Class 11PhysicsWork, Energy, Power and CollisionWork Energy Theorem and Conservation of Mechanical Energy
MediumTwo 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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