Question

Самоделкин скрепил последовательно две пружины равной длины. Растягивая свободные концы получившейся конструкции, он заметил, что первая пружина удлинилась на 5 см, а удлинение второй оказалось равным 1 см. Какова жёсткость второй пружины, если жёсткость первой – 100 Н/м? Ответ ввести в Н/м с точностью до целых.

Answer 1

Вычислим силу, с которой действуют на первую пружину. F=kΔl 5 см=5/100=0.05 (м)
F=100*0.05=5 (Н)
Эта же сила действует и на другую пружину. Вычислим жесткость k1=F/Δl1 1 см=1/100=0.01 (м) k1=5/0.01=500 (Н/м)
Ответ: 500 Н/м

Answer 2

Calculation of the Stiffness of the Second Spring

To calculate the stiffness of the second 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, - k is the stiffness (also known as the spring constant) of the spring, and - x is the displacement of the spring from its equilibrium position.

In this case, we know that the stiffness of the first spring is 100 N/m and the displacement of the first spring is 5 cm (or 0.05 m). We also know that the displacement of the second spring is 1 cm (or 0.01 m).

To find the stiffness of the second spring, we can rearrange the equation as follows:

k2 = F2 / x2

Where: - k2 is the stiffness of the second spring, - F2 is the force applied to the second spring, and - x2 is the displacement of the second spring.

Since the force applied to both springs is the same (as they are connected in series), we can write:

F1 = F2

Substituting the values we know:

k1 * x1 = k2 * x2

100 N/m * 0.05 m = k2 * 0.01 m

Simplifying the equation:

5 N = k2 * 0.01 m

Finally, solving for k2:

k2 = 5 N / 0.01 m = 500 N/m

Therefore, the stiffness of the second spring is 500 N/m.