Answer:
(i) \[u=54kmph=54\times \frac{5}{18}=15\,m/s\] \[v=36\,kmph=36\times \frac{5}{18}=10\,\,m/s\] \[S=20\,cm=\frac{20}{100}=0.2\,m\] \[m=15\,g=15\times {{10}^{-3}}kg\] \[F=?\] We know, \[W=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}}\] \[\Rightarrow F\times S=\frac{1}{2}m({{v}^{2}}-{{u}^{2}})\] \[\Rightarrow F\times S=\frac{1}{2}\times 15\times {{10}^{-3}}[{{10}^{2}}-{{15}^{2}}]\] \[=\frac{1}{2}\times 15\times {{10}^{-3}}\times (-125)\] \[\therefore F=\frac{1}{2}\times \frac{15\times {{10}^{-3}}\times (-125)}{0.2}\] \[=-0.5\times 15\times 125\times {{10}^{-3}}N=-4.687\,\,N\] The negative sign indicates the force is retarding force. (ii) If \[u=15\,m/sv=0\,m/s\] \[S=?F=-4.7\,N\] \[F\times S=\frac{1}{2}m({{v}^{2}}-{{u}^{2}})\] \[\Rightarrow S=\frac{1}{2}\times \frac{15\times {{10}^{-3}}\times (-125)}{-4.687}=360\,m\]
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