A particle which is constrained to move along | vedclass

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

Question

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

Vedclass · Accepted Answer

Vedclass · Answer

(D) The relationship between force $F(x)$ and potential energy $U(x)$ is given by $F(x) = -\frac{dU}{dx}$.Integrating this,we get $U(x) = -\int F(x) dx$.Substituting $F(x) = -kx + ax^3$,we have $U(x) = -\int (-kx + ax^3) dx = \frac{kx^2}{2} - \frac{ax^4}{4} + C$.Assuming $U(0) = 0$,we get $C = 0$,so $U(x) = \frac{kx^2}{2} - \frac{ax^4}{4} = \frac{x^2}{4}(2k - ax^2)$.For $x \ge 0$,$U(x) = 0$ at $x = 0$ and $x = \sqrt{\frac{2k}{a}}$.Between $x = 0$ and $x = \sqrt{\frac{2k}{a}}$,$U(x)$ is positive. For $x > \sqrt{\frac{2k}{a}}$,$U(x)$ becomes negative.The graph starts at the origin,rises to a maximum,and then decreases,crossing the $x-$axis at $x = \sqrt{\frac{2k}{a}}$. This matches Graph $D$.

Similar Questions

The $PE$ of a $2 \, kg$ particle,free to move along the $x$-axis,is given by $V(x) = \left( \frac{x^3}{3} - \frac{x^2}{2} \right) \, J$. The total mechanical energy of the particle is $4 \, J$. The maximum speed (in $m \, s^{-1}$) is ...........

$A$ body at rest may have

$A$ particle moves under the influence of a force $F = -2x$ in one dimension (where $x$ is the distance of the particle from the origin). Assume that the potential energy of the particle at the origin is zero. The schematic diagram of the potential energy $U$ as a function of $x$ is given by:

The potential energy of a particle is given by $U = 8x^2 - 4x + 400 \, J$. Find the position $x$ where the force acting on the particle is zero.

Vedclass Test Series

Exam Paper Generator

Online Exam Module