If a stone is thrown vertically upward and returns to the ground,its potential energy is maximum:

$A$ particle which is constrained to move along the $x-$axis is subjected to a force in the same direction which varies with the distance $x$ of the particle from the origin as $F(x) = -kx + ax^3$. Here $k$ and $a$ are positive constants. For $x \ge 0$,the functional form of the potential energy $U(x)$ of the particle is

The following figure shows the variation of potential energy $V(x)$ of a particle with distance $x$. The particle has

While a person climbs stairs,the gravitational potential energy of the person increases. The source of this energy is

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 $a = \dots \dots \dots \dots$

When a stone is thrown vertically upwards and returns to the ground,at what point will its potential energy be maximum?