{d(x - 4)}}{{dx}} + (x - 4)\dfrac{{dx}}{{dx}}} \right]$
$F = - 5[x(1 - 0) + (x - 4)(1)]$
$F = - 5x - (x - 4)$
$F = - 5x - 5x + 20$
$F = - 10x + 20$ …..(1)
From equation 1 it is clear that force is a function of x. i.e., it is variable.
Hence, option A is correct.
For minimum potential energy
$\dfrac{{dV}}{{dx}}$ should be zero
So, $\dfrac{{dV}}{{dx}} = 0$
$\because F = - \dfrac{{dV}}{{dx}}$
So, $- F = 0$
From equation 1, we get
$- ( - 10x + 20) = 0$
$10x - 20 = 0$
$x = \dfrac{{20}}{{10}}$
$x = 2m$ …..(2)
Hence, at $x = 2m$ potential energy is minimum and when $F = 0$ then $a = 0$ at that time the speed of the particle will be maximum.
Hence at $x = 2m$, the velocity of the particle is maximum.
Hence option C is also correct.
Now, we get that at $x = 2m$ potential energy will be minimum. So, the minimum potential energy can be calculated by substituting the value of $x = 2m$ in potential energy expression given that
$V = 5x(x - 4)$ $[at\,x = 2]$
${V_{\min }} = 5(2)(2 - 4)$
${V_{\min }} = 10( - 2)$
${V_{\min }} = - 20J$ …..(3)
Hence, option B is also correct.
From equation 1, $F = - 10x + 20$
So, $F = ma$
$ma = - 10x + 20$
$a = \dfrac{{ - 10x + 20}}{m}$
Given that $m = 0.1kg$
$a = \dfrac{{ - 10(x - 2)}}{{0.1}}$
$a = - 100(x - 2)$
Let $(x - 2) = x$
So, $a = - 100x$ …..(4)
$\because a = - {w^2}x$ …..(5)
From equation 4 & 5 we get
${w^2} = 100$
$w = \sqrt {100}$ $\Rightarrow w = 10rad/\sec$
Time period is given by
$T = \dfrac{{2\pi }}{w}$
$T = \dfrac{{2\pi }}{{10}}$
$T = \dfrac{\pi }{5}\sec$
So, the period of oscillation of the particle is $\dfrac{\pi }{5}$ sec
So, option D is also correct.
Hence, we conclude the option A, B, C and D all the options are correct.

Note: : In order to solve these type of problems we should remember the maxima & minima concept i.e., $\dfrac{{dy}}{{dx}} = 0$ $($first derivative should be zero$)$
If $\dfrac{{{d^2}y}}{{d{x^2}}} > 0$ $($y is minimum$)$
And If $\dfrac{{{d^2}y}}{{d{x^2}}} < 0$ $($y is maximum$)$