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

Explanation

(a) When the block falls slowely, it comes to rest at a distance `y_(0)`, which is referred to as the equilibrium position. From, condition of equilibrium,
`sum F_(y)=ky_(0)-mg=0 " or " y_(0)=(mg)/(k)`

(b) When the block is released suddenly, it oscillates 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)+KE = 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)` or `y_(m)=(2mg)/(k)=2y_(0)`