QJEEM24Ph7cbe: JEE Main 2024 Physics - Class 11 JEEM24Ph7cbe Solution: A wooden block of mass 5   kg rests on a soft horizontal…

To solve this problem, we'll analyze the forces acting on the combined system of the wooden block and the iron cylinder as they accelerate downward due to the yielding floor.

Given:

Mass of wooden block, $m_1 = 5 \, \text{kg}$

Mass of iron cylinder, $m_2 = 25 \, \text{kg}$

Total mass of the system, $m = m_1 + m_2 = 30 \, \text{kg}$

Acceleration of the system, $a = 0.1 \, \text{m/s}^2$ (downward)

Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

Step 1: Draw the Free Body Diagram (FBD)

For the combined system:

Downward Forces:

Weight of the system: $W = m \times g = 30 \times 9.8 = 294 \, \text{N}$

Upward Forces:

Normal force from the floor: $N$

Step 2: Apply Newton's Second Law

Since the system is accelerating downward, the net force is:

$\text{Net Force} = \text{Total Downward Force} - \text{Total Upward Force}$

According to Newton's second law:

$m \times a = W - N$

Step 3: Solve for the Normal Force ( $N$ )

$N = W - m \times a$

Substitute the given values:

$N = 294 \, \text{N} - 30 \, \text{kg} \times 0.1 \, \text{m/s}^2$

$N = 294 \, \text{N} - 3 \, \text{N}$

$N = 291 \, \text{N}$

Step 4: Interpret the Result

The normal force ( $N$ ) is the force exerted by the floor on the system upward. By Newton's third law, the action force of the system on the floor is equal in magnitude and opposite in direction to the normal force.

Therefore, the action force of the system on the floor is $291 \, \text{N}$ .