Q. 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
A) $\frac{3}{2}$
B) $\frac{2}{3}$
C) $\frac{1}{3}$
D) $\frac{1}{2}$
Explanation
[Solution by Claude - not from source] Work-energy theorem: $F d=KE$. $F_A=100/1000=0.1$ N, $F_B=225/1500=0.15$ N. Ratio $=2/3$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.