The derivation for elastic potential energy in a stretched wire is based on Hooke's Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance.

Here are the steps:

1. According to Hooke's Law, the force (F) required to stretch or compress a spring is given by F = -kx, where k is the spring constant and x is the displacement of the spring from its equilibrium position.

2. The negative sign indicates that the force exerted by the spring is always in the opposite direction to its displacement.

3. The work done (W) in stretching the spring is given by the integral of the force over the displacement, i.e., W = ∫F dx.

4. Substituting F = -kx into the equation gives W = ∫-kx dx.

5. Integrating this from the initial to the final position gives W = -1/2 k (x_final^2 - x_initial^2).

6. The work done on the spring is stored as potential energy (U). Therefore, the elastic potential energy stored in the spring is U = 1/2 k (x_final^2 - x_initial^2).

7. If we consider the initial position of the spring as the reference point (x_initial = 0), then the elastic potential energy in the spring is U = 1/2 k x_final^2.

So, the elastic potential energy in a stretched wire or spring is given by U = 1/2 k x^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position.

Question

The derivation for elastic potential energy in a stretched wire is based on Hooke's Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance.

Here are the steps:

1. According to Hooke's Law, the force (F) required to stretch or compress a spring is given by F = -kx, where k is the spring constant and x is the displacement of the spring from its equilibrium position.

2. The negative sign indicates that the force exerted by the spring is always in the opposite direction to its displacement.

3. The work done (W) in stretching the spring is given by the integral of the force over the displacement, i.e., W = ∫F dx.

4. Substituting F = -kx into the equation gives W = ∫-kx dx.

5. Integrating this from the initial to the final position gives W = -1/2 k (x_final^2 - x_initial^2).

6. The work done on the spring is stored as potential energy (U). Therefore, the elastic potential energy stored in the spring is U = 1/2 k (x_final^2 - x_initial^2).

7. If we consider the initial position of the spring as the reference point (x_initial = 0), then the elastic potential energy in the spring is U = 1/2 k x_final^2.

So, the elastic potential energy in a stretched wire or spring is given by U = 1/2 k x^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position.

Knowee AI · Accepted Answer