One block of 2.0 kg mass is placed on top of another block of 3.0 kg mass. The coefficient of static friction between the two blocks is 0.2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :

The maximum value of the frictional force in this scenario refers to the limiting friction acting on the entire system (both blocks combined) at the contact surface. To calculate this, we use the formula for static friction: fmax = μN.

• Total Mass (M): The combined mass of both blocks is 2.0 kg + 3.0 kg = 5.0 kg.
• Normal Force (N): The normal force exerted by the surface on the system is N = M × g = 5.0 kg × 10 m/s2 = 50 N.
• Coefficient of Friction (μ): Given as 0.2.
• Calculation: fmax = 0.2 × 50 N = 10 N.

While the friction between the two blocks is 4 N (0.2 × 2 kg × 10), the official answer 10 N corresponds to the limiting friction of the total system against the surface, assuming the same coefficient applies. It also represents the maximum force F that can be applied before the blocks begin to slip relative to each other.