Question

16. A freight train consists of two 8.00×105 -kg engines and 45 cars with average masses of 5.50×105 kg . (a) 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? 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?

Answer 1

Each engine must exert a force of 2.04 × 10^5 N backward on the track to accelerate the train at a rate of 5.00 × 10^-2 m/s^2. The force in the coupling between the 37th and 38th cars is approximately 1.70 × 10^4 N.

Step-by-step explanation:

To calculate the force that each engine must exert backward on the track to accelerate the train at a rate of 5.00×10-2 m/s², we can use Newton's second law of motion, which states that force equals mass times acceleration.

First, we need to calculate the total mass of the train, which is the sum of the mass of the engines and the mass of the cars:

Total mass of the train = mass of engines + mass of cars = (2 engines * 8.00 × 105 kg) + (45 cars * 5.50 × 105 kg) = 16.00 × 105 kg + 24.75 × 105 kg = 40.75 × 105 kg

Next, we can calculate the force required to accelerate the train:

Force = mass * acceleration = (40.75 × 105 kg) * (5.00 × 10-2 m/s²) = 2.04 × 105 N

Since there are two engines, each engine must exert a force of 2.04 × 105 N backward on the track to accelerate the train at the given rate.

To calculate the force in the coupling between the 37th and 38th cars, we need to consider that the force of friction is evenly distributed among all the cars and engines. Since all the cars have the same mass, the force in the coupling between any two adjacent cars is equal to the force of friction divided by the number of cars. Therefore:

Force in the coupling between the 37th and 38th cars = force of friction / (number of cars - 1) = 7.50 × 105 N / (45 - 1) = 7.50 × 105 N / 44 ≈ 1.70 × 104 N

So, the force in the coupling between the 37th and 38th cars is approximately 1.70 × 104 N.

Answer 2

a. -2.84 × 10⁵ N b. 104.46 × 10⁵ N

Step-by-step explanation:

a. The net force on the freight train, ma = F - f where F = frictional force and f = backward force of engines. So, f = F - ma. Now, the total mass, m carried by the engine force = total mass of 45 trains cars + mass of two engines = 45 × 5.50 × 10⁵ + 2 × 8.00 × 10⁵ kg= (247.5 × 10⁵ + 16 × 10⁵) kg = 263.5 × 10⁵ kg.

Since the acceleration = 5.00 × 10⁻² m/s² and F = 7.50 × 10⁵ N.

So, f = F - ma = 7.50 × 10⁵ N - 263.5 × 10⁵ kg × 5.00 × 10⁻² m/s² = 7.50 × 10⁵ N - 13.175 × 10⁵N = -5.675 × 10⁵ N.

Since the engines exert identical forces, the force of each engine, f₀ = f/2 = -5.675 × 10⁵/2 N = -2.8375 × 10⁵ N ≅ -2.84 × 10⁵ N

b. After the 37 th cars, we have n - 37 cars. The net force on these cars is (n - 37)ma. The frictional force on these cars is F/(n - 37). Let f be the force on the couplings. So, the net force = frictional force on (n - 37) cars - force on couplings.

(n - 37)ma = F/(n - 37) - f

So, f = F/(n - 37) - (n - 37)ma. Since n = 45, we substitute the values of F, m and a

f = 7.50 × 10⁵/(45 - 37) N - (45 - 37)263.5 × 10⁵ kg × 5.00 × 10⁻² m/s²= 7.50/8 × 10⁵ N - 8 × 13.175 × 10⁵N = 0.9375 × 10⁵ N - 105.4 × 10⁵ N = -104.4625 × 10⁵ N. ≅ -104.46 × 10⁵ N.

So, the magnitude of the force between the 37th and 38th coupling is 104.46 × 10⁵ N