How much will a spring with a stiffness of 500 N / m stretch under a force of 0.2 kN?
| September 16, 2021 | education
Given: k (stiffness coefficient of the considered spring) = 500 N / m; F (the amount of force that stretches the spring) = 0.2 kN (200 N).
The value of tension of the considered spring is determined by the formula (Hooke’s law): | Fcont. | = | F | = k * x and x = F / k.
Calculation: x = 200/500 = 0.4 m (40 cm).
Answer: The spring in question will stretch 40 centimeters.
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.
