1. Initial State Analysis

The system is lowered until block $M$ just touches the table. The cord holding $m$ is then released. At this instant ($t=0$):

• Block $M$ is on the table.
• Block $m$ is suspended above $M$.
• The spring supports the weight of $M$ (from the prior state of being suspended).

Therefore, the initial extension in the spring $x_0$ balances the weight of $M$:

$$ k x_0 = Mg \implies x_0 = \frac{Mg}{k} $$

2. Energy Conservation

The upper block $m$ moves down a distance $d$ and comes to instantaneous rest. We apply the Work-Energy Theorem (or Conservation of Mechanical Energy) between the initial and final states.

Change in Kinetic Energy ($\Delta K$): 0 (Starts from rest, ends at rest).
Work done by Gravity ($W_g$): Positive, as $m$ moves down. $$ W_g = m g d $$ Work done by Spring ($W_s$): The spring length changes from an extension $x_0$ to a new state. Since $m$ moves down (towards $M$), the extension decreases.
Initial potential energy: $U_i = \frac{1}{2} k x_0^2$
Final potential energy: $U_f = \frac{1}{2} k (x_0 – d)^2$
$$ \Delta U_{spring} = U_f – U_i = \frac{1}{2} k (x_0 – d)^2 – \frac{1}{2} k x_0^2 $$

Applying Conservation of Energy ($\Delta K + \Delta U_{grav} + \Delta U_{spring} = 0$):

$$ 0 – m g d + \left[ \frac{1}{2} k (x_0 – d)^2 – \frac{1}{2} k x_0^2 \right] = 0 $$ $$ m g d = \frac{1}{2} k \left[ (x_0 – d)^2 – x_0^2 \right] $$ $$ m g d = \frac{1}{2} k \left[ x_0^2 + d^2 – 2x_0 d – x_0^2 \right] $$ $$ m g d = \frac{1}{2} k (d^2 – 2x_0 d) $$ Dividing by $d$ (since $d \neq 0$): $$ m g = \frac{1}{2} k d – k x_0 $$

3. Solving for Distance d

Substitute $x_0 = \frac{Mg}{k}$ into the equation:

$$ m g = \frac{1}{2} k d – k \left( \frac{Mg}{k} \right) $$ $$ m g = \frac{1}{2} k d – Mg $$ $$ m g + Mg = \frac{1}{2} k d $$ $$ (m + M)g = \frac{1}{2} k d $$ $$ d = \frac{2(m + M)g}{k} $$

Answer: The upper block moves downward by a distance of $\frac{2(m+M)g}{k}$.