A particle of mass m moves in a horizontal ci | vedclass

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

Question

(D) For a particle moving in a circular path,the centripetal force is provided by the given force: $mv^2/r = k/r^2$. From this,the kinetic energy $(K.

Vedclass · Accepted Answer

$-k/(2r)$

Vedclass · Answer

(D) For a particle moving in a circular path,the centripetal force is provided by the given force: $mv^2/r = k/r^2$. From this,the kinetic energy $(K.E.)$ is: $K.E. = 1/2 mv^2 = k/(2r)$. The potential energy $(U)$ is calculated by integrating the force: $U = -\int_{\infty}^{r} F \, dr = -\int_{\infty}^{r} (-k/r^2) \, dr = -k/r$. The total energy $(E)$ is the sum of kinetic and potential energy: $E = K.E. + U = k/(2r) - k/r = -k/(2r)$. The negative sign indicates that the particle is in a bound state.

$A$ particle of mass $m$ moves in a horizontal circle of radius $r$ under the influence of a centripetal force given by $F = -k/r^2$,where $k$ is a constant. Calculate the total energy of the particle.

Similar Questions

If a satellite of mass $400 \, kg$ revolves around the earth in an orbit with speed $200 \, m/s$,then its potential energy is .......... $MJ$.

Let the kinetic energy of a satellite be $x$,then its time of revolution $T$ is proportional to ..............

$A$ satellite of mass $m$ is in a circular orbit of radius $2R_E$ around the Earth. The energy required to transfer it to a circular orbit of radius $4R_E$ is (where $M_E$ and $R_E$ are the mass and radius of the Earth,respectively).

$A$ satellite of mass $m$ is orbiting the Earth (of radius $R$) at a height equal to the radius of the Earth. The total energy of the satellite in terms of $g$ (acceleration due to gravity on Earth's surface) is

In a satellite,if the time of revolution is $T$,then the potential energy $(PE)$ is proportional to ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module