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 \(?\) .............. \(\mathrm{m} / \mathrm{s}^{2}\)

\(\begin{gathered}
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,Here,\,m = 10\,g\, = {10^{ - 2}}\,kg, \hfill \\
  R = 6.4\,cm = 6.4 \times {10^{ - 2}}m,\,{K_f} = 8 \times {10^{ - 4}}\,J, \hfill \\
  {K_i} = 0,\,{a_{t = }}? \hfill \\
  {\text{Using}}\,work\,energy\,theorem, \hfill \\
  Work\,done\,by\,all\,the\,forces = Change\,in\,KE \hfill \\
  {W_{\tan gential\,force\,}} + {W_{centripetal\,force}} = {K_f} - {K_i} \hfill \\
   \Rightarrow {a_t} = \frac{{{K_f}}}{{4\pi Rm}} = \frac{{8 \times {{10}^{ - 4}}}}{{4 \times \frac{{22}}{7} \times 6.4 \times {{10}^{ - 2}} \times {{10}^{ - 2}}}} \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\, = 0.099 \approx 0.1\,m\,{s^{ - 2}} \hfill \\ 
\end{gathered} \)