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http://01.edu-cdn.com/files/static/mcgrawhillprof/9780071623209/NEWTON_S_SECOND_LAW_FNET_MA_PRACTICE_PROBLEMS_01.GIF

Physics
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In the diagram above, a 1.0–kg cart and a 2.0–kg cart are connected by a rope. The spring scale reads 10 N. What is the tension in the rope connecting the two carts? Neglect any friction. 30 N 10 N 6.7 N 5.0 N 3.3 N
As a guess, 10N is being applied on the 3kg system- intuition says 10/3 N is being exerted on the 1kg object to achieve the acceleration, so the tension is 10/3N. Probably incorrect as I didn't include the 2kg mass.
Divide total force into 3 parts (which will give you 3 parts of 3.3 N).. 2 parts pull 2 kg and 1 part pull 1 kg.. 2 parts of 3.3N (total 6.6N) pulls 2 kg.. So, beyond it force applied (on 1 kg via rope)= remaining force=3.3 N which will be the tension in rope.. So, tension= 3.3 N

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Another way of looking at it|dw:1349896310366:dw||dw:1349896342952:dw|
i did not get u uujawal
ok but y did u divid 10/3
To work out acceleration, treat the whole system as a mass m=3kg. F=10N Obviously, a=10/3 (m/s^2) So, the small string's got to pull the small mass enough to make it accelerate at 10/3 (m/s^2). What force can do that? F=ma=1*(10/3) 10/3N
@henpen 's approach is better and more formal.. What i said was more informal and you couldn't write it that way in your exam! Follow henpen's process @ksaimouli
alright thx both of u

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