How the potential energy of an elastically deformed spring changes when its length is doubled.

Given:

k – coefficient of spring stiffness;

dx2 = 2 * dx1 – increased the tension of the spring by 2 times.

It is required to determine W2 / W1 – how the potential energy of an elastically deformed spring will change with an increase in its length by 2 times.

In the first case, the potential energy will be equal to:

W1 = k * dx1 ^ 2/2.

In the second case, the potential energy will be equal to:

W2 = k * dx2 ^ 2/2 = k * (2 * dx1) ^ 2/2 = 4 * k * dx1 ^ 2/2 = 2 * k * dx1 ^ 2.

Then:

W2 / W1 = (2 * k * dx1 ^ 2) / (k * dx1 ^ 2/2) = 2 * 2 * k * dx1 ^ 2 / (k * dx1 ^ 2) = 2 * 2 = 4 times.

Answer: potential energy will quadruple.