The potential energy of a 1 kg particle free to move along the x-axis is given by |
\[V(x)=\left( \frac{{{x}^{4}}}{4}-\frac{{{x}^{2}}}{2} \right)J\] |
The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is [AIEEE 2006] |
A) \[3/\sqrt{2}\]
B) \[\sqrt{2}\]
C) \[1/\sqrt{2}\]
D) 2
Correct Answer: A
Solution :
[a] The variation of potential energy is given as \[V(x)=\left( \frac{{{x}^{4}}}{4}-\frac{{{x}^{2}}}{2} \right)\] ..(i) |
For minimum value of V, \[\frac{dv}{dx}=0\] |
On differentiating Eq. (i), we get |
\[\frac{4{{x}^{3}}}{4}-\frac{2x}{2}=0\] \[\Rightarrow \] \[x=0,\,x=\pm 1\] |
So, \[{{V}_{\min }}(x=\pm 1)=\frac{1}{4}-\frac{1}{2}=\frac{-1}{4}J\] |
\[\therefore \]\[{{K}_{\max }}+{{V}_{\min }}=\]Total mechanical energy |
(from energy conservation) |
\[{{K}_{\max }}=(1/4)+2\Rightarrow {{K}_{\max }}=9/4\] |
\[\Rightarrow \]\[\frac{m{{v}^{2}}}{2}=\frac{9}{4}\] |
\[\Rightarrow \]\[v=\frac{3}{\sqrt{2}}m/s\] \[(\because m=1\,kg)\] |
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