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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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The potential energy of a 1 kg mass free to move along the x-axis is given by $V_{(x)}=\bigg[\large\frac{x^4}{4}-\frac{x^2}{2}\bigg]$$J$ The total mechanical energy of the particle is $2J$ then the maximum speed is?

\[(a)\;\frac{3}{\sqrt 2}\; m/s \quad (b)\;\sqrt 2 \;m/s\quad (c)\; \frac{1}{\sqrt 2}\; m/s \quad (d)\;2\; m/s\]

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1 Answer

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Velocity is maximum when $KE$ of particle is maximum. ie potential energy of particle is minimum
$\large\frac{dV}{dx}$$=0=>x^3-x=0=>x=\pm 1$
minimum $PE$ is when $x=1$
min $PE=\large\frac{1}{4}-\frac{1}{2}=\frac{-1}{4}$$J$
Given $KE_{\large \max}$$+PE_{\large min}$$=2\;J$
$KE_{\large max }=2 +\large\frac{1}{4}=\frac{9}{4}$
$\large\frac{9}{4}=\frac{1}{2} $$mv^2_{\large max} \qquad (m=1 kg)$
$ v^2 _{\large max}=\large\frac{9}{2}$
$v_{\large max}=\large\frac{3}{\sqrt 2}$$\;m/s$
Hence a is the correct answer.

 

answered Aug 6, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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