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 $(U)$ 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 $a = \dots \dots \dots \dots$

If the potential energy of a particle of mass $0.1 \ kg$ moving along $x$-axis is $U(x) = 5x(x-4) \ J$,then the speed of the particle is maximum at a position of

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:

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.

For a system of conservative forces,the potential energy is given by the formula $U = ax^2 - bx$,where $a$ and $b$ are constants. The equilibrium position and the equilibrium potential energy will be,respectively: