Question

Найдите жесткость пружины которая под действием силы 5 Н удлинилась на 8 см

Answer 1

F=kx
k=F/x
k=5 Н / 0,08м = 62,5

Answer 2

Calculation of Spring Stiffness

To calculate the stiffness of a spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, this can be expressed as:

F = k * x

Where: - F is the force applied to the spring (in Newtons) - k is the stiffness constant of the spring (in Newtons per meter) - x is the displacement of the spring (in meters)

In this case, we are given that the spring elongated by 8 cm (or 0.08 meters) under a force of 5 N. We can use this information to calculate the stiffness of the spring.

Plugging the given values into Hooke's Law equation, we have:

5 N = k * 0.08 m

To find the value of k, we can rearrange the equation:

k = 5 N / 0.08 m

Calculating this expression, we find that the stiffness of the spring is approximately 62.5 N/m.

Please note that the search results did not provide a direct answer to the question. However, the calculation is based on the well-known Hooke's Law, which is a fundamental principle in physics.