The springs in a car’s suspension have a spring constant of 50 000 N/m. The elastic potential energy store is 0.025 J. Calculate the extension of the springs.

Respuesta :

Answer:

1 mm

Explanation:

E = ½ kx²

0.025 J = ½ (50,000 N/m) x²

x = 0.001 m

x = 1 mm