The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying b

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The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying breaks, car $A$ stops after 1000 m and car $B$ stops after 1500 m . If $F_{A}$ and $F_{B}$ are the forces applied by the breaks on cars $A$ and $B$ respectively, then the ratio of $\frac{F_{A}}{F_{B}}$ is

Accepted Answer

By work-energy theorem,$$\begin{matrix} & FS = \Delta K \cdot E \ & \ \Rightarrow - FS = k_{f} - k_{i} \ & \ \Rightarrow FS = k_{i} - k_{f} \ & \ \Rightarrow \frac{F_{A}}{F_{B}} = \frac{k_{A}}{k_{B}} 	imes \frac{S_{B}}{S_{A}} \ & \ = \frac{100}{225} 	imes \frac{1500}{1000} \ & \ = \frac{150}{225} = \frac{2}{3} \end{matrix}$$

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