When opening the door, the spring length increased by 0.12m; the spring force was 4N.
| March 28, 2021 | education
When opening the door, the spring length increased by 0.12m; the spring force was 4N. At what elongation of the spring is the elastic force equal to 10N?
These tasks: Δх1 (initial deformation of the door spring) = 0.12 m; Fupr1 (initial elastic force) = 4 N; Fel2 (considered elastic force) = 10 N.
To find the desired extension of the door spring, we use the equality: Fcont1 / Δx1 = k (stiffness of the door spring) = Fupr2 / Δx2, whence Δx2 = Fupr2 * Δx1 / Fupr1.
Calculation: Δх2 = 10 * 0.12 / 4 = 0.3 m.
Answer: An elastic force of 10 N will act on the door spring with an elongation of 0.3 m.
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.
