InterviewSolution
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InterviewSolution
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A paritcal of mass `10 g` moves along a circle of radius `6.4 cm` with a constant tangennitial acceleration. What is the magnitude of this acceleration . What is the magnitude of this acceleration if the kinetic energy of the partical becomes equal to `8 xx 10^(-4) J` by the end of the second revolution after the beginning of the motion?A. `0.2 m//s^(2)`B. `0.1 m//s^(2)`C. `0.15 m//s^(2)`D. `0.18 m//s^(2)` |
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Answer» `K = (1)/(2) mv^(2) implies 8 xx 10^(-4) = (1)/(2) xx 10 xx 10^(-3) v^(2)` `implies v^(2) = 0.16` `v^(2) = u^(2) + 2 a_(t) (2 xx 2 piR)` `0.16 = 2 a_(1) (4pi xx 6.4 pi 10^(-2))` `a_(t) = (10)/(32 pi) = 0.1 m//s^(2)` |
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