Question

A spring with a spring constant of 150 n/m has an equilibrium length of 8. 0 cm. What is the change in spring potential energy when the spring is stretched from a length of 11. 0 cm to 14. 0 cm?.

Answer 1

When the spring is stretched to 11.0 cm - 8.0 cm = 3.0 cm = 0.030 m away from equilibrium, it stores

1/2 (150 N/m) (0.030 m)² = 0.0675 J

of potential energy, while stretching it to 14.0 cm - 8.0 cm = 6.0 cm = 0.060 m from equilibrium stores

1/2 (150 N/m) (0.060 m)² = 0.27 J

Then the change in potential energy is 0.27 J - 0.0675 J = 0.2025 J ≈ 0.21 J