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TS EAMCET · Physics · Work Power Energy

A particle of mass m is moving along a circle of radius R such that its tangential acceleration at varies with distance covered x as at=αx2, where α is a constant. The kinetic energy K of the particle varies with the distance as K=βxc, where β and c are constants. The values of β and c are

  1. A β=mα3;c=3
  2. B β=mα4;c=4
  3. C β=mα2;c=4
  4. D β=mα2;c=3
Verified Solution

Answer & Solution

Correct Answer

(A) β=mα3;c=3

Step-by-step Solution

Detailed explanation

Given, Tangential acceleration, at=αx2 and Kinetic energy, K=βxc. If F is the tangential force acting on the particle thenF=mαx2 Now applying work energy theorem, W=∆KE ⇒  ∫F dx=K ⇒  ∫mαx2dx=K…
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