If the potential energy of a particle of mass $0.1 \ kg$ moving along $x$-axis is $U(x) = 5x(x-4) \ J$,then the speed of the particle is maximum at a position of

$A$ 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:

The potential energy of a particle is related to the $x$ coordinate by the equation $U = x^2 - 2x$. The particle will be in stable equilibrium at:

The decrease in the potential energy of a ball of mass $20 \,kg$ which falls from a height of $50 \,cm$ is ............ $J$.

The potential energy of a particle in a force field is $U(r) = \frac{A}{r^2} - \frac{B}{r}$,where $A$ and $B$ are positive constants and $r$ is the distance of the particle from the centre of the field. For stable equilibrium,the distance of the particle is:

If a stone is thrown vertically upward and returns to the ground,its potential energy is maximum: