Q 11-04-076JEE MainJEE Main 2024 (5 Apr, Shift 1)Easy
A wooden block of mass $5\ \text{kg}$ rests on a soft horizontal floor. When an iron cylinder of mass $25\ \text{kg}$ is placed on top of the block, the floor yields and the block and the cylinder together go down with an acceleration of $0.1\ \text{m s}^{-2}$. The action force of the system on the floor is equal to (take $g = 9.8\ \text{m s}^{-2}$)
Answer: (B) $291\ \text{N}$
For the system of $30\ \text{kg}$ moving down with $a = 0.1\ \text{m s}^{-2}$:
$$mg - N = ma \Rightarrow N = m(g - a) = 30\times(9.8 - 0.1) = 291\ \text{N}$$
By Newton's third law the system pushes on the floor with $291\ \text{N}$.
Solution by Sreeraj P, M.Sc Physics