The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is .................. N. (take g = 10 ms⁻²)

Solution

Physics sir · 4 steps

Only friction drags the 1 kg block along, so its largest possible acceleration sets the limit for both.

1. Force on top block. F acts on the 2 kg block. The 1 kg block on top feels no applied force, so friction from the lower block is the only horizontal force on it.
2. Top block's limit. Static friction can be at most μ m₁ g = 0.5 × 1 × 10 = 5 N. So the top block can accelerate at most at a = 5/1 = 5 m s⁻². Any faster and it slips.
3. Whole system. Moving together, both blocks share this acceleration. The table is smooth, so the only horizontal force on the pair is F.

   Fₘₐₓ = (m₁ + m₂) aₘₐₓ = (1 + 2) × 5

4. So. Fₘₐₓ = 3 × 5 = 15 N, so the answer is 15.

The common slip: 5. Taking the maximum friction itself as the answer gives 5 N. Friction only sets the top block's acceleration, and F must accelerate all 3 kg.

The idea to keep. In stacked blocks, find the block moved only by friction. Its maximum acceleration μ g is the common limit, then multiply by the total mass to get the largest force.

Answer from the NTA final answer key. Solution by PrepWiser.