Given, mass of block A and B are (mA),(mB) be 3kg and 7kg, respectively. Coefficient of friction between block A and B,B and floor (µA),(µB) be 0.4 and 0.55 respectively, Applied force (F) on block B be 50N. Let fA,fB be the friction forces between blocks A,B and B, floor respectively.
Now, from block B diagram, fB=µBNB . . . (i) where, NB be the normal reaction on block B ‌‌=µB(mA+mB)g ‌‌=0.55(3+7)10=55N Similarly, fA=µANA where, NA is normal reaction on block A. ∴‌‌fA‌‌=µAmAg fA‌‌=0.4×3×10 ‌‌=12N Now, since, fB>f ∴ the block B will not move. Hence, fA=0N