Question

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

Its
total energy will be:

(a)$\rm{-\frac{3k}{2a^2}}$


(b)$\rm{-\frac{k}{4a^2}}$


(c)$\rm{\frac{k}{2a^2}}$


(d) Zero

$\blue\implies$Explanation needed ​

Answer 1

Answer:

Correct option is

A

Zero

We know, F

x

=

dx

−dU

where

x

and

F

are in same direction.

Here, radius vector

r

and and centripital force

F

r

are in opposite direction.

hence F

r

=

dr

dU

F=

dr

dU

F=

r

3

k

=

r

mv

2

K.E.=1/2mv

2

=

2r

2

k

T.E.=P.E.+K.E.=

2r

2

−K

+

2r

2

K

=0

Answer 2

Answer:

click the photo

Explanation:

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