Question

The potential energy of a 1kg particle free to move along the X-axis is given by \[V(x) = (\dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2})J\]. If the total mechanical energy of the particle is $2J$, then the maximum speed of the particle is (in $m/s$):
A) $\dfrac{3}{{\sqrt 2 }}$
B) $\sqrt 2 $
C) $\dfrac{1}{{\sqrt 2 }}$
D) $2$

Answer 1

Hint:We know that ${E_T} = {E_k} + {V_{\min }}$ (Where, ${E_T}$ is the total mechanical energy, ${E_k}$ is the kinetic energy, and ${V_{\min }}$ is the minimum potential energy). $\dfrac{1}{2}m{v^2}$ is regarded as the measure of the kinetic energy and for minimum potential energy, $\dfrac{{dV}}{{dx}} = 0$.

Complete step by step solution:
The ability of a body to do work due to its speed, position, or configuration, is called its mechanical energy. Now, in the given question, total mechanical energy is given that ${E_T} = 2J$ (which is fixed).
Now, mechanical energy is of two types. One is kinetic energy (${E_k}$) and the other one is potential energy ($V$). Now, the kinetic energy of a body is defined as the ability of a body to do work due to its speed alone and the potential energy of a body is defined as the ability of a body to do work due to its special position, or configuration. Now, for maximum speed, the kinetic energy will be maximum and therefore the potential energy should minimum.
Now, er are given that $V(x) = (\dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2})$ Differentiating the potential energy with respect to $x$, we get, $\dfrac{{dV}}{{dx}} = \dfrac{{4{x^3}}}{4} - \dfrac{{2x}}{2} = {x^3} - x = x\left( {{x^2} - 1} \right)$
Now, for minimum potential energy, $\dfrac{{dV}}{{dx}} = 0$
$\therefore x({x^2} - 1) = 0$ or, $x = 0, \pm 1$
For, $x = 0$, $V(x) = 0$ For, $x = \pm 1$, $V(x) = - \dfrac{1}{4}J$
Now, we know that ${E_T} = {E_k} + {V_{\min }}$ (Where, ${E_T}$ is the total mechanical energy, ${E_k}$ is the kinetic energy, and ${V_{\min }}$ is the minimum potential energy) Now, if $m$ is the mass of a body, and ${v_m}$ is the maximum speed, then the kinetic energy will be ${E_k} = \dfrac{1}{2}m{v_m}^2$
So, $2 = \dfrac{1}{2}m{v_m}^2 + ( - \dfrac{1}{4})$ or, $\dfrac{1}{2}m{v_m}^2 = 2 + \dfrac{1}{4} = \dfrac{9}{4}$ or, ${v_m}^2 = \dfrac{{9 \times 2}}{{4 \times m}} = \dfrac{9}{{2 \times 1}}$ [The mass of the particle is $1kg$] or, ${v_m} = \dfrac{3}{{\sqrt 2 }}$
So, the maximum speed of the particle is $\dfrac{3}{{\sqrt 2 }}m/s$.

Note:If a body moves with mass $m$ and velocity $v$, and a constant force $F$ is applied against the motion of the body, a retardation $a$ is produced, and after a further displacement $s$, the body comes to rest. So, work done against the force, until the body stops = $Fs = mas$ As the final velocity of the body is zero, $0 = {v^2} - 2as$ or, $mas = \dfrac{{m{v^2}}}{2} = \dfrac{1}{2}m{v^2}$ This expression, $\dfrac{1}{2}m{v^2}$ is regarded as the measure of the kinetic energy.