1. Tardigrade
  2. Question
  3. Physics
  4. A particle, which is constrained to move along 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) =-kx+ax3. Here, k and a are positive constants. For x ge 0, the functional form of the potential energy U (x) of the particle is

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

IIT JEEIIT JEE 2002Work, Energy and Power

Solution:

$F=-\frac{dU}{dx}$
$\therefore dU=-F.dx$
or $U(x)=-\int\limits_0^x(-kx+ax^3)dx$
$U(x)=\frac{kx^2}{2}-\frac{ax^4}{4}$
$U(x)=0 \, at \, x=0$, and $x=\sqrt{\frac{2k}{a}}$
$U(x)$=negative for $x=\sqrt{\frac{2k}{a}}$
From the given function, we can see that
$F = 0$ at $x = 0$ i.e. slope of $U-x$ graph is zero at
$x = 0$.