Question

Поезд массой 4*10^5 кг, двигавшийся по горизонтальному прямолинейному участку дороги, начал тормозить так, что его координата с течением времени изменялась в соответствии с уравнением х=А+Вt+Ct^2, где А=0,8км, В=8м/с, С=-0,5м/с^2. Определите а) модуль силы трения, действующей на поезд; б)коэффициент трения; в)время торможения поезда до полной остановки

Answer 1

Mdv/dt=|Fтр|
a=dv/dt=d2x/dt2=2C
2mC=|Fтр|=4*10^5 Н
F=kmg
k=1/g=0,1
В начальный момент времени t=0:
X=A
A=vt-at^2/2=at^2/2
V=at
(2A/a)^1/2=t
a=2C
t=40 c

Answer 2

Given Information:

- Mass of the train: 4 * 10^5 kg - Equation of motion: x = A + Bt + Ct^2, where A = 0.8 km, B = 8 m/s, and C = -0.5 m/s^2

a) Modulus of the friction force:

To determine the modulus of the friction force acting on the train, we need to consider the forces acting on the train during its deceleration. The main force opposing the motion of the train is the friction force.

The equation of motion can be rewritten as: x = A + Bt + Ct^2

Taking the second derivative of x with respect to time (t), we get: a = C

Since the acceleration (a) is equal to the second derivative of the position (x) with respect to time (t), we can conclude that the acceleration of the train is constant and equal to C.

The force (F) acting on an object is given by Newton's second law of motion: F = ma

Substituting the value of acceleration (C) and mass (m) of the train into the equation, we get: F = m * C

Substituting the given mass of the train (4 * 10^5 kg) and the value of C (-0.5 m/s^2), we can calculate the modulus of the friction force.

The modulus of the friction force acting on the train is 2 * 10^5 N.

b) Coefficient of friction:

The coefficient of friction (μ) is a dimensionless quantity that represents the frictional interaction between two surfaces. It can be calculated using the equation:

μ = F / N

where F is the frictional force and N is the normal force.

In this case, the normal force is equal to the weight of the train, which can be calculated using the equation:

N = mg

where m is the mass of the train and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the given mass of the train (4 * 10^5 kg) into the equation, we can calculate the normal force.

Using the modulus of the friction force calculated in part a) (2 * 10^5 N), we can calculate the coefficient of friction.

The coefficient of friction acting on the train is 0.5.

c) Time of braking until complete stop:

To determine the time it takes for the train to come to a complete stop, we need to find the time (t) when the velocity (v) of the train becomes zero.

The velocity (v) of the train can be obtained by taking the first derivative of the equation of motion: v = B + 2Ct

Setting v = 0, we can solve for t: 0 = B + 2Ct

Solving for t, we get: t = -B / (2C)

Substituting the given values of B (8 m/s) and C (-0.5 m/s^2) into the equation, we can calculate the time it takes for the train to come to a complete stop.

The time it takes for the train to come to a complete stop is 8 seconds.

Please note that the given equation of motion assumes that the train starts from rest at t = 0. If the train has an initial velocity, the equation of motion would be modified accordingly.