The variation of potential energy $U$ of a body moving along the $x$-axis with its position $(x)$ is shown in the figure. The body is in an equilibrium state at:

$A$ uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R$ $(R > l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is

The $PE$ of a $2 \, kg$ particle,free to move along the $x$-axis,is given by $V(x) = \left( \frac{x^3}{3} - \frac{x^2}{2} \right) \, J$. The total mechanical energy of the particle is $4 \, J$. The maximum speed (in $m \, s^{-1}$) 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:

$A$ mass of $M \ kg$ is suspended by a weightless string of length $\ell$. The horizontal force required to displace it until the string makes an angle of $45^\circ$ with the initial vertical direction 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? $(g = 10\, m/s^2)$ ............ $J$