## College Physics (7th Edition)

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.
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.