A particle which is constrained to move along $x -$ axis is subjected to a force in the same direction which varies with distance $x$ of the particle from the origin as $F (x) = - k x + a x^{3}$ . Here, $k$ and $a$ are positive constant. For $x \geq 0$ , the functional form of the potential energy $U (X)$ of the particle is

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Solution

Given:
$F (x) = - k x + a x^{3}$

As we know that,

$F = - \frac{d U}{d x}$

$∴ d U = - F d x$

taking integration both side

$U (x) = - x \int 0 (- k x + a x^{3}) d x$

$U (x) = \frac{k x^{2}}{2} - \frac{a x^{4}}{4}$

So, $U (x) = 0$ at,

$\frac{k x^{2}}{2} - \frac{a x^{4}}{4} = 0 \Rightarrow x^{2} (\frac{k}{2} - \frac{a x^{2}}{4}) = 0$

$∴ x = 0$ and $x = \pm \sqrt{\frac{2 k}{a}}$

So, we can see that $U (x)$ will be negative for $x > \sqrt{\frac{2 k}{a}}$

From the given function, we observe that $F = 0$ at $x = 0$ i.e., the slope of $U - x$ graph is zero at $x = 0$ .

Therefore, the most appropriate option is $(D)$ .

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A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as $F (x) = - k x + a x^{3}$ . Here k and a are positive constants. For $x \geq 0$ , the functional form of the potential energy U(x) of the particle is

Q. A particle, which is constrained to move along the

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F (x) = - k x + a x^{3}

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A particle which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as $F (x) = - k x + a x^{3}$ . Here k and a are positive constants. For $x \geq 0$ , the functional from of the potential energy $U_{(x)}$ of the particle is