COMEDK · Physics · 6. Work Power Energy
A particle of mass \(m\) is moving in a horizontal circle of radius \(r\) under a centripetal force given by \(\left(\dfrac{-K}{r^2}\right)\), where \(K\) is a constant. Then
- A the total energy of the particle is \(\left(\dfrac{-K}{2 r}\right)\)
- B the kinetic energy of the particle is \(\left(\dfrac{K}{r}\right)\)
- C the potential energy of the particle is \(\left(\dfrac{K}{2 r}\right)\)
- D the kinetic energy of the particle is \(\left(\dfrac{-K}{r}\right)\)
Answer & Solution
Correct Answer
(A) the total energy of the particle is \(\left(\dfrac{-K}{2 r}\right)\)
Step-by-step Solution
Detailed explanation
The centripetal force acting on the particle is given by \(F = \dfrac{-K}{r^2}\). The magnitude of the centripetal force is \(F_c = \dfrac{mv^2}{r} = \dfrac{K}{r^2}\). From this, the kinetic energy \(K.E.\) is calculated as…
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