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An object starting from rest has two forces acting on it: one performing 40 J of work and the other (friction) performing -20J. What is the final kinetic energy of this object? |
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Answer» Solution :The total work DONE is (40J)+(-20J)=20J. So, by the work-energy THEOREM, `W_("total")=DeltaK`, we have 20J=`DeltaK`. Since `DeltaK=K_(f)-K_(f)-K_(i)`, we FIND that `K_(f)=K_(i)+DeltaK=0J+20J=20J`. |
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