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
B
12
C
2
D
2

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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