Answer:
E = 30.37 J
Step-by-step explanation:
Given that,
The spring constant of the spring, k = 300 N/m
The spring is stretched by 0.45 m, x = 0.45 m
We need to find the elastic energy stored in the spring. The formula for the elastic energy is given by :
[tex]E=\dfrac{1}{2}kx^2\\\\=\dfrac{1}{2}\times 300\times (0.45)^2\\\\=30.37\ J[/tex]
So, the required elastic energy is equal to 30.37 J.
