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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 2 J. Then, the maximum speed (in m/s) is

Option 1)
$2$
Option 2)
$3/\sqrt{2}\;$

Option 3)
$\sqrt{2}\;$
Option 4)
$1/\sqrt{2}$

Answers (1)

best_answer

As we learnt in

If only conservative forces act on a system, total mechnical energy remains constant -

$K+U=E\left ( constant \right )$

$\Delta K+\Delta U=0$

$\Delta K=-\Delta U$

-

Potential energy $V=\frac{x^{4}}{4}-\frac{x^{2}}{2}.$

First we neet to calculate minimum potential energy.

When potential energy is minimum

$\frac{dV}{dx}=0\ \Rightarrow\ x^{3}-x=0$

$\therefore\ x=0\ or\ x=1$

V(x = 0) = 0

$V(x=1)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}J$

So minimum potential energy $=\frac{-1}{4}J$

K.E. + P.E. = Total mechanical energy.

$K.E.|_{max}+P.E.|_{min}=Total\ mechanical\ energy$

$K.E._{max}-\frac{1}{4}=2J$

$K.E._{max}=\frac{9}{4}J$

$\Rightarrow\ \frac{1}{2}mV_{max}^{2}=\frac{9}{4}\ or\ V_{max}^{2}=\frac{9}{2}$

$N_{max}=\frac{3}{\sqrt{}2}$

Correct answer is 2

Option 1)

$2$

This is an incorrect option.

Option 2)

$3/\sqrt{2}\;$

This is the correct option.

Option 3)

$\sqrt{2}\;$

This is an incorrrect option.

Option 4)

$1/\sqrt{2}$

This is is an incorrect option.

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