A particle is moving in a circular path of ra | vedclass

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

Question

(B) The force $F$ is given by the negative gradient of the potential energy $U$:$F = -\frac{dU}{dr} = -\frac{d}{dr} \left( -\frac{k}{2r^2} \right) = -

Vedclass · Accepted Answer

Zero

Vedclass · Answer

(B) The force $F$ is given by the negative gradient of the potential energy $U$:$F = -\frac{dU}{dr} = -\frac{d}{dr} \left( -\frac{k}{2r^2} \right) = -\frac{k}{r^3}$.Since the particle is moving in a circular path of radius $a$,the magnitude of the centripetal force must equal the magnitude of the attractive force:$\frac{mv^2}{a} = \frac{k}{a^3} \Rightarrow mv^2 = \frac{k}{a^2}$.The kinetic energy $(K.E.)$ is given by:$K.E. = \frac{1}{2}mv^2 = \frac{1}{2} \left( \frac{k}{a^2} \right) = \frac{k}{2a^2}$.The potential energy $(P.E.)$ at $r = a$ is:$P.E. = -\frac{k}{2a^2}$.The total energy $(E)$ is the sum of kinetic and potential energy:$E = K.E. + P.E. = \frac{k}{2a^2} + \left( -\frac{k}{2a^2} \right) = 0$.

$A$ particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = - \frac{k}{2r^2}$. Its total energy is

Similar Questions

$A$ man is slipping on a frictionless inclined plane and a bag falls down from the same height. Then the velocity of both is related as

$A$ locomotive of mass $m$ starts moving so that its velocity varies according to the law $v = k \sqrt{S}$,where $k$ is a constant and $S$ is the distance covered. Find the total work performed by all the forces acting on the locomotive during the first $t$ seconds after the beginning of motion.

$A$ man of mass $60 \ kg$ gets $10^5 \ calories$ of energy from food. If his efficiency is $28\%$,up to what height in $m$ can he climb?

Explain the different forms of energy.
$(a)$ Heat Energy
$(b)$ Chemical Energy
$(c)$ Electrical Energy

Vedclass Test Series

Exam Paper Generator

Online Exam Module