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:

$A$ uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R$ $(R > l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is

The potential energy of a body of mass $m$ is given by $U = ax + by$,where $x$ and $y$ are the position coordinates of the particle. The acceleration of the particle is:

$A$ particle has a total mechanical energy that is small and negative. It is under the influence of a one-dimensional potential $U(x) = \frac{x^4}{4} - \frac{x^2}{2} \, J$,where $x$ is in meters. At time $t = 0 \, s$,it is at $x = -0.5 \, m$. Then,at a later time,it can be found:

The potential energy of a body is given by $U = A - Bx^2$ (where $x$ is the displacement). The magnitude of the force acting on the particle is:

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.