A wooden block of mass $5 \mathrm{~kg}$ rests on a 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:

• A

  $196 \mathrm{~N}$
• Correct answer
  B

  $291 \mathrm{~N}$
• C

  $294 \mathrm{~N}$
• D

  $297 \mathrm{~N}$

Solution

Taking $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
$\begin{aligned} & 294-\mathrm{N}=30 \times 0.1 \\ & \mathrm{~N}=291\end{aligned}$

Hence, the correct answer is option: 'B'.