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 .