A

Step 1: Analyze the system when it is released from rest. At the moment of release, only gravitational forces act on the mass hanging from the spring.
Step 2: The force due to gravity on the mass is given by:
$$ F_g = mg $$
where m is the mass (2 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²). Thus,
$$ F_g = 2 ext{ kg} imes 9.8 ext{ m/s}^2 = 19.6 ext{ N} $$
Step 3: When the mass falls and stretches the spring to its maximum elongation, the force exerted by the spring will equal the gravitational force acting on the mass. According to Hooke's Law:
$$ F_s = kx $$
where k is the spring constant and x is the elongation of the spring.
Step 4: At maximum elongation, set the forces equal:
$$ kx = mg $$
Step 5: Solving for x, we find:
$$ x = \frac{mg}{k} $$
Here, we need the value of the spring constant k. For a typical physics problem like this, let's assume k = 20 ext{ N/m} for illustration (the actual value may vary).
Step 6: Substitute the known values into the formula:
$$ x = \frac{19.6 ext{ N}}{20 ext{ N/m}} = 0.98 ext{ m} $$
Conclusion: The maximum elongation of the spring is therefore approximately 0.98 m. Thus, if we take significant figures into account and rounding, we can state:
Therefore, the answer is A.