What is the tension in the coupling

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SUMMARY

The discussion centers on calculating the tension in the coupling between the third and fourth cars of an intercity Express locomotive pulling 14 cars, each with a mass of 53,000 kg, while accelerating at 0.26 m/s². Participants utilize Newton's second law (F=ma) to derive the tension, specifically focusing on the mass of the three cars involved in the calculation. The correct approach involves summing the masses of the three cars (159,000 kg) and applying the acceleration to find the tension in the coupling, resulting in a tension of 41,340 N.

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An inter city Express locomotive pulls 14 cars, each with a mass of 53,000 kg. What is the tension in the coupling between the thrid and fourth cars when the acceleration of the train is 0.26m/s.


ok so I am looking for T.
F=ma
53000(0.26) = T?
i don't think this is correct because they mentioned 3,4 cars
or was that just random?


I don't know at all.




Added:
a) What is the minimum time in which one can hoist a 1.00-kg rock height of 10.0m if the string used to pull the rock up has a breaking strength of 10.8 N? Assume the rock to be initially at rest.
b) If the string is replaced by one that is 50% stronger, by what percentage will the minimum time for the hoist be reduced.


alright a) i just used a= 2(x-xo)/t^2 and then i got 2(10)/t^2 = 10.8/1
so i got 10.8=20/t^2 and so t would be root 10.8 but the answer is 4.5s... why??

b) i need to get a) to solve this...
 
Last edited:
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Don't you have an idea about the whole system?

Your whole system is 14 cars, being pulled by a train with an acceleration of 0.26 m/s^2. Think about the problem of the 3 blocks (between each block there's a string) being pulled by a force F.
 
so there is T1 T2 and T3 and i have to find T3.
isnt T3
F=ma
F= (m1+m2+m3).26
so it would be 3(53000)(0.26) = T?
 

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