The work done in moving a body of mass $2 \,kg$ to a height of $4 \,m$ from the surface of the earth is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,J$)

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:

The potential energy of a particle is given by $U = A - Bx^2$,where $x$ is the displacement. What is the magnitude of the force acting on the particle?

The potential energy of a particle of mass $1 \ kg$ moving freely along the $x$-axis is given by $V(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. If the total mechanical energy of the particle is $2 \ J$,what is its maximum speed in $m/s$?

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 potential energy $(U)$ of a diatomic molecule is a function dependent on $r$ (interatomic distance) as $U = \frac{\alpha}{r^{10}} - \frac{\beta}{r^5} - 3$,where $\alpha$ and $\beta$ are positive constants. The equilibrium distance between two atoms will be $\left(\frac{2\alpha}{\beta}\right)^{\frac{a}{b}}$,where $a = \dots \dots \dots \dots$