Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 20.0 kg and Box B has mass 5.0 kg. A horizontal force of 100 N is exerted on box A. What is the magnitude of the force that box A exerts on box B?
I know the answer is 20 N (back of the book)... I am just trying to understand the process to get there. We literally just learned f=ma and no other complex equations, so I'm quite lost as to how to get 20 N. I would really appreciate it if someone could help me asap!
I know the answer is 20 N (back of the book)... I am just trying to understand the process to get there. We literally just learned f=ma and no other complex equations, so I'm quite lost as to how to get 20 N. I would really appreciate it if someone could help me asap!

Both will accelerate at the same magnitude if they stay in contact. Find the rate at which both boxes will accelerate using F = ma...
a = F/m = (100 N)/(20.0 kg + 5.0 kg) = 4 m/s²
So you know Box B's mass and acceleration, and since the applied force is on A, the only force acting on Box B is the force Box A is exerting on it, so solve for it.
F = ma
F = (5.0 kg)(4 m/s²)
F = 20 N
You could also figure it out through box A. You know it's mass, and you know how much it accelerates. You know a 100 N force is being applied, but another force exists in the opposite direction of the motion: the force box B is exerting onto box A. So...
(100 N  F) = ma
100 N  F = (20.0 kg)(4 m/s²)
100 N  F = 80 N
F = 20 N
By Newton's Third Law, this is also the force A exerts onto B, so 20 N is the answer.
Either way works.
a = F/m = (100 N)/(20.0 kg + 5.0 kg) = 4 m/s²
So you know Box B's mass and acceleration, and since the applied force is on A, the only force acting on Box B is the force Box A is exerting on it, so solve for it.
F = ma
F = (5.0 kg)(4 m/s²)
F = 20 N
You could also figure it out through box A. You know it's mass, and you know how much it accelerates. You know a 100 N force is being applied, but another force exists in the opposite direction of the motion: the force box B is exerting onto box A. So...
(100 N  F) = ma
100 N  F = (20.0 kg)(4 m/s²)
100 N  F = 80 N
F = 20 N
By Newton's Third Law, this is also the force A exerts onto B, so 20 N is the answer.
Either way works.