Question
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?
Answers
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
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
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