In the figure shown,the potential energy (U) | vedclass

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

Question

(D) For a particle to be in equilibrium,the net force must be zero,which implies $\frac{dU}{dx} = 0$. This corresponds to the points where the slope o

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) For a particle to be in equilibrium,the net force must be zero,which implies $\frac{dU}{dx} = 0$. This corresponds to the points where the slope of the $U-x$ graph is zero.At $x_2$,the potential energy $U$ is at a local maximum. For a potential energy maximum,the equilibrium is unstable because any small displacement results in a force that pushes the particle further away from the equilibrium position.At $x_1$ and $x_3$,the slope is not zero,so the particle is not in equilibrium at these points.Therefore,none of the given options correctly describe the state of the particle at the points shown.

In the figure shown,the potential energy $(U)$ of a particle is plotted against its position $x$ from the origin. The particle at:

Similar Questions

$A$ body of mass $2\, kg$ is situated at a height $5\, m$ above the ground. Its potential energy is reported as $70\, J$. The potential energy of a $3\, kg$ body situated at a height $1\, m$ above the ground will be reported as ................ $J$.

The following figure shows the variation of potential energy $V(x)$ of a particle with distance $x$. The particle has

When a stone is thrown vertically upwards and returns to the ground,at what point will its potential energy be maximum?

$A$ rod of mass $m$ and length $l$ is lying on a table. What is the work done to make it stand vertically?

$A$ particle of mass $1 \ kg$ is free to move along the $x$-axis. Its potential energy is given by $U(x) = (\frac{x^2}{2} - x) \ J$. If the total mechanical energy of the particle is $2 \ J$,find the maximum speed of the particle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module