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

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$0.15 m s^{- 2}$

$0.2 m s^{- 2}$

$0.18 m s^{- 2}$

$0.1 m s^{- 2}$

answer is D.

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Detailed Solution

$W_{Tangential} = Δ K E = K_{f} - K_{i}$

$\frac{1}{2} M V^{2} = 8 \times 10^{- 4} \frac{1}{2} \times 10 \times 10^{- 2} \times V^{2} = 8 \times 10^{- 4} V^{2} = 16 \times 10^{- 2} - - - (1) V^{2} - U^{2} = 2 a S$

$a_{T} = \frac{16 \times 10^{- 2}}{2 (2 π r \times 2)} = \frac{16 \times 10^{- 2}}{8 π \times 6.4 \times 10^{- 2}} = 0.1 m s^{- 2}$ .

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