Answer
The work done in stretching a spring from its equilibrium position is directly proportional to $(\Delta x)^{2}$, $\Delta x$ being the amount of stretch.
Hence, the work done to stretch 2cm is $2^{2}=4$ times the work done to stretch it by 1cm.
Work Step by Step
The work done in stretching a spring from its equilibrium position is directly proportional to $(\Delta x)^{2}$, $\Delta x$ being the amount of stretch.
Hence, the work done to stretch 2cm is $2^{2}=4$ times the work done to stretch it by 1cm.