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NEET UGPhysicsModerateNEET 2025

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 FAF_A and FBF_B are the forces applied by the breaks on cars A and B respectively, then the ratio of FAFB\frac{F_A}{F_B} is

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Sol. By work-energy theorem,
FS=ΔK.EFS = \Delta K.E
⇒−FS=kf−ki\Rightarrow -FS = k_f - k_i
⇒FS=ki−kf\Rightarrow FS = k_i - k_f
⇒FAFB=kAkB×SBSA\Rightarrow \frac{F_A}{F_B} = \frac{k_A}{k_B} \times \frac{S_B}{S_A}
=100225×15001000= \frac{100}{225} \times \frac{1500}{1000}
=150225=23= \frac{150}{225} = \frac{2}{3}

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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