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JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion

A wooden block of mass \(5 \mathrm{~kg}\) rests on soft horizontal floor. When an iron cylinder of mass \(25\) \(\mathrm{kg}\) is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of \(0.1 \mathrm{~ms}^{-2}\). The action force of the system on the floor is equal to _______.

  1. A \(297 \mathrm{~N}\)
  2. B \(294 \mathrm{~N}\)
  3. C \(291 \mathrm{~N}\)
  4. D \(196 \mathrm{~N}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(291 \mathrm{~N}\)

Step-by-step Solution

Detailed explanation

Taking \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\) \(294-N=30 \times 0.1\) \(N=291\)
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