Answer
$39.2J$
Work Step by Step
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:
$E_{p}=\dfrac {kx^{2}}{2}=\dfrac {19.6\times 10^{2}\dfrac {N}{m}\times \left( 0.2m\right) ^{2}}{2}=39.2J$
