Solving Newton's Second Law: 3 Blocks, 667 N Force

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SUMMARY

The discussion focuses on solving a physics problem involving three blocks on a frictionless surface subjected to a 667 N force. The blocks have masses of 2.8 kg, 6 kg, and 7.7 kg, and they all experience the same acceleration due to the applied force. The net force on the 2.8 kg block can be calculated using Newton's second law (F=ma), leading to the conclusion that the acceleration of the system is uniform across all blocks. Additionally, to determine the force between the 6 kg and 7.7 kg blocks, one must subtract the net force acting on the 6 kg block from the total force acting on the system.

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Momentum09
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Can somebody please give me hints as to how to approach this problem?

Three blocks are in contact with each other on a frictionless horizontal surface. [order: 2.8kg, 6kg, 7.7kg] A 667 N horizontal force is applied to the block with mass of 2.8kg. What is the net force on the block with mass 2.8kg? What is the resultant force on the block with mass 6kg?

Thank you!
 
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Whats similar about all three blocks? They all have the same acceleration. So
F=ma ---> \frac{F}{block_{1}+block_{2}+bock_{3}}=a
You can use that information to find the answers to your other questions.
 
Thanks!
And to find the magnitude of the force between the block with mass 6kg and 7.7kg, do I subtract the net force for 6kg from that of 7.7kg?
 

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