## Fundamentals of Physics Extended (10th Edition)

$39.2J$
Elastic potential energy stored in spring can be defined as: $E_{p}=\dfrac {kx^{2}}{2}$ Given the numbers, $k=19.6\dfrac {N}{cm}=19.6\dfrac {N}{10^{-2}m}=19.6\times 10^{2}\dfrac {N}{m};x=20cm=0.2m$ We get: $E_{p}=\dfrac {kx^{2}}{2}=\dfrac {19.6\times 10^{2}\dfrac {N}{m}\times \left( 0.2m\right) ^{2}}{2}=39.2J$