Answer:
The spring constant of the spring is 8888.8 N/m
Explanation:
Given that,
Elastic potential energy of the spring, U = 25 J
Compression in the spring, x = 7.5 cm = 0.075 m
We need to find the spring constant of the spring. The elastic potential energy of the spring is given by the below formula as :
[tex]E=\dfrac{1}{2}kx^2[/tex]
k is the spring constant
[tex]k=\dfrac{2E}{x^2}[/tex]
[tex]k=\dfrac{2\times 25}{(0.075)^2}[/tex]
k = 8888.8 N/m
So, the spring constant of the spring is 8888.8 N/m. Hence, this is the required solution.
