$A$ 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?

$A$ 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:

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 ................ $J$.

$A$ 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 a body of mass $200\, g$ falls from a height $200\, m$ and its total $P.E.$ is converted into $K.E.$ at the point of contact of the body with the earth's surface,then what is the decrease in $P.E.$ of the body at the contact? $(g = 10\, m/s^2)$ ............ $J$