Question

A spring whose stiffness is 830N/m has a relaxed length of 0.49m. If the length of the spring changes from 0.22m to 0.8m, what is the change in the potential energy of the spring

Answer 1

The change in potential energy of the spring is 125.91 J.

Step-by-step explanation:

To find the change in the potential energy of the spring, we can use the formula for elastic potential energy: PE = 1/2 k x^2, where PE is the potential energy, k is the stiffness of the spring, and x is the change in length of the spring.

In this case, the stiffness of the spring is 830 N/m and the change in length is from 0.22 m to 0.8 m. So, the change in potential energy is:


PE = 1/2 (830 N/m) ((0.8 m)^2 - (0.22 m)^2) = 125.91 J.