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

\frac{2 a s^{2}}{R}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 a S [1 + \frac{s^{2}}{R^{2}}]^{\frac{1}{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 a s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 a [\frac{R^{2}}{s}]^{\frac{1}{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{2 a s^{2}}{R}$
$G i v e n K = a s^{2} a n d r a d i u s = R K i n e t i c e n e r g y K = \frac{1}{2} m v^{2} = a s^{2} m v^{2} = 2 a s^{2} v^{2} = \frac{2 a s^{2}}{m} - - - - - - - (a) N o w d i f f e r e n t i a t i n g w i t h r e s p e c t t o t w e g e t 2 m v \frac{d v}{d t} = 4 a s \frac{d s}{d t} 2 m v a_{t} = 4 a s v w h e r e a_{t} = t a n g e n t i a l a c c e l e r a t i o n a n d a = c o n s t a n t a_{t} = \frac{2 a s}{m} - - - - - - - - - (b) T a n g e n t i a l f o r c e F_{T} = m a_{t} F_{T} = m \frac{2 a s}{m} {f r o m b} F_{T} = 2 a s A l s o N o r m a l f o r c e F_{N} = m \frac{v^{2}}{R} = \frac{m}{R} (\frac{2 a s^{2}}{m}) = \frac{2 a s^{2}}{R} N o w t h e n e t f o r c e F_{n e t} = \sqrt{{(F}_{T}^{2} +} {F_{N}}^{2}) = \sqrt{(2 a s)^{2} +} ({\frac{2 a s^{2}}{R})}^{2} = 2 a s \sqrt{1 + \frac{s^{2}}{R}}$

Suggest Corrections

Similar questions

s^{2}

K = a s^{2}

R

(s)

K = A s^{2}

A

K

R

K = a s^{2},

α

R

(s)

K = A s^{2}

A

Join BYJU'S Learning Program