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 FA and FB are the forces applied by the breaks on cars A and B respectively, then the ratio of F_A/F_Bis

To find the ratio of the braking forces applied to the two cars, 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.

When breaks are applied to stop a car, the final kinetic energy is zero, so the work done by the retarding force (braking force) equals the initial kinetic energy of the car:
$W=F\cdot d=K$
where:
- $F$ is the braking force applied,
- $d$ is the stopping distance, and
- $K$ is the initial kinetic energy of the car.

Therefore, the braking force can be expressed as:
$F=\frac{K}{d}$

Let us write this relation for both cars A and B:
For car A:
$F_A=\frac{K_A}{d_A}$
Given: $K_A=100\text{J}$ and $d_A=1000\text{m}$.
$F_A=\frac{100}{1000}=0.1\text{N}$

For car B:
$F_B=\frac{K_B}{d_B}$
Given: $K_B=225\text{J}$ and $d_B=1500\text{m}$.
$F_B=\frac{225}{1500}=0.15\text{N}$

Now, we find the ratio of the braking forces $F_A/F_B$:
$\frac{F_A}{F_B}=\frac{K_A/d_A}{K_B/d_B}=\frac{0.1}{0.15}=\frac{10}{15}=\frac{2}{3}$

Thus, the ratio of the forces $F_A/F_B$ is 2/3.