It takes 1.81 J of work to strech a Hooke's-law spring 7.32 cm from its unstressed length.?

Question

How much the extra work is required to stretch it an additional 11.5 cm? Answer in units of J.

Answer 1

Fin the spring constant using the given information
U = k x²
1.81 J = k (7.32/100)² m²
k = 337.8 N/m
Now find the answer using the same formula for energy in a spring and the spring constant found above:
U = k x²
U = 337.8 N/m * (7.32 + 11.5)²/100² m² = 11.96 J

Extra work = 11.96J - 1.81J = 10.15 J

Answer 2

A shorter way; you don't need to find k.
PE is proportional to x^2
PE2 = PE1 * ((11.5+7.32)/7.32)^2
ΔPE = PE2 - PE1