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:

$A$ chain of mass $M$ and length $L$ is placed on a smooth table such that $1/n$ of its length is hanging over the edge of the table. The hanging part of the chain is pulled back onto the surface of the table. Find the work done to pull it up.

The potential energy between two atoms is given by the function $U(r) = a/r^{12} - b/r^{6}$. Find the equilibrium distance between them.

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity,the work required to pull the hanging part onto the table is:

Potential energy of a particle varies with position $r$ as,$U = \left( \frac{\alpha}{r^4} - \frac{\beta}{r^5} \right) \text{ J}$,where $\alpha$ and $\beta$ are positive constants. The particle will be in equilibrium at $r = \dots$

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: