Question

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m.What is the displacement of the spring ?

Answer 1

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