A block of mass m is suspened through a spring of spring constant k and is in equlibrium. A sharp blow gives the block an initial downward velocity v. How far below the equilibrium psitin, the block comes to an instantaneous rest?

Explanation

`l_0`: Natural length of spring when the block is in position (1),
The block is in equilibrium `mg=kx_0impliesx_0=(mg)/(k)`
Now the block is given velocity `v_0` in upward direction.
Taking equilibrium position as the reference line between (1) and (2) mechanical energy conservation
`(1)/(2)mv_0^2+(1)/(2)kx_0^2=(1)/(2)k(h-x_0)^2+mgh`
`=(1)/(2)kh^2+(1)/(2)kx_0^2-khx_0+mgh`
`(1)/(2)mv_0^2=(1)/(2)kh^2-khxx(mg)/(k)+mgh`
`=(1)/(2)kh^2`
`h=v_0sqrt((m)/(k))`
(b)
`(1)/(2)mv_0^2+(1)/(2)kx_0^2=(1)/(2)k(h+x_0)^2-mgh`
Solving as above
`h=v_0sqrt((m)/(k))`