SATHEE: Work Energy and Power - Result Question 13

Work Energy and Power - Result Question 13

15. A particle of mass $10 g$ moves along a circle of radius $6.4 c m$ 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?

(a) $0.1 m / s^{2}$

(b) $0.15 m / s^{2}$

(c) $0.18 m / s^{2}$

(d) $0.2 m / s^{2}$

[2016]

Show Answer

Answer:

Correct Answer: 15. (a)

Solution:

(a) Given: Mass of particle, $M = 10 g =$ $\frac{10}{1000} k g$

radius of circle $R = 6.4 c m$

Kinetic energy E of particle $= 8 \times 10^{- 4} J$ acceleration $a_{t} =$ ?

$\frac{1}{2} m v^{2} = E \Rightarrow \frac{1}{2} (\frac{10}{1000}) v^{2} = 8 \times 10^{- 4}$

$\Rightarrow v^{2} = 16 \times 10^{- 2}$

$\Rightarrow v = 4 \times 10^{- 1} = 0.4 m / s$

Now using

$v^{2} = u^{2} + 2 a_{t} s (s = 4 π R)$

$(0.4)^{2} = 0^{2} + 2 a_{t} (4 \times \frac{22}{7} \times \frac{6.4}{100})$

$\Rightarrow a_{t} = (0.4)^{2} \times \frac{7 \times 100}{8 \times 22 \times 6.4} = 0.1 m / s^{2}$

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