Question

NEET & ALIMS
A particle moves under an attractive force F = - K/r^2 in a circle of radius r. The total energy
of revolving particle is


options

- K/r
K/ r
- K / 2r
K/ 2r​

Answer 1

answer : -K/2r

explanation : given, a particle moves under an attractive force, F = - K/r² in a circle of radius r.

so, To move particle in circular motion, a centripetal force is acting to balance the attractive force.

i.e., centripetal force = attraction force

or, mv²/r = K/r²

or, mv² = K/r

so, kinetic energy of particle , K.E = 1/2 mv² = 1/2 (K/r) = K/2r

again, potential energy , U = - negative of workdone

= - F.dr

= -∫(-K/r²).dr

= K ∫dr/r²

= K [-1/r]

= -K/r

so, total force = K.E + U

= K/2r - K/r

= -K/2r

hence, total energy of revolving particle is -K/2r