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Topic: Kinetic and potential energy, work-energy theorem· Work, Energy and Power

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…

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

1. \(\frac{3}{2}\)

2. \(\frac{2}{3}\)

3. \(\frac{1}{3}\)

4. \(\frac{1}{2}\)

1 answer

Best answer

Answer: 2. \(\frac{2}{3}\)

Key idea: by the work-energy theorem, the brake force does work equal to the kinetic energy lost: \(F\,d = K\).

$$F_A=\frac{100}{1000}=0.1\ \mathrm{N},\qquad F_B=\frac{225}{1500}=0.15\ \mathrm{N}$$$$\frac{F_A}{F_B}=\frac{0.1}{0.15}=\frac23$$
Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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