Two blocks (m = 0.5 kg and M = 4.5 kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is 3/7. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is ___________ N. (Round off to the Nearest Integer) [Take g as 9.8 ms⁻²]

Solution

Physics sir · 4 steps

Friction is the only thing pulling the small block, so it sets the top speed-up for both blocks.

1. Who drives m. F acts only on the lower block M. The upper block m moves only because static friction from M pulls it forward.
2. Maximum acceleration. Static friction on m can be at most μ m g. So the largest acceleration m can get is aₘₐₓ = (μ m g)/m = μ g = 3/7 × 9.8 = 4.2 m s⁻².
3. Whole system. Moving together, both blocks share this acceleration. The table is frictionless, so F alone accelerates the total mass m + M = 5 kg.

   Fₘₐₓ = (m+M) μ g = 5 × 4.2 = 21 N

4. So. The maximum force on the larger block is 21 N.

The common slip: 2. Stopping at the friction limit μ m g = 2.1 N gives 2. That is the force on m alone, not the force F needed for both blocks.

The idea to keep. When a force pushes only the lower block, friction must drag the upper one along. Get the upper block's limit a = μ g, then use F = (total mass) × a.

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