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 [IIT JEE 2002]

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Solution

The correct option is D

$F = - \frac{d U}{d x} \Rightarrow d U = - F . d x \Rightarrow U = - \int_{0}^{x} (- k x + a x^{3}) d x \Rightarrow U = \frac{k x^{2}}{2} - \frac{a x^{4}}{4}$
$∴ We get U=0 at x = 0 and x = \sqrt{\frac{2 k}{a}} Also we get U = negative for x$ > $\sqrt{\frac{2 k}{a}}$

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

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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 [IIT JEE 2002]

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