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Question

The potential energy of 1 kg particle free to move along the X-axis is given by $$U=\left (\dfrac{x^4}{4}-\dfrac{x^2}{2}  \right )J$$
The total mechanical energy of the particle is 2 J. Then Maximum speed of the particle is :


A
32
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B
12
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C
2
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D
2
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Solution

The correct option is A $$\dfrac{3}{\sqrt{2}}$$
$$U=\dfrac{x^4}{4}-\dfrac{x^2}{2}$$ (Given)
$$\because F=\dfrac{dU}{dx}=(x^3-x)$$
$$\therefore x(x^2-1)=0\Rightarrow x=0, or \pm 1$$
$$\dfrac{d^2U}{dx^2}=3x^2-1$$ at $$\dfrac{d^2U}{dx}=+ve$$ i.e., at $$x=\pm 1$$
P.E. is minimum when kinetic energy is maximum
$$Min U=-\dfrac{1}{4}$$
$$ K_{max}+U_{min}\Rightarrow 2$$
$$ K_{max}=\dfrac{9}{4}$$
$$\therefore \dfrac{1}{2}mV^2_{max}=\dfrac{9}{4}$$
$$\because m=1kg$$ (given) $$\Rightarrow V_{max}=\dfrac{3}{\sqrt{2}}m/s$$

Physics

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