Elastic Potential Energy Increases with Deformation

Steps:

• Elastic potential energy is given by U = (1/2)kx², where x is displacement from equilibrium.
• Initially, the spring is at equilibrium (x=0), so U=0.
• Extension means x>0; compression means x<0.
• In both cases, x²>0, so U>0 and increases from initial value.

Why D is correct:

• The formula U = (1/2)kx² depends on x², which is positive for both extension and compression, storing energy regardless of deformation direction.

Why the others are wrong:

• A: Incorrect; energy increases, not decreases, in both cases.
• B: Incorrect; energy increases when extended, not decreases.
• C: Incorrect; energy increases when compressed, not decreases.

Final answer: D