Question: 5. 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
(1) \dfrac{3}{2}
(2) \dfrac{2}{3}
(3) \dfrac{1}{3}
(4) \dfrac{1}{2}
Answer: Option (2)
Explanation:
Use the work–energy principle: the work done by the braking force equals the initial kinetic energy. For car A,
F_{A}\times d_{A} = K_{A}so
F_{A} = \dfrac{K_{A}}{d_{A}} = \dfrac{100\ \text{J}}{1000\ \text{m}}For car B,
F_{B}\times d_{B} = K_{B}so
F_{B} = \dfrac{K_{B}}{d_{B}} = \dfrac{225\ \text{J}}{1500\ \text{m}}Now compute the ratio:
\dfrac{F_{A}}{F_{B}} = \dfrac{100/1000}{225/1500} = \dfrac{100}{1000}\times\dfrac{1500}{225} = \dfrac{1}{10}\times\dfrac{1500}{225} = \dfrac{1}{10}\times\dfrac{1500\div75}{225\div75} = \dfrac{1}{10}\times\dfrac{20}{3} = \dfrac{2}{3}Therefore \dfrac{F_{A}}{F_{B}} = \dfrac{2}{3}, which is Option (2).