Question:
The kinetic energy of a particle moving along a circle of radius \(R\) depends on distance as \(K = \frac{as}{R}\). Find force.
The kinetic energy of a particle moving along a circle of radius \(R\) depends on distance as \(K = \frac{as}{R}\). Find force.
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Circular motion $\Rightarrow$ total force = tangential + centripetal components.
Updated On: Apr 23, 2026
- \(2a\frac{s}{R}\)
- \(2as\left(1+\frac{s^2}{R^2}\right)^{1/2}\)
- \(2as\)
- \(2a\frac{R^2}{s}\)
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The Correct Option is B
Solution and Explanation
Concept:
\[
F = \frac{dK}{ds}
\]
Given:
\[
K = \frac{as}{R}
\]
Step 1: Tangential force
\[ F_t = \frac{dK}{ds} = \frac{a}{R} \]
Step 2: Centripetal force
\[ F_c = \frac{mv^2}{R} = \frac{2K}{R} = \frac{2as}{R^2} \]
Step 3: Resultant force
\[ F = \sqrt{F_t^2 + F_c^2} = 2as\left(1+\frac{s^2}{R^2}\right)^{1/2} \] Conclusion: \[ {(B)} \]
Step 1: Tangential force
\[ F_t = \frac{dK}{ds} = \frac{a}{R} \]
Step 2: Centripetal force
\[ F_c = \frac{mv^2}{R} = \frac{2K}{R} = \frac{2as}{R^2} \]
Step 3: Resultant force
\[ F = \sqrt{F_t^2 + F_c^2} = 2as\left(1+\frac{s^2}{R^2}\right)^{1/2} \] Conclusion: \[ {(B)} \]
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