Question

Тело массой 200 грамм подвесили на пружину, при этом пружина удлинилась на 6 см, определите жесткость пружины. Ответ укажите в СИ.

Answer 1

Дано:

g= 10 м/с²

x= 6 см= 0,06 м

m= 200 г= 0,2 кг

________

k-?

Розв'язок:

$F= kx→k = \frac{F}{x}$

$F = mg$

Тоді формула набуває вигляду:

$k = \frac{mg}{x}$

$k = \frac{0.2 \times 10}{0.06} ≈ 33,3 Н/кг$

Відповідь: k≈ 33,3 Н/кг

Answer 2

Calculation of the Spring Constant

To determine the spring constant, 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 spring constant, - x is the displacement of the spring from its equilibrium position.

In this case, we know that the mass suspended on the spring is 200 grams, which is equivalent to 0.2 kilograms. We also know that the spring elongates by 6 centimeters, which is equivalent to 0.06 meters.

To find the spring constant, we need to calculate the force applied to the spring. The force can be calculated using Newton's second law of motion:

F = m * g

Where: - m is the mass of the object, - g is the acceleration due to gravity.

The acceleration due to gravity is approximately 9.8 m/s^2.

Let's calculate the force applied to the spring:

F = 0.2 kg * 9.8 m/s^2 = 1.96 N

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

k = F / x

Substituting the values we have:

k = 1.96 N / 0.06 m = 32.67 N/m

Therefore, the spring constant of the given spring is approximately 32.67 N/m.