Answer
Spring 1 is stiffer.
Work Step by Step
To find the force constant using work, use the work done on a spring formula $$W=\frac{1}{2}k\Delta x^2$$ Solving for $k$ yields $$k=\frac{2W}{\Delta x^2}$$ Substituting known values of $W_1=180J$ and $\Delta x_1=0.20m$ yields a force constant of $$k_1=\frac{2(180J)}{(0.20m)^2}=9000N/m$$ Substituting known values of $W_2=210J$ and $\Delta x_2=0.30m$ yields a force constant of $$k_2=\frac{2(210J)}{(0.30m)^2}=4700N/m$$ The spring with the higher force constant, spring 1, is stiffer.
