Question:
A wooden block of mass 5kg rests on soft horizontal floor. When an iron cylinder of mass 25 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 ms–2 . The action force of the system on the floor is equal to:
A wooden block of mass 5kg rests on soft horizontal floor. When an iron cylinder of mass 25 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 ms–2 . The action force of the system on the floor is equal to:
Updated On: Jan 13, 2026
- 297 N
- 294 N
- 291 N
- 196 N
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The Correct Option is C
Approach Solution - 1
Given:
Total mass = \( 5 \, \text{kg} + 25 \, \text{kg} = 30 \, \text{kg} \)
Acceleration due to gravity (\( g \)) is taken as \( 9.8 \, \text{m/s}^2 \).
The weight of the system is:
\[ W = 30 \times 9.8 = 294 \, \text{N} \]
Considering the downward acceleration of the system:
Net force = \( W - N = 30 \times 0.1 \)
Rearranging:
\[ 294 - N = 3 \implies N = 291 \, \text{N} \]
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Approach Solution -2
Step 1: Given data
Mass of wooden block, \( m_1 = 5\,\text{kg} \)
Mass of iron cylinder, \( m_2 = 25\,\text{kg} \)
Total mass, \( m = m_1 + m_2 = 30\,\text{kg} \)
Acceleration of the system downward, \( a = 0.1\,\text{m/s}^2 \)
Acceleration due to gravity, \( g = 9.8\,\text{m/s}^2 \)
Step 2: Forces acting on the system
The forces acting on the combined system are:
1. Weight of the system \( W = mg = 30 \times 9.8 = 294\,\text{N} \) (downward)
2. Normal reaction (action force) of the floor on the system \( N \) (upward).
Step 3: Apply Newton’s second law
Taking downward direction as positive:
\[ W - N = ma \] \[ 294 - N = 30 \times 0.1 \] \[ 294 - N = 3 \] \[ N = 294 - 3 = 291\,\text{N}. \]
Step 4: Interpretation
The action force of the system on the floor (which equals the normal reaction by the floor on the system) is \( 291\,\text{N} \).
Final answer
Mass of wooden block, \( m_1 = 5\,\text{kg} \)
Mass of iron cylinder, \( m_2 = 25\,\text{kg} \)
Total mass, \( m = m_1 + m_2 = 30\,\text{kg} \)
Acceleration of the system downward, \( a = 0.1\,\text{m/s}^2 \)
Acceleration due to gravity, \( g = 9.8\,\text{m/s}^2 \)
Step 2: Forces acting on the system
The forces acting on the combined system are:
1. Weight of the system \( W = mg = 30 \times 9.8 = 294\,\text{N} \) (downward)
2. Normal reaction (action force) of the floor on the system \( N \) (upward).
Step 3: Apply Newton’s second law
Taking downward direction as positive:
\[ W - N = ma \] \[ 294 - N = 30 \times 0.1 \] \[ 294 - N = 3 \] \[ N = 294 - 3 = 291\,\text{N}. \]
Step 4: Interpretation
The action force of the system on the floor (which equals the normal reaction by the floor on the system) is \( 291\,\text{N} \).
Final answer
291 N
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