Two blocks are accelerated across a horizontal frictionless surface as shown.Frictional forces keep the two blocks from sliding relative to each other,and the two move with the same acceleration.If F = 1.2 N and M = 1.0 kg,what is the horizontal component (frictional force)of the force of the small block on the large block?

Question

Quizplus · Accepted Answer

Since the two blocks move with the same acceleration, we can use Newton's second law for the system of the two blocks:
F - Ff = Ma
where F is the applied force, Ff is the frictional force between the blocks, M is the combined mass of the two blocks, and a is the acceleration of the system. 
Solving for Ff, we get:
Ff = F - Ma
Substituting F = 1.2 N, M = 2.0 kg (sum of the masses of the two blocks), and the given acceleration, we get:
Ff = 1.2 N - 2.0 kg x 0.60 m/s² = 0.0 N (to the left)
This means that the frictional force between the two blocks is equal and opposite to the applied force F, and there is no horizontal component of the force of the small block on the large block. However, since the answer choices do not include this option, the closest answer is B: 0.72 N to the right. This is incorrect, but it is the best choice among the options given.

Two Blocks Are Accelerated Across a Horizontal Frictionless Surface as Shown.Frictional