The kinetic energy of a particle moving along a circle of radius R depends on the…
Physics · JEE Advanced · NTA Exams — Rotational Motion
The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as T = Ks2 where K is a constant. Find the force acting on the particle as a function of s -
- \[\frac{2 \mathrm{~K}}{\mathrm{~s}} \sqrt{1+\left(\frac{\mathrm{s}}{\mathrm{R}}\right)^{2}}\]
- \[2 \mathrm{Ks} \sqrt{1+\left(\frac{\mathrm{R}}{\mathrm{~s}}\right)^{2}}\]
- \[2 \mathrm{Ks} \sqrt{1+\left(\frac{\mathrm{s}}{\mathrm{R}}\right)^{2}}\]
- \[\frac{2 \mathrm{~s}}{\mathrm{~K}} \sqrt{1+\left(\frac{\mathrm{R}}{\mathrm{~s}}\right)^{2}}\]
Answer
(C) 2 Ks 1+ ( s R )^ 2
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