Two blocks having masses 8 kg and 16 kg are connected to the two ends of a light spring. The system is placed on a smooth horizontal floor. An inextensible string also connects B with ceiling as shown in figure at the initial moment. Initially the spring has its natural length.A constant horizontal force F is applied to the heavier block as shown. What is the maximum possible value of F so that lighter block doesn't loose contact with ground. (##NAR_NEET_PHY_XI_P2_C06_E11_012_Q01.png" width="80%">

Two blocks having masses 8 kg and 16 kg are connected to the two ends

1.	Two blocks having masses 8 kg and 16 kg are connected to the two ends of a light spring. The system is placed on a smooth horizontal floor. An inextensible string also connects B with ceiling as shown in figure at the initial moment. Initially the spring has its natural length.A constant horizontal force F is applied to the heavier block as shown. What is the maximum possible value of F so that lighter block doesn't loose contact with ground. (##NAR_NEET_PHY_XI_P2_C06_E11_012_Q01.png" width="80%">
Answer» 10 N 20 N 30 N 40 N Solution :Draw FBD to B to get extension in spring. When block B just looses contact with ground resultant force on it is zero. ` (##NAR_NEET_PHY_XI_P2_C06_E08_012_S01.png" width="80%"> `Kx - T COS THETA = 0` `Rightarrow T = Kx/(cos theta), T sin theta + N -mg = 0` When N = 0 then `T sin theta = mg Rightarrow Kx/(cos theta)sin theta = mg ` `x = mg /(K TAN theta) = (80)/(K XX(4/3)) = 60/K` If spring has to just extend till this value then from WORK energy theorem we get `Fx = 1/2Kx^(2) Rightarrow F = 30N`