The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying b

Question

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

Professor PC Thomas Classes · Answer

(d) : According to work-energy theorem, \ Work done $=$ change in kinetic energy $ W=\Delta K E $ \ For car $A, W_A=(K E)_A$ $ F_A S_A=(K E)_A $ \Similarly for car $B, F_B S_B=(K E)_B$ \ 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} 	imes \frac{1000}{1500} \Rightarrow \frac{F_A}{F_B}=\frac{2}{3} $