Physics - W.E.P. Question with Solution | TestHub

Total energy ${\mathrm{E}}_{\mathrm{T}}= 2 \mathrm{J}$ It is fixed. For maximum speed, kinetic energy is maximum The potential energy should, therefore, be minimum

$\because \mathrm{V}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^4}{4}-\frac{{\mathrm{x}}^2}{2}$  

or $\frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{x}}=\frac{4{\mathrm{x}}^3}{4}-\frac{2\mathrm{x}}{2}=\mathrm{x}({\mathrm{x}}^2-1)$

For $\mathrm{V}$ to be minimum,$\frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{x}}=0$  

$\therefore \mathrm{x}({\mathrm{x}}^2-1)=0\mathrm{,}\mathrm{or} \mathrm{x}=0\mathrm{,}\pm 1$  

$$at $\mathrm{x}=0$, $\mathrm{V}\left(\mathrm{x}\right)=0$

At $\mathrm{x}=\pm 1\mathrm{,}\mathrm{V}\left(\mathrm{x}\right)=-\frac{1}{4} \text{J}$  

$\therefore (\text{Kinetic energy}{)}_{\text{max}}=\mathrm{E}{}_{\mathrm{T}}-\mathrm{V}{}_{\text{min}}$  

or $\therefore (\text{Kinetic energy}{)}_{\text{max}}=2+\frac{1}{4}=\frac{9}{4} \text{J}$ 

or $\frac{1}{2}\mathrm{m}{\mathrm{\nu }}_{\mathrm{m}}^2=\frac{9}{4}$ 

${\mathrm{\nu }}_{\text{m}}^2=\frac{9\times 2}{\text{m}\times 4}$

or ${\mathrm{\nu }}_{\mathrm{m}}^2=\frac{9\times 2}{\mathrm{m}\times 4}=\frac{9\times 2}{2\times 4}$

$\therefore {\mathrm{\nu }}_{\mathrm{m}}=1.5 \mathrm{m} {\mathrm{s}}^{-1}$