Determine the potential energy of a spring stretched by 10 cm, if it is known that under

Determine the potential energy of a spring stretched by 10 cm, if it is known that under the action of a force of 30 N, the spring is stretched by 1 cm

According to Hooke’s law, the deformation of the spring x is proportional to the applied force F:

x = F / k,

where k is the coefficient of spring stiffness, N / m.

Let us determine the value of stiffness k, proceeding from the fact that with a force of F = 30 N, the spring is stretched by x1 = 1 cm = 0.01 m:

k = F / x = 30 / 0.01 = 3000 H / m.

Potential energy of a spring with rigidity k, stretched by x2 = 10 cm = 0.1 m:

A = k x2 ^ 2/2 = 3000 × 0.12 / 2 = 15.0 J.

Answer: 15 J.