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?
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
To determine the force exerted by the 2.72 kg block on the 4.05 kg block, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
In this case, as the 1.10 kg block exerts a force on the 2.72 kg block, the 2.72 kg block will exert an equal and opposite force on the 1.10 kg block. Similarly, the 2.72 kg block will also exert a force on the 4.05 kg block, and this force will be the same as the force exerted by the 1.10 kg block on the 2.72 kg block.
Therefore, we just need to determine the force exerted by the 1.10 kg block on the 2.72 kg block, which is 14.3 N, and this will be the force exerted by the 2.72 kg block on the 4.05 kg block.
Hence, the force exerted by the 2.72 kg block on the 4.05 kg block is 14.3 N.