Thread: can someone check my work on this "work" question (hooke's law- springs)

sorry to ask another of these, but i got an exam tomorrow, and im not at all confident im doing these right... all the different units are killing me...

a spring has a natural length of 20 cm. if a 25 N force is required to stretch it to 30 cm, how much work is required to stretch it from 20 to 25 cm.

i found k to be 250 (after i realized to convert the 10 cm to .1 m)
so i evaluated the integral of 250x dx from .05 to .1 ...
then i found te answer to be 15/16 N-m, which I believe is equal to 15/16 J ?

unfortunately that answers not in the book, so i have no idea if im right or not..any help appreciated!

solution for the spring constant, k, is fine.

limits of integration should be from 0 to 0.05 to calculate the work done in stretching it from 20 cm (natural length, no stretch) to 25 cm (5 cm beyond its natural length).