Work Energy and Power - Result Question 13
15. A particle of mass $10 g$ 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?
(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]
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Answer:
Correct Answer: 15. (a)
Solution:
- (a) Given: Mass of particle, $M=10 g=$ $\frac{10}{1000} kg$
radius of circle $R=6.4 cm$
Kinetic energy E of particle $=8 \times 10^{-4} J$ acceleration $a_t=$ ?
$\frac{1}{2} mv^{2}=E \Rightarrow \frac{1}{2}(\frac{10}{1000}) v^{2}=8 \times 10^{-4}$
$\Rightarrow v^{2}=16 \times 10^{-2}$
$\Rightarrow v=4 \times 10^{-1}=0.4 m / s$
Now using
$v^{2}=u^{2}+2 a_t s \quad(s=4 \pi R)$
$(0.4)^{2}=0^{2}+2 a_t(4 \times \frac{22}{7} \times \frac{6.4}{100})$
$\Rightarrow a_t=(0.4)^{2} \times \frac{7 \times 100}{8 \times 22 \times 6.4}=0.1 m / s^{2}$
