$A$ body of mass $2\, kg$ is situated at a height $5\, m$ above the ground. Its potential energy is reported as $70\, J$. The potential energy of a $3\, kg$ body situated at a height $1\, m$ above the ground will be reported as ................ $J$.

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:

$A$ 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

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 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:

$A$ carpet of mass $M$ made of a material is rolled along its length in the form of a cylinder of radius $R$ and kept on a rough floor. If the carpet is unrolled without sliding to a radius $R/2$,the change in potential energy is ($g=$ acceleration due to gravity).