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 2.8 m/s2 The masses of blocks A, B and C, are 5.8 kg, 7.8 kg and 2.7 kg respectively.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 2.8 m/s2 The masses of blocks A, B and C, are 5.8 kg, 7.8 kg and 2.7 kg respectively.What is the magnitude of the force of Block C on Block B?

BlackTom AI · Accepted Answer

To analyze the force between blocks, I’ll consider the system as a whole and then the interaction between specific blocks.
First, sum of external forces on the three-block system equals total mass times acceleration. Since the surface is frictionless, the only external horizontal force is F_a. The total mass is m_A + m_B + m_C = 5.8 kg + 7.8 kg + 2.7 kg = 16.3 kg. With acceleration a = 2.8 m/s^2, the required pushing force is F_a = M a = 16.3 × 2.8 = 45.64 N. This confirms the applied push is sufficient to move all three blocks together with the given acceleration.
Now, focus on the contact force between Block B and Block C, i.e., the force that Block B exerts on Block C. For Block C alone, the only horizontal force acting on it is the contact force from Block B, call it F_{B on C}. This force must produce the same acceleration a for Block C as for the system, so F_{B on C} = m_C × a = 2.7 kg × 2.8 m/s^2 = 7.56 N. This is the magnitude of the force of Block C on Block B but in the opposite direction; by Newton’s third law, Block B experiences an equal and opposite force from Block C (i.e., 7.56 N backward on B).
To assess the other options: if you use 38.08 N, that would correspond to m_A + m_B portion times a or could be inferred from different breakdowns, but it does not equal the required m_C a for Block C; thus it’s inconsistent with the acceleration of the system. The value 29.40 N would imply a different mass-times-acceleration product that doesn’t match m_C or the net force on the connecting interface. The 23.80 N option is similarly not aligned with the direct requirement F = m_C a for Block C’s acceleration. In short, since the acceleration is uniform for all blocks and Block C has mass 2.7 kg, the only force from B that accelerates C is F = m_C a = 7.56 N, which is consistent with Newton’s laws for connected blocks on a frictionless surface.
Thus, the reasoning hinges on treating Block C as a system driven solely by the contact force from Block B and applying F = m a to its own mass to find the contact force.

MUF0121 Physics Unit 1 - Semester 2, 2025

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 2.8 m/s2 The masses of blocks A, B and C, are 5.8 kg, 7.8 kg and 2.7 kg respectively.What is the magnitude of the force of Block C on Block B?

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