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 and are the forces applied by the brakes on cars A and B, respectively, then the ratio is

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Solution

Q.
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 $F_{A} / F_{B}$ is [2025]

1 $\frac{1}{3}$

2 $\frac{1}{2}$

3 $\frac{3}{2}$

4 $\frac{2}{3}$

Ans.
(4)

According to work - energy theorem,

Work done = change in kinetic energy

$W = Δ K E$

For car A, $W_{A} = {(K E)}_{A}$

$F_{A} S_{A} = {(K E)}_{A}$ ...(i)

Similarly for car B, $F_{B} S_{B} = {(K E)}_{B}$ ...(ii)

From eqn. (i) and (ii)

$\frac{{(K E)}_{A}}{{(K E)}_{B}} = \frac{F_{A} S_{A}}{F_{B} S_{B}} \Rightarrow \frac{100}{225} = \frac{F_{A}}{F_{B}} \times \frac{1000}{1500} \Rightarrow \frac{F_{A}}{F_{B}} = \frac{2}{3}$

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