Data:
\(E_{ep} = ?\) (The SI unit of potential energy is the joule)
\(k = 53 N/m\)
\(x = 0,21m\) → \(x = 2,1 * 10^{-1}m\)
Formula:
\(E_{ep} = \frac{k*x^2}{2}\)
Solving:
\(E_{ep} = \frac{k*x^2}{2}\)
\(E_{ep} = \frac{53*(2,1 * 10^{-1})^2}{2}\)
\(E_{ep} = \frac{53*4,41 * 100^{-1}}{2}\)
\(E_{ep} = \frac{233,73 * \frac{1}{100} }{2}\)
\(E_{ep} = \frac{ \frac{233,73}{100} }{2}\)
\(E_{ep} = \frac{ {2.3373} }{2}\)
\(\boxed{E_{ep} = 1.16865J}\)
