Answer:
Potential energy of the spring is 0.024 Joules.
Explanation:
It is given that,
Spring constant of the spring, k = 120 N/m
The spring is stretched by 0.02 m
We have to find the potential energy of the spring. Mathematically, it is given by :
[tex]PE=\dfrac{1}{2}kx^2[/tex]
Putting the values of k and x in above equation we get :
[tex]PE=\dfrac{1}{2}\times 120\ N/m\times (0.02\ m)^2[/tex]
PE = 0.024 J
Hence, the potential energy of the spring is 0.024 J
