A

Given:
Mass of block A, m_A = 5 kg
Mass of block B, m_B = 2 kg
Speed of mass A when B leaves contact, v = 5 m/s
Spring constant, k = 40 N/m
Gravitational acceleration, g = 10 m/s²

When block B leaves the ground, it can be considered that the motion of block A is also an increase in spring potential energy.
The gravitational potential energy lost by block B plus the kinetic energy gained by block A will equal the elastic potential energy stored in the spring.

1. Calculate the gravitational potential energy lost by block B as it descends by distance x: \[ PE_B = m_B imes g imes x \]

2. The kinetic energy gained by block A is given by: \[ KE_A = \frac{1}{2} m_A v^2 \]

3. The elastic potential energy stored in the spring is: \[ PE_{spring} = \frac{1}{2} k x^2 \]

Using conservation of energy, we can equate the energies:
\[ m_B g x + \frac{1}{2} m_A v^2 = \frac{1}{2} k x^2 \]
Substitute the values:
\[ 2 kg \times 10 m/s² \times x + \frac{1}{2} \times 5 kg \times (5 m/s)^2 = \frac{1}{2} \times 40 N/m \times x^2 \]
\[ 20x + \frac{1}{2} \times 5 \times 25 = 20x^2 \]
\[ 20x + 62.5 = 20x^2 \]
Rearranging gives: \[ 20x^2 - 20x - 62.5 = 0 \]
Dividing by 5 leads to: \[ 4x^2 - 4x - 12.5 = 0 \]
Solving using the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-12.5)}}{2 \cdot 4} \]
\[ x = \frac{4 \pm \sqrt{16 + 200}}{8} \]
\[ x = \frac{4 \pm 14}{8} \]
Taking the positive root: \[ x = \frac{18}{8} = 2.25 \text{ m} \]
Therefore, the value of x is 2.25 m.