Question

A particle of mass \( m \) is moving in a horizontal circle of radius \( r \) under a centripetal force given by \( \left( \frac{-K}{r^2} \right) \), where \( K \) is a constant. Then

• A. the total energy of the particle is \( \left( \frac{-K}{2r} \right) \)
• B. the kinetic energy of the particle is \( \frac{K}{r} \)
• C. the potential energy of the particle is \( \frac{K}{2r} \)
• D. the kinetic energy of the particle is \( \frac{-K}{r} \)

Hint:

For a particle moving in a circle under a centripetal force, use the relationship between force, velocity, and energy to solve for the total energy.

Answer 1

The Correct Option is A

The total energy of a particle moving in a circular path is the sum of its kinetic and potential energies. The centripetal force is given by: \[ F = \frac{-K}{r^2} \] The work done by this force leads to the potential energy. The kinetic energy is given by: \[ K.E. = \frac{1}{2} m v^2 \] Using the relation between centripetal force and velocity, we get the kinetic energy and potential energy. The total energy of the particle is: \[ E = K.E. + P.E. = \frac{-K}{2r} \] Thus, the correct answer is (A).