JEE Main 2024 Physics question
A wooden block of mass 5kg rests on soft horizontal floor. When an iron cylinder of mass 25 kg is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of 0.1 ms. The action force of the system on the floor is equal to:
Answer: (C)
Solution
N = m(g − a) = 30(9.8 − 0.1) = 291 N. (This needs g = 9.8 m/s², which the question
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