$A$ 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

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? $(g = 10\, m/s^2)$ ............ $J$

The potential energy of a particle of mass $5 \ kg$ moving in the $XY$ plane is given by $V = -7x + 24y$ joules,where $x$ and $y$ are in metres. Initially at $t = 0$,the particle is at the origin $(0, 0)$ moving with a velocity of $\vec{v}_0 = 6[0.24 \hat{i} + 0.7 \hat{j}] \ m/s = [1.44 \hat{i} + 4.2 \hat{j}] \ m/s$. Then:

The potential energy of a body is given by $U = A - Bx^2$ (where $x$ is the displacement). The magnitude of the force acting on the particle is:

$A$ 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:

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: