The potential energy function of a particle in a region of space is given as $U=(2x^2+3y^3+2z) \mathrm{J}.$ Here $x\text{,} y$ and $z$ are in metres. Find the force acting on the particle at point $P\mathit{ }(1 \mathrm{m}, 2 \mathrm{m}, 3 \mathrm{m}).$

Force acting on a particle in a conservative force field is:

$\text{(}\mathrm{i}\text{)} \overrightarrow{F}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}$    $\text{(}\mathrm{ii}\text{)} \overrightarrow{F}=2x\hat{\mathrm{i}}+2y\hat{\mathrm{j}}$   $\text{(}\mathrm{iii}\text{)} \overrightarrow{F}=y\hat{\mathrm{i}}+x\hat{\mathrm{j}}$

Find the potential energy function, if it is zero at the origin.

The potential energy function for a particle executing linear simple harmonic motion is given by $U\left(x\right)=\frac{1}{2}k{x}^2 \mathrm{J},$ where $k$ is the force constant. For $k=\mathrm{0.5 }\mathrm{N} {\mathrm{m}}^{–1},$ the graph of $U\left(\mathrm{x}\right)$ verses $x$ is shown in the figure. Show that a particle of total energy $1 \mathrm{J}$ moving under this potential ‘turns back’ when it reaches $x=\pm 2 \mathrm{m}.$