Answer
$W2=3W1$
Work Step by Step
The spring has a natural length of 20cm
$W1$ stretches the spring from $20cm$ to $30cm$
$W2$ stretches the spring from $30cm$ to $40cm$
Convert from $cm$ to $m$
Remember that $F=kx$
$W1=\int^{0.1}_{0}(kx)dx$
$W1=[\frac{kx^{2}}{2}]_{0}^{0.1}=\frac{k(0.01)}{2}=0.005k$
$W2=\int^{0.2}_{0.1}(kx)dx$
$W2=[\frac{kx^{2}}{2}]_{0.1}^{0.2}=\frac{k(0.04)}{2}-\frac{k(0.01)}{2}=0.02k-0.005k=0.015k$
$\frac{W2}{W1}=\frac{0.015k}{0.005k}=3$
$W2$ is three times the work of $W1$
