Question

Тело массой 20 кг подвесили на пружину, жесткостью 100 Н/м. На сколько при этом удлини, ась пружина? Ответ дать в метрах. Ускорение свободного падения принять равным 10 м/с2. ПОМОГИТЕ ПОЖАЛУЙСТА!!!

Answer 1

Ответ:

Объяснение:

Дано:

m = 20 кг

k = 100 Н/м

_________

Δx - ?

Закон Гука:

F = k·Δx

Удлинение:

Δx = F / k = m·g / k = 20·10 / 100 = 2 м

Answer 2

Calculation of Spring Elongation

To calculate the elongation of the spring when a 20 kg mass is suspended from it, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its elongation. 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 (in newtons per meter) - x is the displacement of the spring from its equilibrium position (in meters)

We can rearrange this formula to solve for the displacement x:

x = F / k

Given: - Mass (m) = 20 kg - Spring constant (k) = 100 N/m - Acceleration due to gravity (g) = 10 m/s^2

First, we need to calculate the force exerted by the mass:

F = mg F = 20 kg * 10 m/s^2 F = 200 N

Now, we can use Hooke's Law to find the displacement of the spring:

x = F / k x = 200 N / 100 N/m x = 2 meters

Answer

The spring elongates by 2 meters when a 20 kg mass is suspended from it.

I hope this helps! If you have any other questions, feel free to ask!