Work Energy And Power Question 249
Question: . 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 1{0^{-4}}J $ by the end of the second revolution after the beginning of the motion?
Options:
A) $ 0.1 m/s^{2} $
B) $ 0.15 m/s^{2} $
C) $ 0.18 m/s^{2} $
D) $ 0.2 m/s^{2} $
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Answer:
Correct Answer: A
Solution:
[a] Given: Mass of particle,
$ M=10g=\frac{10}{1000}kg $
radius of circle $ R=6.4cm $
Kinetic energy E of particle $ =8\times 1{0^{-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}+2a _{t}S~~~~~( s=4\pi R ) $
$ {{( 0.4 )}^{2}}=0^{2}+2a _{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}=0.1m/s^{2} $
