$A$ particle is constrained to move along the $x$-axis. $A$ force is applied to it in the same direction,which varies with the distance $x$ from the origin as $F(x) = -kx + ax^3$,where $k$ and $a$ are positive constants. What will be the graph of the potential energy $U(x)$ of the particle for $x \ge 0$?

The potential energy of a particle of mass $5 \ kg$ moving in the $XY$ plane is given by $V = -7x + 24y$ joules,where $x$ and $y$ are in metres. Initially at $t = 0$,the particle is at the origin $(0, 0)$ moving with a velocity of $\vec{v}_0 = 6[0.24 \hat{i} + 0.7 \hat{j}] \ m/s = [1.44 \hat{i} + 4.2 \hat{j}] \ m/s$. Then:

The $PE$ of a $2 \, kg$ particle,free to move along the $x$-axis,is given by $V(x) = \left( \frac{x^3}{3} - \frac{x^2}{2} \right) \, J$. The total mechanical energy of the particle is $4 \, J$. The maximum speed (in $m \, s^{-1}$) is ...........

The potential energy of a particle in a force field is $U(r) = \frac{A}{r^2} - \frac{B}{r}$,where $A$ and $B$ are positive constants and $r$ is the distance of the particle from the centre of the field. For stable equilibrium,the distance of the particle is:

$A$ particle is placed at the origin and a force $F = kx$ is acting on it (where $k$ is a positive constant). If $U(0) = 0$,the graph of $U(x)$ versus $x$ will be (where $U$ is the potential energy function):

The potential energy function (in $J$) of a particle in a region of space is given as $U = (2x^2 + 3y^3 + 2z)$. Here $x, y$ and $z$ are in meters. The magnitude of the $x$-component of the force (in $N$) acting on the particle at point $P(1, 2, 3) \ m$ is: