The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying brakes, car A stops after 1000 m a…

Question

The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying brakes, car A stops after 1000 m and car B stops after 1500 m. If FA and FB are the forces applied by the brakes on cars A and B respectively, then the ratio FA/FB is:

1. A. 3/2
2. B. 2/3 (Correct answer)
3. C. 1/3
4. D. 1/2

Correct Answer

Option B — 2/3

Detailed Solution & Explanation

The correct answer is $\dfrac{2}{3}$.

Key Points

• By the work-energy theorem the braking force does work equal to the car's kinetic energy: $F\,S=\Delta KE=KE$ (final KE is zero).
• So $F=\dfrac{KE}{S}$, giving $\dfrac{F_A}{F_B}=\dfrac{KE_A}{KE_B}\times\dfrac{S_B}{S_A}$.
• $\dfrac{F_A}{F_B}=\dfrac{100}{225}\times\dfrac{1500}{1000}=\dfrac{150}{225}=\dfrac{2}{3}$.