Question

C помощью пружины жесткостью 96 Н/м брусок массой 2,4 кг равномерно тянут по столу.Определите удлинение пружины если коэффициент трения между бруском и столом равен 0,2

Answer 1

Сила реакции опоры в данной задаче равна силе тяжести:
N = mg
Если брусок движется равномерно, то сила упругости пружины равна силе трения:
Fy = Fтр
Fy = kx - сила упругости пружины
Fтр = мN - сила трения
из этой системы 4 уравнений:
kx = мmg
x = мmg/k=0,2 * 2,4 * 10 / 96 = 0,05м = 5 см

Answer 2

Fтр=Fудл
μmg=kx ,где х это удлинение
0,2*2,4*10=96х
4,8=96х
х=4.8/96=5см
Ответ: 5 см

Answer 3

Calculation of the Spring Extension

To determine the extension of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the extension or compression of the spring. 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 stiffness of the spring), and - x is the extension or compression of the spring.

In this case, we are given the spring constant (k = 96 N/m) and the mass of the block (m = 2.4 kg). We need to find the extension of the spring.

To find the force applied to the spring, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. In this case, the block is being pulled with a constant velocity, so the acceleration is zero. Therefore, the force applied to the spring is also zero.

Since the force applied to the spring is zero, we can set up the equation as follows:

0 = k * x

Solving for x, we find:

x = 0

Therefore, the extension of the spring is zero. The spring is not stretched or compressed when the block is pulled with a constant velocity.

Please let me know if you need any further assistance!