A system consisting of two movable blocks and an unstretched spring with a spring constant of 98.1 N/m are connected as shown. At the instant shown, the spring's length is 20 m more than its unstretched length. The coefficient of friction between the two blocks is 0.6 and that between the floor and the lower block is 0.1 . The mass of the blocks A  and B  are 10 Kg and 20 Kg respectively. Find the accelerations of block A and block  B  when the spring's length is 10 m more than its unstretched length

a ) 3.92 m / s ² , 1.47 m / s ²
b ) 3.92 m / s ² , 0.98 m / s ²
c ) 13.73 m / s ² , 0.98 m / s ²
d ) 13.73 m / s ² , 1.47 m / s ²

Let N_A and N_B be the normal reactions on blocks A and B respectively; M_A and M_B the mass of the blocks ; m₁ and m₂ be the coefficients of friction between the blocks and, the block B and the floor respectively ; k , the spring constant (stiffness constant) and x be the elongation of the spring.

Since there is no vertical motion ,
   N_A = M_A g    and   N_B = ( M_A + M_B ) g .

The equation of motion of the block A when x =10 meters is ,
   k x - m₁ N_A = M_A a_A .
Substituting the values and solving we get,
   a_A = 3.92 m / s ²

The equation of motion of the block B when x =10 meters is ,
   m₁ N_A - m₂ N_B = ( M_A + M_B ) a_B .
Substituting the values and solving we get,
   a_B = 0.98 m / s ²

Hence the accelerations of the blocks A and B are
3.92 m / s ² and 0.98 m / s ² respectively .