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 $m_3 = 7 \text{ kg}$.

The maximum static friction force between the $2 \text{ kg}$ and $3 \text{ kg}$ blocks is:

$f_{1,\max} = \mu_1 m_1 g = 0.2 \times 2 \times 10 = 4 \text{ N}$

The maximum static friction force between the $3 \text{ kg}$ and $7 \text{ kg}$ blocks is:

$f_{2,\max} = \mu_2 (m_1 + m_2) g = 0.3 \times (2 + 3) \times 10 = 15 \text{ N}$

Assume all three blocks move together as a single system. The common acceleration $a$ is given by the total applied force divided by the total mass:

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

Now, check the required friction forces to keep the blocks moving together with this acceleration.

For the $2 \text{ kg}$ block, the required friction force $f_1$ to accelerate it at $\dfrac{5}{6} \text{ m/s}^2$ is:

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

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

For the $7 \text{ kg}$ block, the required friction force $f_2$ to accelerate it at $\dfrac{5}{6} \text{ m/s}^2$ is:

$f_2 = m_3 a = 7 \times \dfrac{5}{6} = \dfrac{35}{6} \text{ N}$

Since $f_2 < f_{2,\max}$ ($\dfrac{35}{6} \text{ N} < 15 \text{ N}$), the $7 \text{ kg}$ block does not slip on the $3 \text{ kg}$ block.

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

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