In this problem, we are tasked with calculating the speed of a 3-kilogram block just before it hits the ground, while considering a system of connected objects. The scenario involves two blocks, labeled A and B, where block B (3 kg) falls vertically, and block A (4 kg) slides horizontally due to the connection between them. As the system is released, both blocks will have the same final speed, denoted as \( v_{\text{final}} \).

To find the final speed, we apply the principle of conservation of energy, which states that the total mechanical energy in a closed system remains constant. The energy conservation equation can be expressed as:

\[ K_{\text{initial}} + U_{\text{initial}} + W_{\text{non-conservative}} = K_{\text{final}} + U_{\text{final}} \]

Here, \( K \) represents kinetic energy, \( U \) represents potential energy, and \( W \) represents work done by non-conservative forces (like friction). The initial kinetic energy for both blocks is given by:

\[ K_{\text{initial}} = \frac{1}{2} m_A v_{A,\text{initial}}^2 + \frac{1}{2} m_B v_{B,\text{initial}}^2 \]

Since both blocks start from rest, their initial velocities are zero, leading to zero initial kinetic energy. The potential energy for block B, which is falling, is given by:

\[ U_{\text{initial}} = m_B g y_{B,\text{initial}} \]

Block A does not change height, so its potential energy does not contribute to the energy change. The work done by friction, which acts on block A, is expressed as:

\[ W_{\text{non-conservative}} = -F_k \cdot d = -\mu_k m_A g \cdot d \]

As block B falls a distance \( \Delta y \), block A moves horizontally the same distance \( d = \Delta y \). The final kinetic energy for both blocks, which are moving at the same speed, can be combined as:

\[ K_{\text{final}} = \frac{1}{2} (m_A + m_B) v_{\text{final}}^2 \]

Substituting these expressions into the energy conservation equation, we simplify to find:

\[ m_B g y_{B,\text{initial}} - \mu_k m_A g \Delta y = \frac{1}{2} (m_A + m_B) v_{\text{final}}^2 \]

Plugging in the values, where \( m_B = 3 \, \text{kg} \), \( m_A = 4 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( y_{B,\text{initial}} = 2 \, \text{m} \), and \( \mu_k = 0.5 \), we calculate:

\[ 3 \cdot 9.8 \cdot 2 - 0.5 \cdot 4 \cdot 9.8 \cdot 2 = \frac{1}{2} (4 + 3) v_{\text{final}}^2 \]

This simplifies to:

\[ 58.8 - 39.2 = 3.5 v_{\text{final}}^2 \]

Solving for \( v_{\text{final}} \), we find:

\[ v_{\text{final}}^2 = \frac{19.6}{3.5} \approx 5.6 \]

Thus, taking the square root gives:

\[ v_{\text{final}} \approx 2.37 \, \text{m/s} \]

This final speed represents the velocity of both blocks just before block B hits the ground. Understanding the interplay of kinetic and potential energy, along with the effects of friction, is crucial in solving problems involving systems of connected objects.