Step 1: Velocity Analysis

Let the tube move down with velocity $v$. Consider the segment of rope of mass $M$ currently inside the horizontal part of the tube.

• Vertical Motion: The rope segment is carried downwards by the tube. $v_y = v$.
• Horizontal Motion: For the rope to pass through the tube, it must flow horizontally relative to the tube. The rate of flow matches the tube’s speed. $v_x = v$.

The total kinetic energy of this mass $M$ is determined by its absolute velocity squared:

$$v_{abs}^2 = v_x^2 + v_y^2 = v^2 + v^2 = 2v^2$$

Step 2: Conservation of Energy

We apply the work-energy theorem. The only work done is by gravity on the descending mass $M$.

For a small downward displacement $dy$:

• Work Done: $dW = Mg \, dy$
• Change in Kinetic Energy: $dK = d\left(\frac{1}{2} M v_{abs}^2\right) = d\left(\frac{1}{2} M (2v^2)\right) = d(Mv^2)$

Step 3: Solve for Acceleration

Equating Work and Change in KE:

$$Mg \, dy = M (2v \, dv)$$ $$g \, dy = 2 v \, dv$$

We know that acceleration $a = v \frac{dv}{dy}$, so $v \, dv = a \, dy$. Substituting this into the equation:

$$g \, dy = 2 (a \, dy)$$ $$g = 2a$$ $$a = 0.5g$$