CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much does the elastic potential energy of the spring change during this process?

The elastic potential energy of the spring change during this process is 21.6 J.    

Explanation:

Given that,

Spring constant of the spring,

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

So, the elastic potential energy of the spring change during this process is 21.6 J.

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