A train consists of an engine and three cars tied closely together. The mass of

Question

A train consists of an engine and three cars tied closely together. The mass of the whole train is 1.5 times 10^5 kg; the engine alone has a mass of 80,000 kg and each car has the same mass. Through its wheels, the engine can exert a horizontal force of 2.0 times 10^4 N on Earth. (a) When the engineer wishes to accelerate forward, in which direction should the engine exert its force on Earth? backward forward What is the magnitude of the force exerted by Earth on the engine ? N (b) When the engineer acts to accelerate the train, what is the force on the set of three cars and what is the subsequent acceleration of these cars? (Treat the three cars as a unified whole.) N m/s^2 (c) When the engineer acts to accelerate the train, what are all the horizontal forces on, and subsequent acceleration of, the second car of the train? force from car ahead N force from car behind N resulting acceleration m/s^2

Explanation / Answer

(A) from newton's third law,

every action has reaction force of same magnitude in opposite direction.

F = 2 x 10^4 N

(B) a = F / M

a = (2 x 10^4) / (1.5 x 10^5) = 0.133 m/s^2 ......Ans

mass of three car = 1.5 x 10^5 - 80000 = 70000 kg

F = m a = 70000 x 0.1333 = 9.33 x 10^3 N ........Ans

(C) net force on middle car = m a = (70,000 / 3) (0.133) = 3111.11 N

net force on last car = 3111.11 forward

]net force on eangine and first car = (80000 + 70000/3) (0.1333)

= 13777.78 N

hence force due to middle car on this system = (2 x 10^4) - 13777.78

= 6222.22 N

So force on middle car from car ahead = 6222.22 N .....Ans

from car behind = 3111.11 N .......Ans

a = 0.133 m/s^2 .......Ans

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