Question

The kinetic energy of a particle moving along a circle of radius $R$ depends on distance $\left(s\right)$ as $K=A{s}^2$ where $A$ is a constant.

The net force acting on the particle is

• A. $2As(1+\frac{s}{R})$
• B. $As{(1+\frac{s^2}{R^2})}^{1/2}$
• C. $2As{(1+\frac{s^2}{R^2})}^{1/2}$
• D. zero

Answer 1

The correct option is C $2As{(1+\frac{s^2}{R^2})}^{1/2}$

As we know, $F_{net}=\sqrt{{F_c}^2+{F_t}^2}$
First calculate centripetal force ($F_c$)
$F_c=\frac{m{v}^2}{R}$
As we know, $K.E.=\frac{1}{2}m{v}^2$
Given, $K=A{s}^2$
$A{s}^2=\frac{1}{2}m{v}^2$
$v=s\sqrt{\frac{2A}{m}}$
$F_c=\frac{m{v}^2}{R}$
$F_c=m{s}^2\frac{2A}{m}\frac{1}{R}$
$F_c=\frac{2A{S}^2}{R}$

For tangential force
Tangential acceleration is given as, $a_t=\frac{dv}{dt}$
Tangential force $F_t=m{a}_t$
$\begin{array}{}{F}_t=m\frac{dv}{dt}& \text{(1)}\end{array}$
put $v$ in equation 1
$F_t=m\sqrt{\frac{2A}{m}}\frac{ds}{dt}=m\sqrt{\frac{2A}{m}}v$
$F_t=m\sqrt{\frac{2A}{m}}s\sqrt{\frac{2A}{m}}$
$F_t=ms\frac{2A}{m}$
$F_t=2As$

$F_{net}=\sqrt{{\left[\frac{2A{s}^2}{R}\right]}^2+{\left[2As\right]}^2}$
$F_{net}=2AS\sqrt{1+\frac{S^2}{R^2}}$