The potential energy of a particle in a force | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given potential energy: $U(r) = \frac{A}{r^2} - \frac{B}{r}$.For equilibrium,the force $F = -\frac{dU}{dr} = 0$,which implies $\frac{dU}{dr} = 0$.

Vedclass · Accepted Answer

$\frac{2A}{B}$

Vedclass · Answer

(B) Given potential energy: $U(r) = \frac{A}{r^2} - \frac{B}{r}$.For equilibrium,the force $F = -\frac{dU}{dr} = 0$,which implies $\frac{dU}{dr} = 0$.$\frac{dU}{dr} = \frac{d}{dr}(Ar^{-2} - Br^{-1}) = -2Ar^{-3} + Br^{-2} = 0$.$\frac{B}{r^2} = \frac{2A}{r^3} \implies r = \frac{2A}{B}$.For stable equilibrium,the second derivative must be positive: $\frac{d^2U}{dr^2} > 0$.$\frac{d^2U}{dr^2} = \frac{d}{dr}(-2Ar^{-3} + Br^{-2}) = 6Ar^{-4} - 2Br^{-3}$.Substituting $r = \frac{2A}{B}$:$\frac{d^2U}{dr^2} = 6A(\frac{B}{2A})^4 - 2B(\frac{B}{2A})^3 = 6A(\frac{B^4}{16A^4}) - 2B(\frac{B^3}{8A^3}) = \frac{3B^4}{8A^3} - \frac{B^4}{4A^3} = \frac{3B^4 - 2B^4}{8A^3} = \frac{B^4}{8A^3}$.Since $A, B > 0$,$\frac{B^4}{8A^3} > 0$. Thus,the condition for stable equilibrium is satisfied at $r = \frac{2A}{B}$.

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:

Similar Questions

$A$ rod of length $4 \ m$ and mass $20 \ kg$ is lying horizontally on the ground. Calculate the work done in $J$ to keep it in a vertical position such that one end remains in contact with the ground.

The potential energy of a particle is related to the $x$ coordinate by the equation $U = x^2 - 2x$. The particle will be in stable equilibrium at:

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

$A$ uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $(1/n)^{th}$ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface,the work done should be

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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module