The potential energy versus displacement graph of a system is represented in the figure below. The force action on the system will be represented by

Potential energy function along $x-$axis, in a certain force field, is given as $U\left(x\right)=\frac{x^4}{4}-2x^3+\frac{11}{2}{x}^3-6x$. For the given force field,

$\left(\mathrm{i}\right)$ the points of equilibrium are $x=1, x=2$ and $x=3$.

$\left(\mathrm{ii}\right)$ the point $x=2$ is a point of unstable equilibrium.

$\left(\mathrm{iii}\right)$ the points $x=1$ and $x=3$ are points of stable equilibrium.

$\left(\mathrm{iv}\right)$ there exists no point of neutral equilibrium.

Which of the above statements is correct?

A particle of mass $5 \mathrm{kg}$ moving in the $x\mathit{-}y$ plane has its potential energy given by $U=\left(-7x+24y\right) \mathrm{J}$. The particle is initially at origin and has a velocity $\overrightarrow{u}=\left(14.4\hat{\mathrm{i}}+4.2\hat{\mathrm{j}}\right)$. Then,

The potential energy curve $U$ of a particle as a function of its position is shown. The particle has total mechanical energy $E$ of $3.0 \mathrm{J}$. Which of the following is correct?