$A$ 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 $g = 10 \ m/s^2$)

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity,the work required to pull the hanging part onto the table is:

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 $r = \dots$

Consider a one-dimensional motion of a particle with total energy $E$. There are four regions $A, B, C$ and $D$ in which the relation between potential energy $V$,kinetic energy $K$,and total energy $E$ is as given below:
Region $A: V > E$
Region $B: V < E$
Region $C: K < E$
Region $D: V > E$
State with reason in each case whether a particle can be found in the given region or not.

$A$ 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?

In the figure shown,the potential energy $(U)$ of a particle is plotted against its position $x$ from the origin. The particle at: