Answer:
0.47 J
Explanation:
The elastic potential energy of a spring is given as,
E = 1/2ke²........................ Equation 1
Where E = Elastic potential energy, k = spring constant, e = extension/compression.
Given: k = 15 N/m, e = 0.25 m.
Substitute into equation 1.
E = 1/2(15)(0.25)²
E = 0.46875
E ≈ 0.47 J.
Hence the elastic potential energy stored in the spring = 0.47 J
