Two bodies m₁ and m₂ are kept on a table with coefficient of friction ‘μ’ and are joined by a spring.  Initially, the spring is in its relaxed state. The minimum constant force F which will make the other block m₂ move is ( k is the spring constant).

Motion of  m₂ starts when,  kx = μ m₂.g    

Where  x = elongation in the spring

The minimum force will be such that m₁ has no kinetic energy.
    Applying work energy principle for m₁

Alternative solution :

Minimum spring force required to move the block m₂ = µ m₂ g

.: kx = µ m₂ g ....................(1)

For minimum firce F , kinetic energy of the block will be negligibly small.

.: F X x = work done against frictional force on m₁ + Elastic P.E strored .

= µ m₁g X x + 1/2 kx²

.: kx = 2 F - 2 µ m₁ g ....................(2)

From (1) and (2), F =(µ m₁ g + 1/2 µ m₂ g)