# Step 1: Calculate Initial Potential Energy in the Spring When the spring is compressed, all the energy is stored as potential energy: \[ E_{\text{spring}} = \frac{1}{2} k x^2 \] Where: - \(k = 2341\, \text{N/m}\) (spring constant) - \(x = 1.63\, \text{m}\) (compression from equilibrium) Plug in the values: \[ E_{\text{spring}} = \frac{1}{2} \times 2341 \times (1.63)^2 = \frac{1}{2} \times 2341 \times 2.6569 \approx 3107.23\, \text{J} \] **Explanation:** This is the total energy stored in the spring when compressed, which will convert to kinetic energy as the spring passes through equilibrium. --- # Step 2: Set Potential Energy Equal to Kinetic Energy at Equilibrium At \(x = \), all the potential energy becomes kinetic: \[ E_{\text{spring}} = E_{\text{kinetic}} = \frac{1}{2} m v^2 \] Set the energies equal and solve for \(v\): \[ 3107.23 = \frac{1}{2} \times 20.7 \times v^2 \] \[ 3107.23 = 10.35 \times v^2 \] \[ v^2 = \frac{3107.23}{10.35} \approx 300.34 \] \[ v = \sqrt{300.34} \approx 17.33 \] **Explanation:** By equating the elastic potential energy to kinetic energy, we solve for the velocity when the spring passes through equilibrium. --- # Final Answer \[ \boxed{17.33} \]