The potential energy of a particle of mass $1 \ kg$ moving freely along the $x$-axis is given by $V(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. If the total mechanical energy of the particle is $2 \ J$,what is its maximum speed in $m/s$?

The potential energy of a particle is given by $U = A - Bx^2$,where $x$ is the displacement. What is the magnitude of the force acting on the particle?

The following figure shows the variation of potential energy $V(x)$ of a particle with distance $x$. The particle has

The potential energy of a particle oscillating along the $x-$axis is given by $U = 20 + (x - 2)^2$,where $U$ is in joules and $x$ is in meters. The total mechanical energy of the particle is $36\,J$. The maximum kinetic energy of the particle is ................ $J$.

$A$ particle has a total mechanical energy that is small and negative. It is under the influence of a one-dimensional potential $U(x) = \frac{x^4}{4} - \frac{x^2}{2} \, J$,where $x$ is in meters. At time $t = 0 \, s$,it is at $x = -0.5 \, m$. Then,at a later time,it can be found:

If the potential energy of a particle is given by $U = A - Bx^2$,then the force is proportional to which of the following?