$A$ uniform chain of length $2 \ m$ is placed on a table such that $60 \ cm$ of it hangs freely from the edge of the table. The total mass of the chain is $4 \ kg$. What is the work done in pulling the chain completely onto the table (in $J$)? (Take $g = 10 \ m/s^2$)

$A$ rod of length $4 \ m$ and mass $20 \ kg$ is lying horizontally on the ground. Calculate the work done in $J$ to keep it in a vertical position such that one end remains in contact with the ground.

Potential energy of a particle varies with position $r$ as,$U = \left( \frac{\alpha}{r^4} - \frac{\beta}{r^5} \right) \text{ J}$,where $\alpha$ and $\beta$ are positive constants. The particle will be in equilibrium at $r = \dots$

If the potential energy between two molecules is given by $U = -\frac{A}{r^6} + \frac{B}{r^{12}}$,then at equilibrium,the separation between the molecules and the potential energy are:

$A$ particle moves under the influence of a force $F = -2x$ in one dimension (where $x$ is the distance of the particle from the origin). Assume that the potential energy of the particle at the origin is zero. The schematic diagram of the potential energy $U$ as a function of $x$ is given by:

Potential energy as a function of $r$ is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$,where $r$ is the interatomic distance,and $A$ and $B$ are positive constants. The equilibrium distance between the two atoms will be