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