Find the maximum energy stored in the spring shown in figure, for which the block remains stationary on the rough horizontal surface.

Explanation

Let the spring be compressed by `x`. The potential energy stored in the spring `=U=(1//2)Kx^2` for maximum potential energy in the spring, this deformation has to be maximum to ensure maximum spring force `Kx`. Since the block is in equilibrium along horizontal, `sumf_x=0impliesf-Kx=0` where `f=fimpliesfor Kx` to be maximum the frictional force must be maximum. The maximum frictional force required to keep the block in equilibrium=limiting frictional force.
`N-mg=0impliesN=mg`
Therefore, `f_(max)=mumg`
`kx_(max)=mumgimpliesx_(max)=(mumg)/(k)`
The maximum potential energy stored in the spring
`1/2kx_(max)^2=1/2k((mumg)/(k))^2=(mu^2m^2g^2)/(2k)`