Question

What is the Potential Energy of a spring stretched 5.0 cm that has a

spring constant of 79 N/m?

Answer 1

The Potential Energy of the stretched spring is = 1.975 J

Step-by-step explanation:

Recall that the potential energy of a spring of constant k stretched x\a distance x is given by the formula:

U = 1/2 k x^2

Therefore in our case, converting the stretch from cm of meters (0.05 m), this is

U = 1/2 * 79 * 0.05 J = 1.975 J