Answer to Question #144125 in Mechanics | Relativity for Fahad Gujjar

Question

figure below). The train is moving horizontally with a constant speed of 20 m/s (see the figure below), and the carriage is L = 20 m long. Suddenly, the train starts to accelerate uniformly with a = 2 m/s2 horizontally.
(a) Is the reference frame of the train now an inertial frame? Design an experiment that allows you to find out. (3 marks)
(b) Draw a free-body diagram of the man. Label the forces appropriately and explain their physical origin (one line each will suffice). (3 marks)
(c) If the coefficient of friction between the man’s shoes and the ‘floor’ of the train is μ = 0.1, what is the maximum time that the man takes to hit the back of the train carriage? For simplicity, you can assume that the coefficient of static friction is the same as the coefficient of kinetic friction. Assume that there are no obstacles between the man and the back of the train. (4 marks)
(d) How does your answer change if the coefficient of friction is 0.3 instead? Ex-

Answer 1

(a) The reference frame of the train is not an inertial frame now because the train is accelerating.

Experiment: Use the glass with tea. The glass with tea remains exactly the same as it would if the train were stopped in the station. This occurs when the train is moving with constant speed. If all the curtains of the train are pulled, you would have no way to know whether the train was moving or not. As the train starts to accelerate, you can notice the tea will be spilled out. Thus you know that the train is accelerating. A frame that is accelerating is not an inertial frame.

b)

W = mg is the weight of the man, m is the mass and g is the acceleration due gravity.

N is the normal reaction force due to the weight of the man. The normal reaction force is applied on the man by the floor and it is equal to the net weight of the man but the direction is opposite.

As the train starts with acceleration a, the man inside the train also starts to accelerate with acceleration a. Thus man fells a force F = ma.

F = μN is the horizontal force on the man due to friction between the floor and the man’s shoe. μ is the coefficient of friction.

c) m = 80 kg

a = 2 m/s²

F = ma

F = 160 N

f = μN = μmg

μ = 0.1

$f = 0.1 \times 80 \times 9.8 = 78.4 \;N$

$F_{net} = (160 – 78.4) = 81.6 \;N$

$a_{man} = \frac{F_{net}}{m} = \frac{81.6}{80} = 1.02\; m/s^2$

a_man is the acceleration of the man to reach the back of the train in a particular time t.

Path length S

$S = \frac{train\; length}{2}$

$S = \frac{20}{2} = 10\;m$

Initial velocity is 0

$S = \frac{1}{2}a_{man}t^2$

$t = \sqrt{\frac{2S}{a_{man}}}$

$t = \sqrt{\frac{2\times 10}{1.02}} = 4.43 \;sec$

d) If μ = 0.3

$f = 0.3 \times 80 \times 9.8 = 235.2 \;N$

$F_{net} = (160 – 235.2) = -75.2\; N$ (backward direction)

Acceleration of the man in the backward direction

$a_{man} = \frac{F_{net}}{m} = \frac{75.2}{80} = 0.94\; m/s^2$

$t = \sqrt{\frac{2\times 10}{0.94}} = 4.61 \;sec$