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NEET 2025 · Work-Energy Theorem

Work-Energy Theorem question from NEET 2025

Q5NEET 2025Easy

Work-Energy Theorem

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 $F_A$ and $F_B$ are the forces applied by the brakes on cars $A$ and $B$ respectively, then the ratio of $\frac{F_A}{F_B}$ is

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Answer: (d)

Work-energy theorem: braking force $\times$ stopping distance $=$ initial KE, so $F = \frac{K}{d}$ .
$F_A = \frac{100}{1000} = 0.1\text{ N}$ , $F_B = \frac{225}{1500} = 0.15\text{ N}$
$\frac{F_A}{F_B} = \frac{0.1}{0.15} = \frac{2}{3}$
So the answer is (d) $\frac{2}{3}$ .

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