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If a body of mass m suspended by a spring comes to rest after a downward displacement `y_(0)` find (a) the force constant of the spring, (b) loss in gravitational potential energy and (c) gain in elastic potential energy. |
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Answer» (a) For equilibrium `ky_(0)=mg` i.e, `k=mg//y_(0)` (b) As the mass m has descended a distance `y_(0)` loss in potential energy `=mgy_(0)` (c) As the spring is stretched by `y_(0)` gain in elastic energy `=1/2ky_(0)^(2)=1/2mgy_(0)` [as from Eqn. (a)`k=(mg)/y_(0)`] |
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