Question

Two blocks of mass m and M are placed on a table with coefficient of friction μ. The blocks are joined by a spring of spring constant k. The minimum force F applied to B which just makes A to move (See fig) is

(1) μ(M + m/2)g

(2) μ(M - m)g

(3) μMg - μmg/2

(4) μg(M + m/2)

Answer 1

(4) μg(M + m/2)

Mass m (or body A) will be on brink of moving when the force applied by the spring (i.e. kx) is just equal to the force of limiting friction between A and surface in contact. Therefore

kx = μmg .....(1)

x is the elongation in the spring

x = μmg/k .....(2)

The force will be minimum for M when it has no kinetic energy. Applying work energy theorem to M, we get 

Work done = Change in kinetic energy