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 NA and NB be the normal reactions on blocks A and B respectively; MA and MB 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 ,    NA = MA g    and   NB = ( MA + MB ) g . The equation of motion of the block A when x =10 meters is ,    k x - m1 NA = MA aA . Substituting the values and solving we get,    aA = 3.92 m / s 2 The equation of motion of the block B when x =10 meters is ,    m1 NA - m2 NB = ( MA + MB ) aB . Substituting the values and solving we get,    aB = 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 .