A wire is secured at one end and a force $F_1$ is applied at the other end so that the wire stretches. In this situation, the wire stores $0.15\,\text{J}$ of elastic potential energy. The applied force is then replaced by a force $F_2$. The wire then stores $0.60\,\text{J}$ of elastic potential energy. The wire obeys Hooke’s law. What is the relationship between $F_1$ and $F_2$?
- A$F_1 = 2F_2$
- B$F_1 = 4F_2$
- C$2F_1 = F_2$
- D$4F_1 = F_2$