hey ramyani

wen both blocks are given a gentle push towards right,the frictional force on the upper block acts left and not right..

 

 

let f1 be frictioanl force on upper block.it is towards left

-(mg) = ma1 (force is in opp direction)

 

  a1 = - g

 

f1 acts towards right for the lower block of mass 2m

f2 due to ground acts leftwards

 

(mg) - (m+2m)g/2 = 2ma2

 

a2 = (g)/2 - (3g)4 = a2

 

a1 - a2 = - g -- g/2 +3 g/4

 

 =--3 g/2 +3 g/4

= - 3 g/4

3 g/4 in opp direction

 

rate me if im rite