Question

The kinetic energy \[k\] of a particle moving along a circle of radius \[R\] depends on the distance covered. It is given as $K = a{s^2}$, where a is a constant. The force acting on the particle is
A. $2a\dfrac{{{s^2}}}{R}$
B. $2as{\left( {1 + \dfrac{{{s^2}}}{R}} \right)^{1/2}}$
C. $2as$
D. $2a\dfrac{{{R^2}}}{s}$a

Answer 1

Hint:Kinetic energy is the energy stockpiled in the body under its motion. Energy is the capacity to get some work done. It means that the quantity of work done on the body is stored in that body in potential form, and that establishes the energy of the body. If a force F yields a very small displacement, the work done will be also very small and we can obtain the force due to tangential acceleration and the centripetal force which gives us the net force.

Complete step by step solution:
From the question, we get the kinetic energy of the body is, $K = a{s^2}$.
In a non-uniform circular motion, the two forces will operate on a particle ${F_c}$​ and ${F_t}$
So, the net force \[{F_{net}} = \,\sqrt {{F_c}^2 + {F_t}^2} \]which is equation 1
The centripetal force ${F_c} = \dfrac{{m{v^2}}}{R} = \dfrac{{2a{s^2}}}{R}$ which is equation 2
[Given $\dfrac{1}{2}m{v^2} = a{s^2}$]
Again from $\dfrac{1}{2}m{v^2} = a{s^2}$
${v^2} = \dfrac{{2a{s^2}}}{m}$
Therefore $v = s\sqrt {\dfrac{{2a}}{m}} $
Tangential acceleration ${a_t} = \dfrac{{dv}}{{dt}} = \dfrac{{dv}}{{ds}}.\dfrac{{ds}}{{dt}}$
${a_t} = v\sqrt {\dfrac{{2a}}{m}} = \dfrac{{2as}}{m}$
and ${F_t} = m{a_t} = 2as$ equation 3
Now on substituting the value of ${F_c}$and ${F_t}$ in equation 1
we get
${F_{net}} = \sqrt {{{\left( {\dfrac{{2a{s^2}}}{R}} \right)}^2} + {{\left( {2a{s^2}} \right)}^2}} = 2as$
$ = 2as{\left( {1 + \dfrac{{{s^2}}}{R}} \right)^{1/2}}$

Note:
Force and displacements are both vector quantities, but the work and energy are the scalar quantities. Work done is called the dot product of both Force and displacement.
$W = F.s \Rightarrow W = F \times s \times \cos \theta $. When the body moves in a straight line, the angle between the force and displacement are in the same one line, which indicates the angle between the force and the displacement is \[0\] degrees.