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A train consists of 50 cars, each of which has a mass of $6.8 \times 10^{3} \mathrm{kg} .$ The train has an acceleration of $+8.0 \times 10^{-2} \mathrm{m} / \mathrm{s}^{2} .$ Ignore friction and determine the tension in the coupling (a) between the 30th and 31st cars and (b) between the 49th and 50th cars.

$1.1 \times 10^{4} \mathrm{N}$

$5.4\times 10^{2} \mathrm{N}$

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this question is not difficult to solve, but we have. You have to be quite smart about how can you think about it? So what I'm doing here is the following We wanted at her mind, for instance, in the first item, the tension in the coupling between the car number 30 on the car number 31. So what happens in the situation is that we have 30 cars pulling 20 cars, and this is the connection that happens between the car number 30 on the car number 31 and then to complete distention is quite a straightforward exercise. We just have to applying Newton's second law on either this car set or these car set. I choose to apply Newton's second law on these cars set by doing that to get the following. It says that the Net Force I'm also applying Newton's second law on this direction, which I'm calling the axe direction. So the net force in that direction is the course to the mass times acceleration off the object. There is only one force acting on that object to the right direction, and that only force is this tension. So the tension force is the course. True, the mass off that set off 20 cars so 20 times the mass off each car times the acceleration off that ah sent off cars. Using the information that we're given in the problem, we get attention. That is equals to 20 times 6.8 climb standard. The third times eight times 10 to the minus Truth. This is 20 times eight times 6.8 times 10. So the tension is approximately 1.1 time Stand to the fourth new terms in the situation. On the other situation, we have only one car that is push it by 49 cars. Applying Newton's second law of that single car, we get the following and that forces because of the mass off that single car times its acceleration, the net force is attention force. And then these easy questions. 6.8 times 10 to the third times eight times Stand to the miners Truth. These is 6.8 times eight times 10. These results, in attention off approximately 5.4 time stand to the second new terms

Brazilian Center for Research in Physics