A particle of mass 10g moves along a circle of radius 6.4cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8×10-4 J by the end of the second revolution after the beginning of the motion?

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0.18 m/s2

0.2 m/s2

0.1 m/s2

0.15 m/s2

answer is C.

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Detailed Solution

Here, m=10 g=10-2 kgR=6.4 cm=6.4×10-2 m,Kf=8×10-4 J,Ki=0,at=?Using work-energy theorem, Work done by all the forces = Change in KEWtangential force +Wcentripetal force =Kf-Ki⇒ Ft×s+0=Kf-0⇒mat×(2×2πR)=Kf⇒at=Kf4πRm=8×10-44×227×6.4×10-2×10-2=0.099≈0.1 m s-2

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