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
Answer: B! The right to vote!
Explanation:
just took test 100%
