Medium
Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying a retarding force of the same magnitude. How would the distance moved by them before coming to rest compare?
(C) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the force: $\Delta K = W$.
Since the bodies are brought to rest,the work done by the retarding force $F$ over a distance $d$ is $W = F \cdot d$.
Given that the initial kinetic energy $K$ is the same for both bodies and the final kinetic energy is $0$,we have $\Delta K = K - 0 = K$.
Thus,$K = F \cdot d$,which implies $d = K / F$.
Since both the kinetic energy $K$ and the retarding force $F$ are the same for both bodies,the distance $d$ moved by them before coming to rest will be equal.
Since the bodies are brought to rest,the work done by the retarding force $F$ over a distance $d$ is $W = F \cdot d$.
Given that the initial kinetic energy $K$ is the same for both bodies and the final kinetic energy is $0$,we have $\Delta K = K - 0 = K$.
Thus,$K = F \cdot d$,which implies $d = K / F$.
Since both the kinetic energy $K$ and the retarding force $F$ are the same for both bodies,the distance $d$ moved by them before coming to rest will be equal.
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