**Answer:**

1) Case A: a = 2.0 m/s^2

2) Case B, D, and E: a = 0.5 m/s^2

3) Case C: a = 0.16 m/s^2

4) Case F: a = 0.1 m/s^2

**Explanation:**

**Given:**

- A: 5N/m spring constant, 0.4m stretched, 1kg mass

- 5N/m spring constant, 0.2m stretched, 2kg mass

- 4N/m spring constant, 0.2m stretched, 5kg mass

- 1N/m spring constant, 0.5m stretched, 1kg

- 4N/m spring constant, 0.5m stretched, 4kg mass

- 1N/m spring constant, 0.5m stretched, 5kg mass

**Find:**

- Rank these systems on the basis of the magnitude of the inital acceleration of the blocks from greatest to least.

**Solution:**

- Develop a free body diagram of the block. There is only one force acting on the block the moment it is released i.e springs restoring force.

F_sp = k*x

- Apply Newton's second law of motion:

F_net = m*a

k*x = m*a

** a = k*x / m**

- After obtaining the general expression for acceleration we will compute for each case:

Case A: 5N/m spring constant, 0.4m stretched, 1kg mass

a = (5)*(0.4) / 1 = 2.0 m/s^2

Case B: 5N/m spring constant, 0.2m stretched, 2kg mass

a = (5)*(0.2) / 2 = 0.5 m/s^2

Case C: 4N/m spring constant, 0.2m stretched, 5kg mass

a = (4)*(0.2) / 5 = 0.16 m/s^2

Case D: 1N/m spring constant, 0.5m stretched, 1kg mass

a = (1)*(0.5) / 1 = 0.5 m/s^2

Case E: 4N/m spring constant, 0.5m stretched, 4kg mass

a = (4)*(0.5) / 4 = 0.5 m/s^2

Case F: 1N/m spring constant, 0.5m stretched, 5kg mass

a = (1)*(0.5) / 5 = 0.1 m/s^2

**Rank:**

1) Case A: a = 2.0 m/s^2

2) Case B, D, and E: a = 0.5 m/s^2

3) Case C: a = 0.16 m/s^2

4) Case F: a = 0.1 m/s^2