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Q.

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 the 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 (in N) so the lighter block does not loose contact with ground.

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answer is 30.

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Detailed Solution

Draw FBD of B to get extension in spring. Instant when block B just looses contact with ground, the net force on it becomes zero. kx−Tcos⁡θ=0    ........(i)Tsin⁡θ+N−mg=0 .......(ii)To lift the block B, N = 0, from (i) and (ii), we get kxcos⁡θsin⁡θ=mgx=mgktan⁡θ=80k×(4/3)=60kIf spring has to just extend till this value at their extension it should be at rest.Now we apply work energy theorem to get Fx=12kx2⇒F=30 N
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