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 $$g = 10$$ m s$$^{-2}$$)

We need to find the maximum horizontal force $$F$$ that can be applied to the lower block such that both blocks move together without slipping between them.

1. Analyze the Maximum Friction Force

The static friction between the two blocks keeps the upper $$1\text{ kg}$$ block moving along with the lower $$2\text{ kg}$$ block. The maximum possible static friction force ($$f_{\text{max}}$$) that can act on the upper block is given by:

$$f_{\text{max}} = \mu \cdot N$$

Where:

• $$\mu = 0.5$$ (Coefficient of static friction)
• $$N$$ is the normal reaction force acting on the $$1\text{ kg}$$ block due to the weight of the block itself:

  $$N = m_1 \cdot g = 1\text{ kg} \times 10\text{ m s}^{-2} = 10\text{ N}$$

Substituting the values into the maximum friction formula:

$$f_{\text{max}} = 0.5 \times 10\text{ N} = 5\text{ N}$$

2. Find the Maximum Acceleration of the System

The upper block is accelerated purely due to this frictional force. Therefore, its maximum possible acceleration ($$a_{\text{max}}$$) before it begins to slip is:

$$f_{\text{max}} = m_1 \cdot a_{\text{max}}$$

$$5\text{ N} = 1\text{ kg} \times a_{\text{max}} \implies a_{\text{max}} = 5\text{ m s}^{-2}$$

3. Calculate the Maximum Applied Force ($$F$$)

Since the table is completely smooth, there is no external retarding friction force acting on the bottom surface of the lower block. To prevent slipping, both blocks must move together as a single composite system with a maximum common acceleration of $$a_{\text{max}} = 5\text{ m s}^{-2}$$.

Using Newton's second law for the total mass ($$m_1 + m_2$$):

$$F = (m_1 + m_2) \cdot a_{\text{max}}$$

$$F = (1\text{ kg} + 2\text{ kg}) \times 5\text{ m s}^{-2}$$

$$F = 3\text{ kg} \times 5\text{ m s}^{-2} = 15\text{ N}$$

Final Answer: 15