Question

Пружина в покое имела длину 8 см. Силой 24 H ее сдавили до длины 5 см Чему равна жесткость пружины

Answer 1

Δx = 5 - 8 = - 3 см или  Δx = - 0,03 м

Закон Гука:

F = -k*Δx

Отсюда k:
k = - F/Δx = - 24/(-0,03) = 800 Н/м

Answer 2

Calculation of the Spring Constant

To determine the spring constant of the given 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.

Hooke's Law can be expressed as:

F = k * x

Where: - F is the force applied to the spring, - k is the spring constant, - x is the displacement of the spring from its equilibrium position.

In this case, we are given the following information: - The spring was initially at rest with a length of 8 cm. - It was compressed to a length of 5 cm by a force of 24 H.

To find the spring constant, we need to calculate the displacement of the spring. The displacement can be calculated by subtracting the final length from the initial length:

x = 8 cm - 5 cm = 3 cm

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

24 H = k * 3 cm

To convert the displacement from centimeters to meters, we divide by 100:

24 H = k * 0.03 m

Finally, we can solve for the spring constant:

k = 24 H / 0.03 m = 800 H/m

Therefore, the spring constant of the given spring is 800 H/m.