A row of blocks are lined up with masses of 1.10 kg, 2.72 kg and 4.05 kg. The masses are then pushed forward by a 14.3 N force applied to the 1.10 kg block. If the table is frictionless, how much force does the 2.72 kg block exert on the 4.05 kg block?

Answer 1

The three blocks are accelerated in the amount of a = F/m = 14.3/(1.1+2.72+4.05) = 1.817 m/s^2

They all have that same acceleration. So, since the 1.1kg block is only pushing 2.72+4.05 kg, its force is F=ma

Similarly, the last block is pushed with a force of 4.05a N