Question

Ученик достал из механической ручки пружину. Оказалось, что, если поставить пружину на стол вертикально и положить на неё ластик массой 27 г, то эта пружина сожмётся на 6 мм. Пружину можно считать невесомой. Ускорение свободного падения равно 10 Н/кг. 12. Какова жёсткость пружины? Ответ дайте в Н/м, округлив до целого числа. Ответ можно сохранять несколько раз 13. Эту пружину разрезали на две части, длины которых относятся как 1:2. Какова жёсткость длинной части? Ответ дайте в Н/м, округлив до десятых долей.

Answer 1

Ответ:

да

Объяснение:

12. 45

13. 67,5

14. Не понял(

Answer 2

Calculation of the Spring Constant (Stiffness) of the Spring

To calculate the spring constant (stiffness) of the 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.

The formula for Hooke's Law is:

F = k * x

Where: - F is the force applied to the spring (in Newtons, N) - k is the spring constant (stiffness) of the spring (in Newtons per meter, N/m) - x is the displacement of the spring from its equilibrium position (in meters, m)

In this case, we know the mass of the eraser (27 g) and the compression of the spring (6 mm). We can use this information to calculate the force applied to the spring.

First, let's convert the mass of the eraser to kilograms:

27 g = 0.027 kg

Next, let's convert the compression of the spring to meters:

6 mm = 0.006 m

Now, we can calculate the force applied to the spring using the formula:

F = m * g

Where: - m is the mass of the eraser (in kilograms, kg) - g is the acceleration due to gravity (in meters per second squared, m/s^2)

Given that the acceleration due to gravity is 10 N/kg, we can calculate the force:

F = 0.027 kg * 10 N/kg = 0.27 N

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

k = F / x

Substituting the values we have:

k = 0.27 N / 0.006 m = 45 N/m

Therefore, the spring constant (stiffness) of the spring is 45 N/m.

Calculation of the Stiffness of the Longer Part of the Spring

To calculate the stiffness of the longer part of the spring after it has been cut into two pieces, we can use the principle of conservation of energy.

The potential energy stored in a spring is given by the formula:

PE = (1/2) * k * x^2

Where: - PE is the potential energy stored in the spring (in Joules, J) - k is the spring constant (stiffness) of the spring (in Newtons per meter, N/m) - x is the displacement of the spring from its equilibrium position (in meters, m)

Since the spring has been cut into two pieces, the potential energy stored in the longer part of the spring should be equal to the potential energy stored in the original spring.

Let's assume the original length of the spring is L. After cutting, the length of the longer part of the spring is L/3 (since the lengths are in a 1:2 ratio).

The displacement of the longer part of the spring is 2/3 of the original displacement, which is 2/3 * 0.006 m = 0.004 m.

Now, we can set up an equation to calculate the stiffness of the longer part of the spring:

(1/2) * k * (0.004 m)^2 = (1/2) * 45 N/m * (0.006 m)^2

Simplifying the equation:

k * (0.004 m)^2 = 45 N/m * (0.006 m)^2

k = (45 N/m * (0.006 m)^2) / (0.004 m)^2

Calculating the value:

k = 168.75 N/m

Therefore, the stiffness of the longer part of the spring is 168.75 N/m.

Please note that the calculations are based on the given information and assumptions.