Question

The kinetic energy $K$ of a particle moving along a circle of radius $R$ depends upon the distance $s$ as $K=a{s}^2.$ The force acting on the particle is:

• A. $2a\frac{s^2}{R}$
• B. $2as[1+\frac{s^2}{R^2}{]}^{1/2}$
• C. $2as$
• D. $2a$

Answer 1

Answer: (B)

$2as[1+\frac{s^2}{R^2}{]}^{1/2}$

Given $K.E=1/2m{v}^2=a{s}^2$

$=m{v}^2=2a{s}^2$

Now differentiating w.r.t to $t$

$2mv=dv/dt=4as$ $ds/dt=4asv$

$mat=2as$

hence, $Ft=2$ as here $ft$= tangential force

Centripetal force= $fc=mac=m{v}^2/R$

$=2ma{s}^2/mR=2a{s}^{2}R$

Net force= $\sqrt{F{t}^2+F^{2}C}=\sqrt{2(as{)}^2+(2a{s}^2/R{)}^2}$

$=2as({1+s^2/R^2)}^{1/2}$