NEET2025Physics-Kinetic Energy
NEET 2025 Physics Work and Energy MCQ Question
Type: MCQ-numerical-Easy-Class 11
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 Fₐ and F_B are the forces applied by the breaks on cars A and B, respectively, then the ratio Fₐ/F_B is
A
3/2
B
2/3
C
1/3
D
1/2
Correct Answer
Option B
Detailed Explanation
Using the work-energy principle, the work done by the brakes is equal to the initial kinetic energy. Therefore, Fₐ × 1000 = 100 J and F_B × 1500 = 225 J. Solving for the ratio Fₐ/F_B, we get (100/1000)/(225/1500) = 2/3.
Found an issue with this question?