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)

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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