Question

A spring with a spring constant of 2.5 N/m is stretched by 2

m. What is the potential energy in the spring?

Answer 1

Eelas = 5 [J]

Step-by-step explanation:

Elastic energy is associated with the ability that has a spring or any other material with elastic properties.

We can calculate the elastic energy by means of the following equation:

`E_(elas)=(1)/(2) *k*x^(2)`

where:

k = constant spring = 2.5 [N/m]

x = distance stretched = 2 [m]

`E_(elas)=(1)/(2) *2.5*(2)^(2)\\E_(elas)=5[J]`