Question

Тело массой 5 кг начинают тянуть в горизонтальном направлении через пружину, коэффициент жёсткости которой равен 100 Н/м. Определить модуль абсолютной деформации пружины через 2 с после начала дывижения тела, если коэффициент трени равен 0,3.

Answer 1

1) F = m*g 
(слева направо: сила тяжести, масса тела, ускорение свободного падения) 
g = F/m 
g = (G*M)/(R+h)^2 
(слева направо: ускорение свободного падения, гравитационная постоянная, масса Земли, радиус Земли, высота над поверхностью) 
(R+h)^2 = G*M/g 
h = sqrt(G*M/g) - R 
(sqrt - квадратный корень) 
3) a=k*g; a=0,3*10=3(m/c^2). 
4) F=kx 
x=F/k 
дано: 
m-2кг 
к=100н/м 
х=? 
Решение 
x=F/k 
F=m*g 
F=2*10=20Н 
отсюда х=20/100=0,2. 
5) Дано: 
v=12 м/с 
μ=0.4 
t-? 
Решение: 
t=v/a; a=μ*g; t=v/μ*g 
t=12/0.4*9.8=2.94 c 
Ответ: t=2.94≈3 с 
6) Fc*S=0,5*m*v^2; Fc=(0,5*m*v^2)/S; S=(v0*t)/2; v=v0-a*t; v=0; a=V0/t; a=10/40=0,25; S=(10*40)/2=200; Fc=(0,5*60*100)/200= ...Fc-k*g; k=Fc/g; 

Answer 2

Calculation of Absolute Deformation of the Spring

To calculate the absolute deformation of the spring after 2 seconds of the body's movement, we need to consider the force applied to the spring and the spring constant.

Given: - Mass of the body (m) = 5 kg - Spring constant (k) = 100 N/m - Coefficient of friction (μ) = 0.3

To find the absolute deformation of the spring, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the displacement of the spring from its equilibrium position.

The force applied to the spring can be calculated using Newton's second law of motion:

F = m * a

where F is the force, m is the mass, and a is the acceleration.

Since the body is being pulled horizontally, the only force acting on it is the force of friction. The force of friction can be calculated using the equation:

F_friction = μ * m * g

where F_friction is the force of friction, μ is the coefficient of friction, m is the mass, and g is the acceleration due to gravity.

The acceleration of the body can be calculated using the equation:

a = F_friction / m

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

F_spring = k * x

where F_spring is the force applied to the spring, k is the spring constant, and x is the displacement of the spring.

Since the force applied to the spring is equal to the force of friction, we can equate the two equations:

k * x = μ * m * g

Solving for x, we get:

x = (μ * m * g) / k

Now, we can substitute the given values into the equation to calculate the absolute deformation of the spring:

x = (0.3 * 5 * 9.8) / 100

Calculating the above expression, we find:

x ≈ 0.147 m

Therefore, the module of the absolute deformation of the spring after 2 seconds of the body's movement is approximately 0.147 meters.