a chain consisting of 5 links, each with the mass of 100 g, is lifted vertically with constant acceleration of 3.00 m/s2. Find force acting between adjacent links 3 & 4.

Question

Count links from top to bottom as 1, 2, 3, 4, 5

Answer 1

The bottom two links exert a gravitational force of 0.200 kg x 9.8 m/s² on link 3. As the bottom two links are also accelerating at 3.00 m/s², there is an upward net force of 3.00 m/s²x 0.200 kg on the bottom two links.

If this is so, the force acting on link 4 by link 3 is F = net + mg = 3.00 m/s²x 0.200 kg + 0.200 kg x 9.8 m/s² = 3.76 newtons.

This seems to be equivalent to a mass of 0.300 kg attached by a massless string to a mass of 0.200 kg below it and the entire arrangement accelerating upward at 3.00 m/s². The net force in this case would be equal to the tension force on the string minus the weight of the bottom mass or tension force = net force + weight of bottom mass.

I would like it if another tutor can confirm t(or refute) this answer.