Hey, guys. So we got these 2 blocks and they're connected by a string like this, but this whole system here is actually being pulled upwards by a rope like this and, you know, this force is 100 newtons. So we've got the masses of both blocks. What we want to do is we want to calculate the acceleration of the blocks in the first part. So this is really going to be a. And then in the second part, we're going to calculate some tension. So we want to calculate the acceleration. We know we can use this shortcut here. So I've got this 2 kilogram block, this 3. So I'm going to call these 1 a and b. And what I want to do is just really quickly draw free-body diagrams for both of these. So I've got the \(m_ag\), that's the weight force. Then I've got the \(T_{rope}\), that's the tension force in the rope, and I actually know that's 100. And what I've also got is this tension force in the connecting string and I'll call this \(T_s\). Now for the bottom, for the bottom object, I've got the weight force \(m_bg\), and then I've got the upward force of the tension in the connecting string. So we can use the shortcut to find the acceleration. All we're doing really is we're just replacing these 2 blocks with one 5 kilogram block, right, by adding the masses together, and then we're just going to ignore any of the internal or the connecting forces like the tension in the string. This one doesn't get ignored because it's not actually between the objects. Remember that. Right? So basically, what happens is that this one block here is being pulled upwards by \(T_{rope}\), which is 100. And then these weight forces, \(m_ag\) and \(m_bg\), can actually combine. Right? They don't cancel. And so they combine to actually produce a total weight which is really just big m times g. So to figure out the acceleration, all we have to do is just use our \(F = ma\). So then, actually, we're going to use \(F = m \times a\). We're going to use the same rules, you know, upward is positive, so any forces along and against get positives and negatives. So our \(T_{rope}\) is positive, and then minus our big \(mg\) is negative and this equals big \(ma\). So this is going to be a \(100 - 5 \times 9.8 = 5a\). So we generally getting this \(51 = 5a\), and so \(a = 10.2\) meters per second squared. And that's your answer. So, you know, this system here is going to accelerate upwards because we got a positive number at 10.2. That's as easy as that. Right? You can just go ahead and lump all these things together to a single object to solve for the acceleration.
So let's move on to part b now. Now what we want to do is we want to find tension in this connecting string over here. So this is actually going to be \(T_s\). So here's the deal. Whenever you are solving for the acceleration in these kinds of problems, remember, you can always use the shortcut. But if you have to go back and you have to solve for a connecting or internal force, then what you're going to have to do is you're going to have to draw a free-body diagram and write \(F = ma\) for the simplest object. So for example here, we've got both of these objects that involve \(T_s\). Right? They both have \(T_s\) in them. So I'm just going to pick the simplest one, which is really our 3 kilogram block. So if I write for the 3 kilogram block, I write \(F = ma\). Remember, I have to use \(F = ma\) for the little object. Right? So I have to use the little m. So this is going to be \(m_b \times a\). So our forces are going to be \(T_s - m_bg = m_ba\). So when you move this to the other side, the \(m_b\) is actually a common factor, so it can kind of be pulled out of the parenthesis. So we've got \(m_b(a + g\)). And now you can solve this. Right? This is going to be the mass of b which is 3 times a which is 10.2 plus 9.8, that's g. And so if you work this out, what you're going to get is the tension is exactly equal to 60 newtons. Alright? So it's as simple as that. That's the answer. Let me know if you guys have any questions.