The kinetic energies of two identical cars $A$ and $B$ are $200 \mathrm{~J}$ and $300 \mathrm{~J}$ r

Question

The kinetic energies of two identical cars $A$ and $B$ are $200 \mathrm{~J}$ and $300 \mathrm{~J}$ respectively. On applying brakes, car $A$ stops after $400 \mathrm{~m}$ and car $B$ stops after $500 \mathrm{~m}$. If $F_A$ and $F_B$ are the retarding forces applied by the brakes on cars $A$ and $B$ respectively, then the ratio $F_A / F_B$ is

Professor PC Thomas Classes · Answer

(c) : According to the work-energy theorem, the magnitude of work done by the retarding force equals the initial kinetic energy: $F 	imes S = KE$. For car $A$, $F_A 	imes S_A = KE_A \implies F_A = \frac{200}{400} = 0.5 \mathrm{~N}$. For car $B$, $F_B 	imes S_B = KE_B \implies F_B = \frac{300}{500} = 0.6 \mathrm{~N}$. Therefore, the ratio of the forces is $\frac{F_A}{F_B} = \frac{0.5}{0.6} = \frac{5}{6}$.