Question

A freight train consists of two 8.00 × 10⁴ , {kg} engines and 45 cars with average masses of 5.50 × 10⁴ , {kg}.

(a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00 × 10² , {m/s}² if the force of friction is 7.50 × 10⁵ , {N}, assuming the engines exert identical forces?
(b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

What force must each engine exert?
a) 1.25 × 10⁵ , {N}
b) 2.50 × 10⁵ , {N}
c) 3.75 × 10⁵ , {N}
d) 5.00 × 10⁵ , {N}

Answer 1

To accelerate the train, each engine must exert a backward force of 3.925 × 10^7N on the track. The force in the coupling between the 37th and 38th cars is also 3.925 × 10^7N.

Step-by-step explanation:

To calculate the force each engine must exert backward on the track to accelerate the train, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = ma). Given that the mass of each engine is 8.00 × 10^4 kg and the acceleration is 5.00 × 10^2 m/s², the force required is:

F = ma = (8.00 × 10^4 kg)(5.00 × 10^2 m/s²) = 4.00 × 10^7 N.

Since the force of friction is 7.50 × 10^5 N, the engines must exert an identical force to overcome the friction. Therefore, each engine must exert a backward force of 4.00 × 10^7 N - 7.50 × 10^5 N = 3.925 × 10^7 N on the track.

To determine the force in the coupling between the 37th and 38th cars, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. Since all cars have the same mass and friction is evenly distributed among them, each car exerts an identical force on the other. Therefore, the force in the coupling between the 37th and 38th cars is 3.925 × 10^7 N.