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If a body of mass m suspended by a spring comed to rest after a downward displecemtn `y_(0)`, claculate (a) force constant of the spring, (b) loss in gravitational potential energy, (c) gain in elastic potential energy of spring. |
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Answer» (a) In equilibrium ,`mg=Ky_(0)` `K=(mg)/(y_(0))` (b) As mass descends through a distance `y_(0)`, `:.` loss in gravitational P.E. `=mg y_(0)`. (c) As the spring is stretched by `y_(0)`, therefore, gain in elastice potenntial energy `P.E. =(1)/(2)Ky_(0)^(2)=(1)/(2)((mg)/(y_(0)))y_(0)^(2) =(1)/(2)mg y_(0)` |
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