The kinetic energies of two similar cars $A$ and $B$ are $100$ J and $225$ J respectively. On applying brakes, 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 brakes on cars $A$ and $B$ respectively, then the ratio of $\frac{F_A}{F_B}$ is

We are asked to find the ratio \(\frac{F_A}{F_B}\) of the braking forces applied to two similar cars A and B.

When a car brakes uniformly, the work done by the braking force equals the loss of kinetic energy:

$W=F_{A}d$_A = K

Similarly for car B:

$W=F_{B}d$_B = K

Given data:

• Car A: \(K_A = 100\ \text{J}\), stopping distance \(d_A = 1000\ \text{m}\).
• Car B: \(K_B = 225\ \text{J}\), stopping distance \(d_B = 1500\ \text{m}\).

Since the cars come to rest, the change in kinetic energy \(\Delta K\) is simply the initial kinetic energy (the final kinetic energy is zero). Therefore:

$F_A=1001000=0.1 N$

$F_B=2251500=0.15 N$

Now compute the ratio:

$\frac{F_A}{F_B}=0.10.15=23$

Therefore, the ratio \(\frac{F_A}{F_B}\) is \(\displaystyle \frac{2}{3}\).