How Much Extra Work Is Needed to Stretch a Spring Further?

[miamirulz29]

Homework Statement

It takes 3.23 J of work to stretch a Hooke's-law spring 12.1 cm from its unstressed length. How much extra work is required to stretch it an additional 5.76cm. Answer in Joules.

Homework Equations

F= -kx
W= FD

The Attempt at a Solution

At first, I try using proportions to do this problem:

3.23/12.1 = x/5.76
I got that answer but it was wrong.

So I tries finding the force using w=fd. Then I multiplied that by 5.76 and I got the same answer as doing it the proportion method.

What am I doing wrong. Do I need K or do I need to use the equation W = 1/2kx^2? Thanks in advance.

 

[miamirulz29]

Oh figured it out, nvm.