As shown in the figure, two blocks A and B , each having a mass of...

Let $x$ be the horizontal displacement of block A when it breaks off the surface.

The length of the spring at this instant is $L = \sqrt{h^2 + x^2}$, where $h = 0.4 \text{ m}$ is the initial height of the spring.

The extension in the spring is $\Delta L = L - h$.

The spring force is $F_s = k(L - h)$.

The vertical component of the spring force is $F_y = F_s \cos\theta = k(L - h) \dfrac{h}{L} = kh\left(1 - \dfrac{h}{L}\right)$.

Block A breaks off the surface when the normal reaction $N$ becomes zero. At this instant, the vertical component of the spring force balances the weight of block A:

$F_y = m_A g$

$kh\left(1 - \dfrac{h}{L}\right) = m_A g$

Substituting the given values ($k = 40 \text{ N/m}$, $h = 0.4 \text{ m}$, $m_A = 0.32 \text{ kg}$, $g = 10 \text{ m/s}^2$):

$40 \times 0.4 \left(1 - \dfrac{0.4}{L}\right) = 0.32 \times 10$

$16 \left(1 - \dfrac{0.4}{L}\right) = 3.2$

$1 - \dfrac{0.4}{L} = 0.2 \Rightarrow \dfrac{0.4}{L} = 0.8 \Rightarrow L = 0.5 \text{ m}$

The extension in the spring is $\Delta L = 0.5 - 0.4 = 0.1 \text{ m}$.

The horizontal displacement of block A is $x = \sqrt{L^2 - h^2} = \sqrt{0.5^2 - 0.4^2} = 0.3 \text{ m}$.

Since block A and block B are connected by a string, block B moves downwards by the same distance $x = 0.3 \text{ m}$.

Applying the principle of conservation of mechanical energy to the system (Block A + Block B + Spring):

Loss in gravitational potential energy of block B = Gain in kinetic energy of the system + Gain in elastic potential energy of the spring

$m_B g x = \dfrac{1}{2}(m_A + m_B)v^2 + \dfrac{1}{2}k(\Delta L)^2$

Substituting the values ($m_B = 0.32 \text{ kg}$):

$0.32 \times 10 \times 0.3 = \dfrac{1}{2}(0.32 + 0.32)v^2 + \dfrac{1}{2} \times 40 \times (0.1)^2$

$0.96 = 0.32 v^2 + 20 \times 0.01$

$0.96 = 0.32 v^2 + 0.2$

$0.32 v^2 = 0.76$

$v^2 = \dfrac{0.76}{0.32} = 2.375$

$v = \sqrt{2.375} \approx 1.54 \text{ m/s}$

Answer: $1.54 \text{ m/s}$