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