Answer:
The elastic potential energy of the spring change during this process is 21.6 J.
Explanation:
Given that,
Spring constant of the spring, [tex]k=1.2\times 10^4\ N/m[/tex]
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 :
[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J[/tex]
So, the elastic potential energy of the spring change during this process is 21.6 J.
