The displacement of the spring is 0.71 m
Explanation:
The elastic potential energy stored in a spring is given by
[tex]U=\frac{1}{2}kx^2[/tex]
where
k is the spring constant
x is the displacement of the spring, measured with respect to the equilibrium position
For the spring in this problem, we have:
U = 50 J is the potential energy
k = 200 N/m is the spring constant
Solving for x, we find the displacement:
[tex]x=\sqrt{\frac{2U}{k}}=\sqrt{\frac{2(50)}{200}}=0.71 m[/tex]
Learn more about potential energy:
brainly.com/question/1198647
brainly.com/question/10770261
#LearnwithBrainly
