Question

A freight train consists of two 8.00×105-kg engines and 45 cars with average masses of 5.50×105kg. (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00×10−2m/s2 if the force of friction is 7.50×105N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the magnitude of 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 backward on the track to accelerate the train at a rate of 5.00 × 10⁻² , {m/s}² if the force of friction is 7.50 × 10⁵ , {N}, assuming the engines exert identical forces?

A. 4.50 × 10⁵ , {N}
B. 5.25 × 10⁵ , {N}
C. 5.75 × 10⁵ , {N}
D. 6.00 × 10⁵ , {N}

Answer 1

The force each engine must exert backward on the track to accelerate the train at a rate of 5.00x10^-2m/s^2 is 6.00x10^5N. The force in the coupling between the 37th and 38th cars is also 6.00x10^5N.

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:

1. Mass of each engine (m): 8.00x105 kg
2. Number of engines (n): 2
3. Mass of each car (mcar): 5.50x105 kg
4. Number of cars (ncar): 45
5. Acceleration (a): 5.00x10-2 m/s2
6. Force of friction (Ffriction): 7.50x105 N

(a) To find the force each engine must exert, we can use the equation:

F - Ffriction = (mengine + n*mcar) * a

Substituting the given values:

F - 7.50x105 = (2*8.00x105 + 45*5.50x105) * 5.00x10-2

F = 7.50x105 + (2*8.00x105 + 45*5.50x105) * 5.00x10-2

F = 6.00x105 N

(b) To find 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 the cars have the same mass and the friction is evenly distributed, each car exerts an equal force on the other car. Therefore, the force in the coupling between the 37th and 38th cars is equal to the force exerted backward by a single car on the track, which can be calculated using the same equation as in part (a).

So, the force in the coupling between the 37th and 38th cars is also equal to 6.00x105 N.