Two blocks A and B of mass 10 kg and 20 kg respectively are placed as shown in figure. Coefficient of friction between all the surfaces is 0.2 (g = 10 m/s2) Then,

Free body diagram of A isN = normal reactionTcos⁡30∘=N               .......(1)Tsin⁡30∘=0.2N+100  .......(2)Solving these two equations we getT≈306Nand  N≈265NFree body diagram of B is a=mBg−02N−02NmB=200−0.4(265)20=4.7m/s2