Question

Жесткость пружины 5×10 3 степени H/м . предельная длина, на которую ее можно растянуть, 16 мм. сохранится ли упругость пружины, есои к ней приложить силу 50H ,100H

Answer 1

Calculation of Spring Stiffness:

The given spring stiffness is 5 × 10^3 H/m. This means that for every meter of extension, the spring exerts a force of 5 × 10^3 H (H = Newtons) in the opposite direction.

Maximum Extension of the Spring:

The maximum extension of the spring is given as 16 mm. To determine if the spring's elasticity will be maintained when a force of 50H or 100H is applied, we need to convert the maximum extension to meters.

Converting 16 mm to meters: 16 mm = 16/1000 m = 0.016 m

Elasticity of the Spring:

To determine if the spring's elasticity will be maintained when a force of 50H or 100H is applied, we need to compare the force applied to the maximum force the spring can withstand without permanent deformation.

If a force of 50H is applied to the spring, we need to calculate the extension it will cause. We can use Hooke's Law to calculate the extension:

F = k * x

Where: F = Force applied (in Newtons) k = Spring stiffness (in N/m) x = Extension (in meters)

For a force of 50H: 50H = (5 × 10^3 H/m) * x

Solving for x: x = (50H) / (5 × 10^3 H/m) = 0.01 m

Similarly, for a force of 100H: x = (100H) / (5 × 10^3 H/m) = 0.02 m

Comparing the calculated extensions to the maximum extension of the spring (0.016 m), we can see that both forces (50H and 100H) will cause the spring to exceed its maximum extension. Therefore, the elasticity of the spring will not be maintained when a force of 50H or 100H is applied.

Please note that the above calculations are based on the given information and assumptions. If there are any additional factors or considerations, please let me know.