Consider the situation as shown. Initially the spring is unstretched when the block of mass m is released from rest assuming the pulley frictionless and light the spring and string massless. Find the extension of sprinig when the block is in equilibrium also find the maximum extension of the spring.

Explanation

when the block is in equilibrium thenet force on it is zero.
In equilibrium `mg=kx_(0)`
`x_(0)=(mg)/(k)`
Where `x_(0)` is an extension of spring when the block is in equilibrium.
Maximum extension of spring let at some isntant the extension of spring be x.
ltbgt `mg-kx=ma`
`a=g-(k)/(m)x` ltbgt `v(dv)/(dx)=g-(k)/(m)x`
`int_(0)^(v)intvdv=underset(0)overset(x)int(g-(k)/(m)x)dx`
`|(v^(2))/(2)|_(0)^(v)=|gx-(kx^(2))/(2m)|_(0)^(x)`
`(v^(2))/(2)=gx-(kx^(2))/(2m)`
`v=(2gx-(kx^(2))/(m))^((1)/(2))`
where v is the velocity of the block as the function of eextension of spring
the block will stop when
`v=0`
`(2gx-(kx^(2))/(m))^(1//2)=0`
`x=x_(m)=(2mg)/(k)`
where `x_(m)` is the maximum extension in the spring.