$A$ uniform chain of mass $M$ and length $L$ is lying on a smooth horizontal table,with half of its length hanging down. The work done in pulling the entire chain up onto the table is

$A$ particle is placed at the origin and a force $F = kx$ is acting on it (where $k$ is a positive constant). If $U(0) = 0$,the graph of $U(x)$ versus $x$ will be (where $U$ is the potential energy function):

If an object of mass $20 \ g$ falls from a height of $200 \ m$,and at the point where the object touches the Earth's surface,its total potential energy is converted into kinetic energy,then what is the decrease in the potential energy of the object in $J$?

The potential energy of a particle oscillating along the $x-$axis is given by $U = 20 + (x - 2)^2$,where $U$ is in joules and $x$ is in meters. The total mechanical energy of the particle is $36\,J$. The maximum kinetic energy of the particle is ................ $J$.

$A$ stretched rubber band has:

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 $r = \dots$