Asked by Nick
                A spring has a spring stiffness constant, k, of 420 N/m . 
1. How much must this spring be stretched to store 30 J of potential energy?
            
        1. How much must this spring be stretched to store 30 J of potential energy?
Answers
                    Answered by
            nammilee
            
    Potential energy = 30 J,     
Spring stiffness constant k = 420 N/m
distance x = ?
Potential energy = 1/2*k*x^2
=> 30 = 1/2*420*x^2
=> x = √(30/210)
= 0.378 m
    
Spring stiffness constant k = 420 N/m
distance x = ?
Potential energy = 1/2*k*x^2
=> 30 = 1/2*420*x^2
=> x = √(30/210)
= 0.378 m
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.