Question

Какова жёсткость пружины, если сила 10Н растягивает её на 5мм?

Answer 1

Из закона Гука следует, что k=F/x=10/0,005=2000 Н/м

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. The formula for Hooke's Law is:

F = kx

Where: - F is the force applied to the spring (in Newtons) - k is the spring constant or stiffness (in N/m) - x is the displacement of the spring from its equilibrium position (in meters)

In this case, we are given that a force of 10 N stretches the spring by 5 mm (which is equivalent to 0.005 m). We can use this information to calculate the spring stiffness.

F = 10 N x = 0.005 m

Using Hooke's Law, we can rearrange the formula to solve for the spring constant:

k = F / x

Substituting the given values:

k = 10 N / 0.005 m

Calculating the result:

k = 2000 N/m

Therefore, the stiffness of the spring is 2000 N/m.