$A$ particle is moving along a horizontal circle of radius $r$ under a centripetal force $F = -\frac{c}{r^2}$,where $c$ is a constant. Then,the total energy of the particle is

  • A
    $\frac{-c}{2 r^2}$
  • B
    $\frac{c}{2 r}$
  • C
    $\frac{-c}{2 r}$
  • D
    $\frac{c}{2 r^2}$

Explore More

Similar Questions

$A$ body is placed as shown in the figure. To complete one full vertical circular loop,the height $h$ must be:

Difficult
View Solution

$A$ particle of mass $m$ is released from a height $H$ on a smooth curved surface which ends into a vertical loop of radius $R$,as shown. Choose the correct alternative$(s)$ if $H = 2R$.

Difficult
View Solution

$A$ stone of mass $m$ is moving in a vertical circle of radius $20 \ cm$. What is the difference in kinetic energy between the lowest and highest points?

The minimum horizontal speed with which a body must be projected, so that it goes around a smooth vertical circular track of radius $4 \,m$ is $(g=9.8 \,ms^{-2})$. (in $\,ms^{-1}$)

As shown in the figure, a mass $m$ and another block of mass $10m$ are connected with a string of length $L$. Friction is sufficient to prevent the slipping of the $10m$ block. Mass $m$ is given a velocity $u$ in the vertical direction. For the complete circular motion of mass $m$:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo