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 = [tex]\frac{1}{2}[/tex] ke²
k is the elastic constant
e is the extension
Insert the parameters;
Elastic potential energy = [tex]\frac{1}{2}[/tex] x 120 x 0.2² = 2.4J
