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(D) The work-energy theorem states that the work done by the braking force $F$ is equal to the change in kinetic energy: $W = \Delta K$. Since the fin

Vedclass · Accepted Answer

$4 S_1 = 3 S_2$

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(D) The work-energy theorem states that the work done by the braking force $F$ is equal to the change in kinetic energy: $W = \Delta K$. Since the final velocity is $0$, the work done by the braking force is $F \cdot S = \frac{1}{2} m v^2$. Thus, the stopping distance $S = \frac{m v^2}{2 F}$. Since $v$ and $F$ are constant, $S \propto m$. In the first case, the total mass $m_1 = 80 \ kg + 100 \ kg = 180 \ kg$. So, $S_1 = \frac{180 v^2}{2 F}$. In the second case, the total mass $m_2 = 80 \ kg + 100 \ kg + 60 \ kg = 240 \ kg$. So, $S_2 = \frac{240 v^2}{2 F}$. Taking the ratio: $\frac{S_1}{S_2} = \frac{180}{240} = \frac{3}{4}$. Therefore, $4 S_1 = 3 S_2$.

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