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 springThe minimum force will be such that m1 has no kinetic energy.    Applying work energy principle for m1Alternative 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)