According to the work-energy theorem, the work done by the braking force is equal to the change in kinetic energy. Thus, for each car, we have:

$ FS = \Delta KE $

This can be rearranged to:

$ -FS = k_f - k_i \quad \Rightarrow \quad FS = k_i - k_f $

For both cars, since the final kinetic energy $ k_f $ is zero when they stop, the equation simplifies to:

$ FS = k_i $

For car $ A $ and car $ B $, the forces $ F_A $ and $ F_B $ applied by the brakes satisfy:

$ F_A \cdot S_A = k_A \quad \text{and} \quad F_B \cdot S_B = k_B $

Solving for the ratio of the forces:

$ \frac{F_A}{F_B} = \frac{k_A}{k_B} \times \frac{S_B}{S_A} $

Substituting the given kinetic energies and stopping distances:

$ \frac{F_A}{F_B} = \frac{100}{225} \times \frac{1500}{1000} $

This simplifies to:

$ = \frac{150}{225} = \frac{2}{3} $

Thus, the ratio of the forces $ \frac{F_A}{F_B} $ is $ \frac{2}{3} $.