## Essential University Physics: Volume 1 (4th Edition) Clone

We know that $W=(\frac{1}{2})kx^2$ Now we can find the required ratio as $\frac{k_B}{k_A}=(\frac{W_B}{W_A})(\frac{x_A}{x_B})^2=(\frac{2}{1})(\frac{2}{1})^2=8$