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The kinetic energies of two similar cars A and B are $100\text{ J}$ and $225\text{ J}$ respectively. On applying brakes, car A stops after $1000\text{ m}$ and car B stops after $1500\text{ m}$. If $F_A$ and $F_B$ are the forces applied by the brakes 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
From work-energy theorem, $F = \frac{KE}{s}$. Therefore, $F_A / F_B = \frac{100/1000}{225/1500} = \frac{0.10}{0.15} = \frac{2}{3}$.
Detailed Solution
According to the Work-Energy Theorem, work done by the braking force equals the change in kinetic energy: $W = F \cdot s = \Delta KE \implies F = \frac{KE}{s}$. For car A: $F_A = \frac{100\text{ J}}{1000\text{ m}} = 0.10\text{ N}$. For car B: $F_B = \frac{225\text{ J}}{1500\text{ m}} = 0.15\text{ N}$. The ratio is $\frac{F_A}{F_B} = \frac{0.10}{0.15} = \frac{2}{3}$.