A block of mass m hangs on a vertical spring. Initially the spring is unstreched, it is now allowed to fall from rest. Find (a) the distance the block falls if the block is released slowely, (b) the maximum distance the block falls before it beigns to move up.

A block of mass m hangs on a vertical spring. Initially the spring is

1.	A block of mass m hangs on a vertical spring. Initially the spring is unstreched, it is now allowed to fall from rest. Find (a) the distance the block falls if the block is released slowely, (b) the maximum distance the block falls before it beigns to move up.
Answer» Solution :When the block FALLS slowely, it comes to rest at a distance `y_(0)`, which is referred to as the equlibrium POSITION. From, condition of equilibrium, `sumF_(y)=ky_(0)-MG=0 ory_(0)=(mg)/(k)` (b) When the block is released suddenly, it oscilates about the equilibrium position. Initially the speed of the block INCREASES then reaches maximum value and then DECREASES to zero at the lowest position. In this situation the block oscillates about the equilibrium position. The block is released from rest, there fore its total mechanical energy initially. `E_(i)=U_(g)+U_(s)+K=0+0+0` Final total mechanical energy, `E_(f)=-mgy_(m)+(1)/(2)ky^(2)+0` From conservation of energy, `E_(i) = E_(f)` `0=-mgy_(m)+(1)/(2)ky^(2)ory_(m)=(2mg)/(k)=2y_(0)`