Q90 / 180 · NEET 2025 · Physics

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 $\mathrm{F}_{\mathrm{A}}$ and $\mathrm{F}_{\mathrm{B}}$ are the forces applied by the breaks on cars A and B , respectively, then the ratio $\mathrm{F}_{\mathrm{A}} / \mathrm{F}_{\mathrm{B}}$ is :

  • A
    $\frac{3}{2}$
  • Correct answer
    B
    $\frac{2}{3}$
  • C
    $\frac{1}{3}$
  • D
    $\frac{1}{2}$

Solution

$\begin{aligned} & \because \text { W.D. }=\Delta \mathrm{KE} \\ \end{aligned}$ $\overrightarrow{\mathrm{F}} . \overrightarrow{\mathrm{S}}=\Delta \mathrm{KE}$ $\begin{aligned} & \frac{(\Delta K E)_A}{(\Delta K E)_B}=\frac{-\mathrm{F}_A S_A}{-\mathrm{F}_{\mathrm{B}} \mathrm{~S}_{\mathrm{B}}} \\ & \frac{-100}{-225}=\frac{-\mathrm{F}_{\mathrm{A}}(1000)}{-\mathrm{F}_{\mathrm{B}}(1500)} \\ & \frac{\mathrm{F}_{\mathrm{A}}}{\mathrm{~F}_{\mathrm{B}}}=\frac{2}{3} \end{aligned}$

Hence, the correct answer is option: 'B'.

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