1. Analyze the Physical Situation

We have a system consisting of two blocks: a large block $A$ (mass $M = 4\,\text{kg}$) and a small block $B$ (mass $m = 1.0\,\text{kg}$).

• Initial State: The system is at rest. Block $B$ is at the top of a frictionless slope of height $h = 40\,\text{cm} = 0.4\,\text{m}$.
• Motion: Block $B$ slides down the frictionless slope and then moves onto the horizontal rough portion of Block $A$.
• Final State: Block $B$ slides a distance $s = 1.0\,\text{m}$ relative to Block $A$ and then stops relative to Block A.

2. Conservation of Momentum

Since the floor is frictionless, there is no external horizontal force acting on the system ($A + B$). Therefore, the linear momentum of the entire system in the horizontal direction is conserved.

$$P_{initial} = 0$$

When Block $B$ stops sliding relative to Block $A$, both blocks move with a common final velocity, let’s call it $v_{final}$.

$$P_{final} = (M + m)v_{final}$$

Equating initial and final momentum:

$$(M + m)v_{final} = 0 \implies v_{final} = 0$$

This is a crucial insight: The entire system comes to a complete halt once the relative motion ceases.

3. Work-Energy Analysis

Since the initial and final kinetic energies are both zero (starts at rest, ends at rest), the change in the total energy of the system is equal to the work done by non-conservative internal forces (friction).

Initial Energy ($E_i$): Purely gravitational potential energy of Block $B$. (Reference level at the flat part of A).

$$E_i = mgh$$

Final Energy ($E_f$): The blocks are at the lower level and at rest.

$$E_f = 0$$

Work done by Friction ($W_f$): Kinetic friction acts between $B$ and the horizontal surface of $A$. The heat generated (energy dissipated) is equal to the friction force multiplied by the relative distance of sliding.

$$W_{friction} = – f_k \cdot s_{relative} = – (\mu N) s$$

Where normal force $N = mg$.

$$W_{friction} = -\mu m g s$$

4. Calculation

Using the Work-Energy Theorem ($\Delta E = W_{nc}$):

$$E_f – E_i = W_{friction}$$

$$0 – mgh = -\mu m g s$$

$$mgh = \mu m g s$$

Canceling $mg$ from both sides:

$$h = \mu s$$

$$\mu = \frac{h}{s}$$

Substitute the given values ($h = 0.4\,\text{m}$ and $s = 1.0\,\text{m}$):

$$\mu = \frac{0.4}{1.0} = 0.4$$