question_answer

A body moving at 2 m/s can be stopped over a distance x. If the kinetic energy of the body is doubled, how long will it go before coming to rest, if the retarding force remains unchanged?

Answer:

               

Case-I	Case-II
\[{{u}_{1}}=2m/s\]b	\[{{u}_{2}}=?\]
\[{{S}_{1}}=x\]	\[{{S}_{2}}=?\]
\[K.E{{.}_{1}}=K.E{{.}_{1}}\]	\[{{S}_{2}}=?\]
\[K.E{{.}_{1}}=K.E{{.}_{1}}\]	\[K.E{{.}_{2}}=2\]
\[K.E{{.}_{1}}\]

We know, \[K.E.=\frac{1}{2}m{{V}^{2}}\] Applying the formula for both cases, we get \[K.E.=\frac{1}{2}m_{1}^{2}\]     ????. (1) \[K.E{{.}_{2}}=\frac{1}{2}m\] \[u_{2}^{2}\]                     ????.(2) On dividing equation (2) by (1), we get \[\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}=\frac{\frac{1}{2}mu_{2}^{2}}{\frac{1}{2}mu_{1}^{2}}\] \[2=\frac{u_{2}^{2}}{{{2}^{2}}}\Rightarrow \,\,u_{2}^{2}=8\] \[{{u}_{2}}=\sqrt{8}\,m/s\] We know, \[{{v}^{2}}-{{u}^{2}}=-2as\](\[\because \]the body is retarding) Applying the formula for both cases, we get \[\Rightarrow {{0}^{2}}-u_{1}^{2}=-2ax\Rightarrow -2ax\Rightarrow u_{1}^{2}=2ax\] \[2a=\frac{u_{1}^{2}}{x}=\frac{{{(2)}^{2}}}{x}=\frac{4}{x}\]                          ????(1) \[{{0}^{2}}-u_{2}^{2}=-2a{{s}_{2}}\] \[{{S}_{2}}=\frac{u_{2}^{2}}{2a}=\frac{u_{2}^{2}}{\frac{u_{1}^{2}}{x}}\]                                [from equation (1)] \[{{S}_{2}}=\frac{{{(\sqrt{8})}^{2}}}{\frac{{{2}^{2}}}{x}}=\frac{8x}{4}\] \[\therefore \,{{S}_{2}}=2x\]