Solution:

Introduction:
In this problem, we need to find the force on the hoop at any point while a particle of mass m oscillates along arc AB on the inside of a smooth circular track of radius 'a' fixed in a vertical plane. We are given that the kinetic energy at any instant is k.

Analysis:
Let's consider the forces acting on the particle when it is at point P on the arc AB. The forces acting on the particle are:
1. Weight of the particle (mg) acting vertically downwards.
2. Normal reaction force (N) acting vertically upwards, which is perpendicular to the tangent to the arc at point P.
3. Force (F) acting tangentially along the arc AB.

The net force (Fnet) acting on the particle is given by the vector sum of the above forces. Since the track is smooth, there is no friction between the particle and the track. Hence, the force of friction is absent.

Now, let's consider the work done by the net force (Fnet) on the particle as it moves from point A to point B on the arc AB. The work done by the net force (Fnet) is equal to the change in kinetic energy of the particle.

We know that the work done by the tangential force (F) is given by FΔs, where Δs is the distance moved by the particle along the arc AB. Since the force (F) is tangential to the arc, it does not contribute to the work done by the normal force (N). Hence, the work done by the net force (Fnet) is given by FΔs.

As the particle moves from point A to point B, its velocity changes from zero to a maximum value and then back to zero. Therefore, the work done by the net force (Fnet) is equal to the change in kinetic energy (Δk) of the particle, which is given by:

Δk = k(max) - k(min)

where k(max) is the maximum kinetic energy of the particle and k(min) is the minimum kinetic energy of the particle.

Since the force (F) is tangential to the arc AB, its component along the vertical direction is zero. Therefore, the net force (Fnet) acting on the particle is given by:

Fnet = N - mg

where N is the normal reaction force and mg is the weight of the particle.

Since the track is smooth, the normal reaction force (N) is perpendicular to the tangent to the arc at point P. Therefore, the work done by the normal force (N) is zero.

Now, let's consider the work done by the weight of the particle (mg) as it moves from point A to point B on the arc AB. The work done by the weight of the particle is given by:

W = mgΔh

where Δh is the change in height of the particle from point A to point B.

Since the arc AB is a horizontal diameter, the height of the particle remains constant. Therefore, the work done by the weight of the particle is zero.

Hence, the work done by the net force (Fnet) is given by:

Wnet = FnetΔs = Δk = k(max) - k(min)

Substituting the value of Fnet, we get:

(N - mg)Δs = k(max) - k(min)

Since