When the elastic spring is compressed by a value of x1 = 0.5 m, work A = 12.5 J is performed.

When the elastic spring is compressed by a value of x1 = 0.5 m, work A = 12.5 J is performed. What force must be applied to compress this spring by an amount of x2 0.03 m?

x1 = 0.5 m.

A = 12.5 J.

x2 = 0.03 m.

F2 -?

According to Hooke’s law, the elastic force F is directly proportional to the change in the length of the spring x: F = k * x, where k is the coefficient of proportionality, which is called the stiffness of the spring.

F2 = k * x2.

When the spring is compressed, the work of the forces A is equal to the change in the potential energy of the spring Ep: A = ΔEp.

The change in the potential energy of the spring is expressed by the formula: ΔEp = k * x1 ^ 2/2.

A = k * x1 ^ 2/2.

Let us express the stiffness of the spring k: k = 2 * A / x1 ^ 2.

F2 = 2 * A * x2 / x1 ^ 2.

F2 = 2 * 12.5 J * 0.03 m / (0.5 m) ^ 2 = 3 N.

Answer: to compress the spring, a force F2 = 3 N.