Question

A spring has a spring constant of 120 N/m. How much energy is stored in the spring as it is stretched a distance of 0.20 meter?

Answer 1

Answer:

2.4J

Explanation:

Given parameters:

Spring constant  = 120N/m

Extension  = 0.2m

Unknown:

Amount of energy  = ?

Solution:

The energy stored in this stretched spring is called the elastic potential energy.

It can be derived using the expression below:

 Elastic Potential energy  = $\frac{1}{2}$ ke²  

k is the elastic constant

e is the extension

  Insert the parameters;

        Elastic potential energy  =  $\frac{1}{2}$  x 120 x 0.2²  = 2.4J