Question

A spring whose stiffness is 980 N/m has a relaxed length of 0.50 m. If the length of the spring changes from 0.25 m to 0.81 m, what is the change in the potential energy of the spring

Answer 1

The change in potential energy of the spring is approximately 230.85 J.

Step-by-step explanation:

The potential energy of a spring can be calculated using the formula:

PE = (1/2)k(xf2 - xi2)

where PE is the potential energy, k is the spring constant, and xf and xi are the final and initial lengths of the spring, respectively.

In this case, the spring constant is 980 N/m, the initial length is 0.25 m, and the final length is 0.81 m. Substituting these values into the formula, we get:

PE = (1/2)(980 N/m)((0.81 m)2 - (0.25 m)2)

Simplifying this expression, we find that the change in potential energy of the spring is approximately 230.85 J.

Answer 2

The change in potential energy of a spring with a stiffness of 980 N/m and a relaxed length of 0.50 m is found by calculating the energy at 0.25 m and 0.81 m, resulting in a change of 16.494 joules.

Step-by-step explanation:

To calculate the change in potential energy of the spring, we can use Hooke's Law, which states that the potential energy stored in a spring is given by the equation PE = (1/2)kx2, where k is the spring constant and x is the displacement from the spring's relaxed length.

In this case, the spring constant k is given as 980 N/m. The displacement from the relaxed length when the spring is 0.25 m is -0.25 m, and when the spring is 0.81 m the displacement is 0.31 m (since the relaxed length is 0.50 m). The change in potential energy can be found by calculating the potential energy at each length and then subtracting the two.

Potential energy when the spring length is 0.25 m: PEinitial = (1/2)(980 N/m)(-0.25 m)2 = 30.625 J

Potential energy when the spring length is 0.81 m: PEfinal = (1/2)(980 N/m)(0.31 m)2 = 47.119 J

The change in potential energy is: ∆PE = PEfinal - PEinitial = 47.119 J - 30.625 J = 16.494 J