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160WA stone is dropped from a height, h. Its potential energy at a particular instant is nine times its kineticenergy. When the velocity of the stone becomes double the velocity at that instant, what will be theratio of its new kinetic energy and new potential energy?[2/31 |
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Answer» As body is dropped from a height h, its energy = m g hWhen its comes down through a distance x then its potential energy = m g ( h - x )So KEattained has to be m g x because of conservation of energyGiven PE = 9 * KE==> m g ( h - x) = 9 * m g xSo x =h/10As velocity is to be two times of the velocity attained at this location then newKEwould be 4 * oldKEHence KE(New) = 4 * m g x = 4 * m g * h/10OR 2/5 * mg hIf so, then PE = 3/5 m g h (as per conservation of energy)Hence the required ratio of new KE and newPE =2 : 3 |
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