# When the spring is stretched by 0.1 m, an elastic force equal to 2.5 N. arises in it.

When the spring is stretched by 0.1 m, an elastic force equal to 2.5 N. arises in it. Determine the potential energy of this spring when stretched to 0.08m.

x1 = 0.1 m.

Fcont1 = 2.5 N.

x2 = 0.08 m.

Wп -?

The potential energy of a stretched spring Wp is determined by the formula: Wp = k * x2 ^ 2/2, where k is the stiffness of the spring, x2 is the change in the length of the spring during stretching.

We express the stiffness of the spring k from Hooke’s law: Fupr1 = k * x1.

k = Fcont1 / x1.

The formula for determining the potential energy of a stretched spring will take the form: Wp = Fupr1 * x2 ^ 2/2 * x1.

Wp = 2.5 N * (0.08 m) ^ 2/2 * 0.1 m = 0.09 J.

Answer: an extended spring at x2 = 0.08 m has a potential energy Wp = 0.09 J. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.