Respuesta :

= 1/2kx2

U = potential energy of a spring at a certain position

k = the spring constant, specific to the spring, with units N/m.

x = distance the spring is stretched or compressed away from equilibrium

Potential Energy: Elastic Formula Questions:

1) A spring, which has a spring constant k = 7.50 N/m, has been stretched 0.40 m from its equilibrium position. What is the potential energy now stored in the spring?

Answer: The spring has been stretched x = 0.40 m from equilibrium. The potential energy can be found using the formula:

U = 1/2kx2

U = 1/2(7.50 N/m)(0.40 m)2

U = 0.60 N∙m

U = 0.60 J