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 tangential force acting on the particle is

• A. $mAs$
• B. $2mAs$
• C. $As$
• D. $2As$

Answer 1

The correct option is D $2As$

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