Question

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

• A. $\frac{2K}{S}\sqrt{1+{\left(\frac{S}{R}\right)}^2}$
• B. $2KS\sqrt{1+{\left(\frac{R}{S}\right)}^2}$
• C. $2KS\sqrt{1+{\left(\frac{S}{R}\right)}^2}$
• D. $\frac{2S}{K}\sqrt{1+{\left(\frac{R}{S}\right)}^2}$

Answer 1

The correct option is C $2KS\sqrt{1+{\left(\frac{S}{R}\right)}^2}$

Given that, $K.E=K{S}^2$
$\Rightarrow \frac{1}{2}m{v}^2=K{S}^2$
Centripetal force is given by
$F_c=\frac{m{v}^2}{R}=\frac{2K{S}^2}{R}$
$v^2=\frac{2K{S}^2}{m}\Rightarrow v=\sqrt{\frac{2K}{m}}S$
Now, acceleration
$a_T=\frac{dv}{dt}=\sqrt{\frac{2K}{m}}v$
$\Rightarrow a_T=\sqrt{\frac{2K}{m}}\times \sqrt{\frac{2K}{m}}\times S$
$\Rightarrow a_T=\frac{2K}{m}\times S$
$\therefore F_T=m{a}_T=2KS$
$F_{\text{net}}=\sqrt{(2KS{)}^2+{\left(\frac{2K{S}^2}{R}\right)}^2}=2KS\sqrt{1+{\left(\frac{S}{R}\right)}^2}$