If the potential energy of a particle of mass $0.1 \ kg$ moving along $x$-axis is $U(x) = 5x(x-4) \ J$,then the speed of the particle is maximum at a position of

$A$ 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 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 $(U)$ of a diatomic molecule is a function dependent on $r$ (interatomic distance) as $U = \frac{\alpha}{r^{10}} - \frac{\beta}{r^5} - 3$,where $\alpha$ and $\beta$ are positive constants. The equilibrium distance between two atoms will be $\left(\frac{2\alpha}{\beta}\right)^{\frac{a}{b}}$,where $a = \dots \dots \dots \dots$

$A$ particle of mass $1 \ kg$ is free to move along the $x$-axis. Its potential energy is given by $U(x) = (\frac{x^2}{2} - x) \ J$. If the total mechanical energy of the particle is $2 \ J$,find the maximum speed of the particle.

$A$ stick of mass $m$ and length $l$ is pivoted at one end and is displaced through an angle $\theta$. The increase in potential energy is ............