Calculating Extra Work for Hooke's Law Spring

[Momentum09]

Homework Statement

It takes 2.11 J of work to stretch a Hooke's law spring 6.08 cm from its unstressed length. How much the extra work is required to stretch it an additional 6.93 cm?

Homework Equations

F = -kx, W = Fd

The Attempt at a Solution

I first solved for Force by dividing 2.11J by 0.0608 m, then I just took that value and multiplied by 0.0693. I am not sure if I'm doing the correct thing. Is there a need to solve for k?

Thank you very much!

 

[Momentum09]

nvm I got it.