Q: Kinetic energy of a particle moving along a circle of radius ‘r’ depends on the distance as KE = cs^{2}, (c is constant, s is displacement). Find the force acting on the particle .

Sol: $\large K.E = \frac{1}{2}m v^2 = c s^2 $

$\large v = \sqrt{\frac{2c}{m}} s $

$\large a_t = \frac{dv}{dt} = \sqrt{\frac{2c}{m}} \frac{ds}{dt}$

$\large a_t = v \sqrt{\frac{2c}{m}} $

$\large F_t = m a_t = m v \sqrt{\frac{2c}{m}} $

$\large F_t = m (\sqrt{\frac{2c}{m}} s) \sqrt{\frac{2c}{m}} $

F_{t} = 2 c s

$\large F_c = \frac{m v^2}{r} $

Net force acting is $\large F = \sqrt{F_t^2 + F_c^2 }$