Three blocks are being pushed by a force Fa , along a frictionless surface as shown in the diagram below. The acceleration experienced by the blocks is 3.1 m/s2 The masses of blocks A, B and C, are 5.2 kg, 7.2 kg and 1.2 kg respectively. If the order of the blocks are now CAB. What is the magnitude of the force of Block C on Block B?

Question

Three blocks are being pushed by a force Fa , along a frictionless surface as shown in the diagram below. The acceleration experienced by the blocks is 3.1 m/s2 The masses of blocks A, B and C, are 5.2 kg, 7.2 kg and 1.2 kg respectively. If the order of the blocks are now CAB. What is the magnitude of the force of Block C on Block B?

BlackTom AI · Accepted Answer

Newton’s second law gives the contact force that block C must exert on the blocks to its right — it must provide the net force to accelerate those blocks at 3.1 m/s^2. Thus the contact force from C on the blocks to its right equals the total mass of those blocks times the common acceleration.

1) Relevant concepts
- Newton’s 2nd law: F_net = m_total · a.
- For a set of blocks in contact and accelerating together, the contact force at an interface equals the mass of the blocks on one side of that interface multiplied by the acceleration.

2) Step-by-step reasoning
- The order is CAB left-to-right, so to the right of block C are blocks A and B.
- The force of C on B (i.e., the push C transmits toward the right) must accelerate both A and B together.

3) Calculations (all intermediate steps)
- Sum the masses to the right of C: m_A + m_B = 5.2 kg + 7.2 kg = 12.4 kg.
- Multiply by the acceleration: F = (m_A + m_B) · a = 12.4 kg · 3.1 m/s^2.
- Compute: 12.4 × 3.1 = 38.44 N.

4) Why this is correct
- C must push on the remainder of the system (A and B) with the exact force required to accelerate their combined mass at the given acceleration. That force transmitted through the contact interfaces is 38.44 N.

5) Why the other options are incorrect
- a. 26.04 N corresponds to (m_B + m_C)·a = (7.2+1.2)·3.1 = 8.4·3.1 = 26.04 N — the wrong grouping of masses for C pushing A and B.
- c. 3.72 N equals m_C·a = 1.2·3.1, which would be the force needed to accelerate C alone, not the force C exerts on the blocks to its right.
- d. 19.84 N equals (m_A + m_C)·a = (5.2+1.2)·3.1 = 6.4·3.1 = 19.84 N — again the wrong combination for the interface in question.

6) Conclusion
Therefore the magnitude of the force of block C on block B is 38.44 N, matching option b.

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