The force acting on a particle is given by th | vedclass

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

Question

(D) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.Integrating this,we get $dU = -F \, dx$,so $U(x) = -\

Vedclass · Accepted Answer

Vedclass · Answer

(D) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.Integrating this,we get $dU = -F \, dx$,so $U(x) = -\int F \, dx$.Substituting $F(x) = -kx + ax^3$,we have $U(x) = -\int (-kx + ax^3) \, dx = \int (kx - ax^3) \, dx$.Performing the integration,we get $U(x) = \frac{1}{2}kx^2 - \frac{1}{4}ax^4 + C$.Assuming $U(0) = 0$,we find $C = 0$,so $U(x) = \frac{1}{2}kx^2 - \frac{1}{4}ax^4 = \frac{1}{4}x^2(2k - ax^2)$.For $x \ge 0$,$U(x) = 0$ at $x = 0$ and $x = \sqrt{\frac{2k}{a}}$.For $0 For $x > \sqrt{\frac{2k}{a}}$,$U(x)$ becomes negative and decreases as $x^4$ dominates.This corresponds to a curve that starts at the origin,rises to a maximum,and then crosses the $x$-axis to become negative,which matches the graph in option $D$.

The force acting on a particle is given by the formula $F(x) = -kx + ax^3$. Find the graph of its potential energy $U(x)$ for $x \ge 0$.

Similar Questions

For the stability of a physical system,what should be its potential energy? Should it be more or less?

Potential energy as a function of $r$ is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$,where $r$ is the interatomic distance,and $A$ and $B$ are positive constants. The equilibrium distance between the two atoms will be

Calculate the work done by pump $P$ to transfer water of density $d$ from one container to another as shown in the figure.

If an object of mass $20 \ g$ falls from a height of $200 \ m$,and at the point where the object touches the Earth's surface,its total potential energy is converted into kinetic energy,then what is the decrease in the potential energy of the object in $J$?

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 $r = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module