Physical Setup:

Block $m_1$ is on a horizontal surface with coefficient of friction $\mu$. It is connected to a spring $k$ and a hanging block $m_2$ via a pulley. The system is released from rest with the spring relaxed.

Step 1: Determine Equilibrium Position

The blocks will reach their maximum speed at the equilibrium position where the net force on the system is zero (acceleration is zero). Let this position be at a distance $x_{eq}$ from the initial position.

At equilibrium, the driving force (gravity on $m_2$) is balanced by the restoring force (spring) and the resistive force (friction on $m_1$).

$$ T = m_2 g $$

$$ T = k x_{eq} + f_k $$

$$ m_2 g = k x_{eq} + \mu m_1 g $$

Solving for $x_{eq}$:

$$ x_{eq} = \frac{g(m_2 – \mu m_1)}{k} $$

Step 2: Work-Energy Theorem

Apply conservation of energy between the start point ($x=0, v=0$) and the equilibrium point ($x=x_{eq}, v=v_{max}$).

$$ \Delta K + \Delta U_{spring} + \Delta U_{grav} + W_{fric} = 0 $$

$$ \frac{1}{2}(m_1+m_2)v_{max}^2 + \frac{1}{2}kx_{eq}^2 – m_2 g x_{eq} + \mu m_1 g x_{eq} = 0 $$

Rearranging for kinetic energy:

$$ \frac{1}{2}(m_1+m_2)v_{max}^2 = m_2 g x_{eq} – \mu m_1 g x_{eq} – \frac{1}{2}kx_{eq}^2 $$

$$ \frac{1}{2}(m_1+m_2)v_{max}^2 = (m_2 g – \mu m_1 g) x_{eq} – \frac{1}{2}kx_{eq}^2 $$

Step 3: Simplify and Solve

Substitute the term $(m_2 g – \mu m_1 g)$ with $k x_{eq}$ (from the equilibrium condition derived in Step 1):

$$ \frac{1}{2}(m_1+m_2)v_{max}^2 = (k x_{eq}) x_{eq} – \frac{1}{2}kx_{eq}^2 $$

$$ \frac{1}{2}(m_1+m_2)v_{max}^2 = k x_{eq}^2 – \frac{1}{2}kx_{eq}^2 $$

$$ \frac{1}{2}(m_1+m_2)v_{max}^2 = \frac{1}{2}kx_{eq}^2 $$

$$ v_{max} = x_{eq} \sqrt{\frac{k}{m_1 + m_2}} $$

Substitute the expression for $x_{eq}$ back into the equation:

$$ v_{max} = \frac{g(m_2 – \mu m_1)}{k} \sqrt{\frac{k}{m_1 + m_2}} $$

$$ v_{max} = \frac{g(m_2 – \mu m_1)}{\sqrt{k(m_1 + m_2)}} $$