Answer:
The required elastic potential energy is 0.068J
Explanation:
Given,
Spring force constant, k= 3.4 N/M
Spring stretch length/ Displacement, x= 0.2m
We know,
Elastic potential energy, V= [tex]\frac{1}{2} kx^{2}[/tex]
=[tex]\frac{1}{2}*3.4*(0.2)^{2}[/tex]
=[tex]\frac{0.136}{2}[/tex]
=0.068
∴The required elastic potential energy is 0.068J
