Three blocks of masses 2 kg , 3 kg , and 7 kg are stacked one over...

Let the masses of the blocks be $m_1 = 2 \text{ kg}$, $m_2 = 3 \text{ kg}$, and $m_3 = 7 \text{ kg}$.

First, we calculate the maximum possible static friction forces between the surfaces.

The normal force between the $2 \text{ kg}$ and $3 \text{ kg}$ blocks is $N_1 = m_1 g = 2 \times 10 = 20 \text{ N}$.

Maximum friction between them: f_{1,\text\max} = \mu_1 N_1 = 0.2 \times 20 = 4 \text{ N}.

The normal force between the $3 \text{ kg}$ and $7 \text{ kg}$ blocks is $N_2 = (m_1 + m_2) g = (2 + 3) \times 10 = 50 \text{ N}$.

Maximum friction between them: f_{2,\text\max} = \mu_2 N_2 = 0.3 \times 50 = 15 \text{ N}.

Let us assume that all three blocks move together as a single system under the applied force $F = 10 \text{ N}$.

The common acceleration would be:

$a = \dfrac{F}{m_1 + m_2 + m_3} = \dfrac{10}{2 + 3 + 7} = \dfrac{10}{12} = \dfrac{5}{6} \text{ m/s}^2$

Now, we verify if the required friction forces to maintain this common acceleration are within their maximum limits.

For the $2 \text{ kg}$ block to accelerate at $a = \dfrac{5}{6} \text{ m/s}^2$, the required friction force is:

$f_1 = m_1 a = 2 \times \dfrac{5}{6} = \dfrac{5}{3} \text{ N}$

Since f_1 \le f_{1,\text\max} ($\dfrac{5}{3} \text{ N} \le 4 \text{ N}$), the $2 \text{ kg}$ block does not slip on the $3 \text{ kg}$ block.

For the combined system of $2 \text{ kg}$ and $3 \text{ kg}$ blocks to accelerate at $a = \dfrac{5}{6} \text{ m/s}^2$, the required friction force from the $7 \text{ kg}$ block is:

$f_2 = (m_1 + m_2) a = (2 + 3) \times \dfrac{5}{6} = \dfrac{25}{6} \text{ N}$

Since f_2 \le f_{2,\text\max} ($\dfrac{25}{6} \text{ N} \le 15 \text{ N}$), the $3 \text{ kg}$ block does not slip on the $7 \text{ kg}$ block.

Therefore, our assumption is correct, and all three blocks move together with the same acceleration.

$a_1 = a_2 = a_3 = \dfrac{5}{6} \text{ m/s}^2$