Answer:
0.275J
Explanation:
Given parameters:
Extension = 0.5m
Spring coefficient = 2.2N/m
Unknown:
Energy stored in the rubberband = ?
Solution:
The elastic potential energy is given as:
Elastic potential energy = [tex]\frac{1}{2}[/tex] k e²
k is the spring constant
e is the extension
Elastic potential energy = [tex]\frac{1}{2}[/tex] x 2.2 x 0.5² = 0.275J