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

To find the ratio of the braking forces, we can use the work-energy theorem. According to this theorem, the work done by the braking force on a car is equal to the change in its kinetic energy.

For a car that is brought to a stop, the initial kinetic energy is completely dissipated by the work done by the braking force:
$W=\Delta K$
Since the final kinetic energy is 0 (as the cars stop), the work done by the braking force $F$ over a stopping distance $d$ is:
$F·d=K_{\mathrm{initial}}$

Let's write this equation for both cars, A and B:
For car A:
$F_A·d_A=K_A$
For car B:
$F_B·d_B=K_B$

We are given the following values:
Kinetic energy of car A, $K_A=100\text{J}$
Kinetic energy of car B, $K_B=225\text{J}$
Stopping distance of car A, $d_A=1000\text{m}$
Stopping distance of car B, $d_B=1500\text{m}$

Now, let's solve for the braking forces:
$F_A=\frac{K_A}{d_A}=\frac{100}{1000}=0.1\text{N}$
$F_B=\frac{K_B}{d_B}=\frac{225}{1500}=0.15\text{N}$

Next, we find the ratio of the braking force of car A to that of car B:
$\frac{F_A}{F_B}=\frac{0.1}{0.15}$
Simplifying the fraction by multiplying the numerator and the denominator by 100:
$\frac{F_A}{F_B}=\frac{10}{15}=\frac{2}{3}$

Thus, the ratio of the forces applied by the brakes is 2/3.