Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed forward by a 12 N force applied to the 1 kg block. (b) How much force does the 2 kg block exert on the 1 kg block?

Video transcript

Hey, everyone. So this problem is dealing with Newton's second law. I'm just gonna draw a diagram of what's going on as we're reading through this. So three wheeled frictionless toys are placed in contact with each other on a frictionless level bench, their masses are 1.5 kg, two kg and 2.5 kg. To tell us the 1.5 kg is on toys on one side, the two kg toy is in the middle and the 2.5 kg toy is on the other side, a child applies a 15 Newton force in the positive direction on the 1.5 kg toy determine the contact force between the 1. kg and two kg toys. So that force right there in between the two, our multiple choice options, answers choices are a 18.8 newtons. B 11.3 newtons C 3. newtons or D 8.75 newtons. Okay. So for this problem, the first thing we're going to do is a free body diagram to some the forces on the 1.5 kg toy. So we're going to take that toy by itself and see what that scenario looks like. So on the 1.5 kg, toy we know we have the force of the child that's given to us as 15 newtons and then we'll have the force of the two kg toy pushing back on it as it is trying to exert a force on the two kg toys. We'll call that F of two. So when we look, when you can recall Newton's second law is F equals M A. If we sum the forces here, we've got the force from the child minus F two equals mass times acceleration. So we're gonna be solving for F two. We from the problem, we do know the force of the child, we know the mass but we don't know the acceleration. However, what we do know is that each toy has the same acceleration when it is pushed because it's frictionless, they're all moving together. So with that information, we can draw a free body diagram of all of the toys together as a single particle and then some those forces. So that looks like one particle there. We've got the force of the child, there is no friction force. And so there is no other, there's no force acting in the negative direction. The only force in this scenario when we look at all the toys together is the force of the child. So when we look at that F equals M A, we've got F C equals All, all of the masses together. So the mass of the 1.5 kilogram toy, right those out 1.5 kg plus two kg plus 2.5 kg times acceleration. So we know that the force, the child is 15 newtons Divided by six kg equals our acceleration And that equals 2. m/s squared. So now we can go back to our first equation And solve for F two. So F two equals F C minus M A Where FC is again, 15 Newtons mass. Now we're just working um with that 1.5 kg toy. So it's just going to be 1.5 kg and acceleration. We sold for 2.5 m per second squared. Let me plug that in. We get a force F two of 11. newtons. So when we go back up to our answer choices, we can see that aligns with answer choice B so that's all we have for this one. We'll see you in the next video.