Question

Определить жесткость дверной пружины, если при закрывании двери происходит работа 250 Дж, а при открывании ее растягивает на 10 см.

Answer 1

F = kx =>
k = F/x = 250 / 0,01 = 25000 (Дж/м)

Answer 2

To answer your question, I will use the formula for the spring stiffness coefficient, which is the ratio of the elastic force to the deformation of the spring. The elastic force is equal to the work done by the spring when it is compressed or stretched. Therefore, the spring stiffness coefficient can be calculated as follows:

$$k = \frac{A}{\Delta l}$$

where $k$ is the spring stiffness coefficient, $A$ is the work done by the spring, and $\Delta l$ is the deformation of the spring.

Given that the work done by the spring when the door is closed is 250 J, and the deformation of the spring when the door is opened is 10 cm, we can plug in these values into the formula and get:

$$k = \frac{250}{0.1} = 2500 \text{ N/m}$$

Therefore, the spring stiffness coefficient of the door spring is 2500 N/m.

If you want to learn more about the spring stiffness coefficient, you can check out these web search results that I found for you:

- [Coefficient of spring stiffness: definition, formulas, measurement](https://molibden-wolfram.ru/koefficient-zestkosti-pruziny-opredelenie-formuly-izmerenie/) - [Spring stiffness: formula, how to find, coefficient, notation](https://wiki.fenix.help/fizika/zhestkost-pruzhiny) - [Spring stiffness, theory and online calculators](https://www.webmath.ru/poleznoe/fizika_45_zhestkost_pruzhiny.php)