A spring with a spring constant of 50 N/m is stretched by an extension of 2 cm. Calculate the energy in its elastic energy potential store. Use the equation: elastic potential energy = 1/2 × spring constant × extension²

Respuesta :

Answer:

0.01 J

Explanation:

E = ½ kx²

E = ½ (50 N/m) (0.02 m)²

E = 0.01 J