Question

A wooden block of mass 10 kg rests on a soft horizontal floor. When an iron cylinder of mass 20 kg is placed on top of the block, the floor yields, and the block and cylinder together go down with an acceleration of 0.2 m/s². The action force of the system on the floor is 49x N. Find the integer value of x. (Take g = 10 m/s² )

Answer 1

6

Answer 2

Total mass $M = 10 \text{ kg} + 20 \text{ kg} = 30 \text{ kg}$. The system accelerates downwards with $a = 0.2 \text{ m/s}^2$. The normal force $N$ exerted by the floor on the system satisfies $Mg - N = Ma$. Thus, $N = M(g-a)$. Substituting values: $N = 30 \text{ kg} \times (10 \text{ m/s}^2 - 0.2 \text{ m/s}^2) = 30 \times 9.8 \text{ N} = 294 \text{ N}$. By Newton's third law, the action force of the system on the floor is equal in magnitude to $N$. Given the action force is $49x$ N, we have $49x = 294$. Solving for $x$: $x = \frac{294}{49} = 6$.