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:

$A$ particle moving along the $X$-axis has potential energy $U = 2 - 20x + 5x^2 \text{ J}$. The particle is released at $x = -3 \text{ m}$. What is the maximum value of $x$ reached by the particle (in $\text{ m}$)? ($x$ is in meters and $U$ is in joules)

Consider a one-dimensional motion of a particle with total energy $E$. There are four regions $A, B, C$ and $D$ in which the relation between potential energy $V$,kinetic energy $K$,and total energy $E$ is as given below:
Region $A: V > E$
Region $B: V < E$
Region $C: K < E$
Region $D: V > E$
State with reason in each case whether a particle can be found in the given region or not.

$A$ 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:

$A$ chain of length $L$ and mass $M$ is placed on a frictionless table with one-fifth of its length hanging over the edge. What is the work required to pull the hanging part back onto the table?

$A$ particle with constant total energy $E$ moves in one dimension in a region where the potential energy is $U(x)$. The speed of the particle is zero where