Тело массой 5 кг начинают тянуть в горизонтальном направлении через пружину, коэффициент жесткости
которой равен 100 Н/м. Определить модуль абсолютной деформации пружины к моменту начала движения тела, если коэффициент трения равен 0,3. Ответ: 0,15. Дайте полное решение пожалуйста , желательно на листеОтветы на вопрос
к моменту начала движения Fупр= Fтр
k*x=umg
x=umg\k=0,3*5*10\100= 0,15 м
Calculation of Absolute Deformation of the Spring
To calculate the absolute deformation of the spring, we can use Hooke's Law, which states that the deformation of an elastic material is directly proportional to the force applied to it. The formula for Hooke's Law is:
F = k * x
Where: - F is the force applied to the spring, - k is the spring constant (also known as the coefficient of stiffness), - x is the deformation or displacement of the spring.
In this case, we are given the mass of the body (5 kg), the spring constant (100 N/m), and the coefficient of friction (0.3). We need to determine the absolute deformation of the spring at the moment the body starts moving.
To solve this problem, we need to consider the forces acting on the body. The force applied to the spring is equal to the force of friction acting in the opposite direction. The force of friction can be calculated using the formula:
F_friction = μ * m * g
Where: - μ is the coefficient of friction, - m is the mass of the body, - g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values, we can calculate the force of friction:
F_friction = 0.3 * 5 kg * 9.8 m/s^2
Now, we can calculate the force applied to the spring:
F = F_friction
Using Hooke's Law, we can rearrange the formula to solve for x:
x = F / k
Substituting the values we have:
x = (0.3 * 5 kg * 9.8 m/s^2) / 100 N/m
Simplifying the equation:
x = 0.147 m
Therefore, the absolute deformation of the spring at the moment the body starts moving is approximately 0.147 m.
Please note that the answer provided is based on the information given and the calculations made.
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