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.

The work done in moving a body of mass $2 \,kg$ to a height of $4 \,m$ from the surface of the earth is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,J$)

$A$ 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:

Calculate the work done in rotating a vertical meter scale of mass $400 \, g$ by an angle of ${60^o}$. (in $, J$)

The potential energy function (in $J$) of a particle in a region of space is given as $U = (2x^2 + 3y^3 + 2z)$. Here $x, y$ and $z$ are in meters. The magnitude of the $x$-component of the force (in $N$) acting on the particle at point $P(1, 2, 3) \ m$ is:

$A$ chain of length $L$ and mass $M$ is placed on a frictionless table with one-fifth of its length hanging over the edge. What is the work required to pull the hanging part back onto the table?