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?

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

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:

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 potential energy between two atoms is given by the function $U(r) = a/r^{12} - b/r^{6}$. Find the equilibrium distance between them.

$A$ particle with constant total energy $E$ moves in one dimension in a region where the potential energy is $U(x)$. The speed of the particle is zero where