Q 11-05-004NEETNEET 2025Top questionEasy
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 $F_A/F_B$ is
Answer: (B) $\dfrac{2}{3}$
By the work-energy theorem, the braking force removes all the kinetic energy: $F \cdot d = K$.
$$F_A = \frac{100}{1000} = 0.1\ \text{N}, \qquad F_B = \frac{225}{1500} = 0.15\ \text{N}$$
$$\frac{F_A}{F_B} = \frac{0.1}{0.15} = \frac{2}{3}$$
Solution by Sreeraj P, M.Sc Physics