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

Answer:

Eelas = 5 [J]

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:

$E_{elas}=\frac{1}{2} *k*x^{2}$

where:

k = constant spring = 2.5 [N/m]

x = distance stretched = 2 [m]

$E_{elas}=\frac{1}{2} *2.5*(2)^{2}\\E_{elas}=5[J]$