## Step 1: Calculate the Change in Gravitational Potential Energy When the block moves from Position A to Position B, it is lowered by 20 cm (.20 m). The change in gravitational potential energy (\(\Delta U_g\)) is: \[ \Delta U_g = mgh \] where \(m = 4.\, \text{kg}\) \(g = 10\, \text{m/s}^2\) (rounded for simplicity) \(h = .20\, \text{m}\) \[ \Delta U_g = 4. \times 10 \times .20 = 8.\, \text{J} \] **Explanation:** The gravitational potential energy decreases by 8. J as the block is lowered. --- ## Step 2: Calculate the Change in Spring Potential Energy The spring is compressed by .20 m at Position B. The elastic (spring) potential energy (\(U_s\)) stored is: \[ U_s = \frac{1}{2} k x^2 \] where \(k = 500\, \text{N/m}\) \(x = .20\, \text{m}\) \[ U_s = \frac{1}{2} \times 500 \times (.20)^2 = .5 \times 500 \times .04 = 10\, \text{J} \] **Explanation:** The spring gains 10 J of potential energy as it is compressed. --- ## Final Step: Find the Change in Total Energy The change in total energy (\(\Delta E\)) of the block-spring-Earth system is: \[ \Delta E = \text{Change in spring energy} + \text{Change in gravitational energy} \] \[ \Delta E = (+10\, \text{J}) + (-8\, \text{J}) = +2\, \text{J} \] So, the energy **increases by approximately 2 J**. The closest answer is: **Final Answer:** **(C) The energy increases by approximately 2. J.**