Data:
\(E_{pe} = 5184J\)
\(K(constant) = 16200N/m\)
\(x(displacement) = ?\)
For a spring (or an elastic), the elastic potential energy is calculated by the following expression:
\(E_{pe} = \frac{k*x^2}{2}\)
Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.
Solving:
\(E_{pe} = \frac{k*x^2}{2}\)
\(5184 = \frac{16200*x^2}{2}\)
\(5184*2 = 16200*x^2\)
\(10368 = 16200x^2\)
\(16200x^2 = 10368\)
\(x^{2} = \frac{10368}{16200}\)
\(x^{2} = 0.64\)
\(x = \sqrt{0.64}\)
\(\boxed{\boxed{x = 0.8m}}\)
Answer:
The displacement of the spring = 0.8m
