Question

The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered $s$ as $k=a{s}^2$ where $a$ is a constant. The force acting on the particle is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

According to given problem

$\frac{1}{2}m{v}^2=a{s}^2\Rightarrow v=s\sqrt{\frac{2a}{m}}$

So $a_R$ = $\frac{v^2}{R}$ = $\frac{2a{s}^2}{mR}$ ..............(i)

Further more as $a_t$ = $\frac{dv}{dt}$ = $\frac{dv}{ds}.\frac{ds}{dt}$ = $v\frac{dv}{ds}$ ......................(ii)

(By chain rule)

Which in light of equation (i) i.e. $v=s\sqrt{\frac{2a}{m}}$ yields

$a_t=\left[s\sqrt{\frac{2a}{m}}\right]\left[\sqrt{\frac{2a}{m}}\right]$ = $\frac{2as}{m}$ ..............(iii)

So that $a=\sqrt{a_R^2+a_t^2}=\sqrt{{\left[\frac{2a{s}^2}{mR}\right]}^2+{\left[\frac{2as}{m}\right]}^2}$

Hence $a=\frac{2as}{m}\sqrt{1+{\left[\frac{s}{R}\right]}^2}$

$\therefore F=ma=2as\sqrt{1+[\frac{s}{R}{]}^2}$