A train engine pulls out of a station along a straight horizontal track with five identical freight cars behind it, each of which weighs 88.0 kN. The train reaches a speed of 13.0 m/s within 6.50 min of starting out. Assuming the engine pulls with a constant force during this interval, and ignore air resistance and friction on the freight cars. Find the tension in the coupling between cars 2 and 3??? Answer in kN please

Question

A train engine pulls out of a station along a straight horizontal track with five identical freight cars behind it, each of which weighs 88.0 kN. The train reaches a speed of 13.0 m/s within 6.50 min of starting out. Assuming the engine pulls with a constant force during this interval, and ignore air resistance and friction on the freight cars. Find the tension in the coupling between cars 2 and 3??? Answer in kN please

Answer 1

coupling is pulling cars 3, 4,5 of total mass 3*88000/9.8 kg

acceleration= 13/(6.5*60sec)

forceat couling = pulled mass*acceleration.

Answer 2

I got 897.96 as answer what would it be as kN???

Answer 3

To find the tension in the coupling between cars 2 and 3, we need to understand the concept of inertia and Newton's second law of motion.

According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration. In this case, the acceleration is the change in velocity divided by the time it takes to change the velocity.

First, let's calculate the mass of each freight car. Since each freight car weighs 88.0 kN, we can use the formula:

Mass = Weight / Acceleration due to gravity

Mass = 88.0 kN / 9.8 m/s^2
= 8.98 metric tons

Now, let's calculate the total mass of the five freight cars:

Total mass = Mass of one freight car * Number of freight cars
= 8.98 metric tons * 5
= 44.9 metric tons

Next, let's determine the acceleration of the train using the given information. The train reaches a speed of 13.0 m/s within 6.50 minutes.

First, convert the time to seconds:
Time = 6.50 minutes * 60 seconds/minute
= 390 seconds

Acceleration = Change in velocity / Time

Change in velocity = Final velocity - Initial velocity
= 13.0 m/s - 0 m/s
= 13.0 m/s

Acceleration = 13.0 m/s / 390 s
= 0.0333 m/s^2

Now, we can find the net force acting on the train. We assume the engine pulls with a constant force throughout this interval. The net force is equal to the product of mass and acceleration:

Net force = Mass * Acceleration
= 44.9 metric tons * 0.0333 m/s^2
= 1.495 kN

Since the train pulls with a constant force, the net force acting on each freight car should be the same. Therefore, the tension in the coupling between cars 2 and 3 is also 1.495 kN.

Therefore, the tension in the coupling between cars 2 and 3 is 1.495 kN.