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) = (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) = (k*x^2)/(2)`

`5184 = (16200*x^2)/(2)`

`5184*2 = 16200*x^2`

`10368 = 16200x^2`

`16200x^2 = 10368`

`x^(2) = (10368)/(16200)`

`x^(2) = 0.64`

`x = √(0.64)`

`\boxed{\boxed{x = 0.8m}}`

Answer:

The displacement of the spring = 0.8m