Answer:
The elastic potential energy of the spring is 0.28 J
Explanation:
Given;
mass of the block, m = 0.05 kg
spring constant, k = 350 N/m
extension of the spring, x = 4 cm = 0.04 m
The elastic potential energy of the spring is calculated as;
[tex]U_x = \frac{1}{2}kx^2\\\\U_x = \frac{1}{2} \times 350 \times (0.04)^2\\\\U_x = 0.28 \ J[/tex]
Therefore, the elastic potential energy of the spring is 0.28 J
