Elastic potential energy from Hooke's law

Steps:

• Hooke's law: F = kx, where k is the stiffness constant.
• Elastic potential energy U = ∫ F dx from 0 to x = ∫ kx dx = (1/2) k x^2.
• Substitute F = kx to get U = (1/2) F x.
• Young's modulus E relates to k via k = (A E)/L, but U remains (1/2) F x.

Why A is correct:

• It matches the standard formula for energy stored in a linear elastic system, equal to the area of the F-x triangle.

Why the others are wrong:

• B and D: Fx is work done by a constant force, not for varying F in Hooke's law.
• C: 1/4 Fx underestimates the triangular area by half.

Final answer: A