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

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:

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

$A$ metal chain of mass $2 \,kg$ and length $90 \,cm$ hangs over a table with $60 \,cm$ on the table. How much work needs to be done to pull the hanging part of the chain back onto the table (in $\,J$)? (Let $g=10 \,m/s^2$)

For a system of conservative forces,the potential energy is given by the formula $U = ax^2 - bx$,where $a$ and $b$ are constants. The equilibrium position and the equilibrium potential energy will be,respectively:

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