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

$A$ 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):

$A$ number of identical cubical blocks of edge $a$ and mass $m$ are lying on a horizontal table. The work done on the blocks in arranging them in a column of height $(n + 1)a$ on the table is

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:

Consider a one-dimensional motion of a particle with total energy $E$. There are four regions $A, B, C$ and $D$ in which the relation between potential energy $V$,kinetic energy $K$,and total energy $E$ is as given below:
Region $A: V > E$
Region $B: V < E$
Region $C: K < E$
Region $D: V > E$
State with reason in each case whether a particle can be found in the given region or not.

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