Newton's Laws of Motion: Conceptual Doubts Cleared (2)

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Question

A block of mass mm rests on a frictionless horizontal surface. A horizontal force FF is applied to the block. A second identical block is now placed on top of the first. If the same force FF is applied to the bottom block, students often think the acceleration “halves because mass doubles”. Two doubts confuse them: (a) does Newton’s second law apply to the system or each block separately? (b) what is the contact force between the two blocks, and which third-law pair acts where?

Solution — Step by Step

For acceleration, we treat both blocks as one system because they move together. Net external horizontal force is FF, total mass is 2m2m, so a=F/(2m)a = F/(2m). The “halves” intuition is correct here, but only because we lumped the system — students must know when this lumping is allowed (only if both bodies have the same acceleration).

The top block has only one horizontal force on it — the friction or contact push from the lower block. Since the surface is frictionless between the lower block and ground, but there must be friction (or a horizontal push if a wall were present) between the blocks for the top one to accelerate. Assuming friction ff between blocks: f=ma=mF/(2m)=F/2f = m \cdot a = m \cdot F/(2m) = F/2.

The top block pushes the bottom block backward with F/2F/2 (reaction to the friction the bottom pushes forward on the top). So on the bottom block: net horizontal =FF/2=F/2= F - F/2 = F/2, and F/2=ma=mF/(2m)=F/2F/2 = m \cdot a = m \cdot F/(2m) = F/2. Both blocks are consistent. The contact friction is F/2F/2.

Why This Works

Newton’s second law is a statement about each body, but if two bodies have the same acceleration, you may treat them as one system to find that common acceleration faster. The trick is knowing when to switch between system view and isolated-body view.

The isolated view always works. The system view is a shortcut. Use the system to find aa, then isolate one body to find internal forces (contact, tension, friction).

Whenever a question asks for an internal force (tension, normal between stacked blocks, contact force), you must isolate one body — the system view hides internal forces.

Alternative Method

Apply Newton’s second law to each block separately from the start. Let ff be the friction between blocks. For the top block: f=maf = ma. For the bottom: Ff=maF - f = ma. Two equations, two unknowns (ff, aa). Adding: F=2maF = 2ma, so a=F/(2m)a = F/(2m) and f=F/2f = F/2. Same answer, more rigorous, longer.

Common Mistake

Students often write "F=maF = ma" with m=m = mass of top block but a=F/(2m)a = F/(2m) unintentionally — they mix system acceleration with single-body force balance. Either always use the system mass with the system force, or always isolate. Don’t mix.

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