So let's check out this problem here. I have a 5.1 kilogram block and what I want to do is I want to calculate the force that I need to get the block moving. And then I want to calculate the force needed to keep it moving at constant speed. So there are really two different situations here that I'm going to draw out the diagrams for. We've got a 5.1 kilogram block on the ground, and I'm going to be pushing it with some mysterious force here to get it moving. And then in the second case, I've got the block like this, except now it's moving with some constant speed. So \(v \) equals constant here, and I want to figure out how hard you need to push it to keep it moving at constant speed. So there are two different forces here. Alright. So we want to draw the free body diagrams. Let's just draw all the other forces that are acting on these blocks. I've got my weight force. That's \( mg \), and then I've got the normal force. Alright. So in this particular case, the first one, where you're trying to push it, you're trying to push it with enough force to get the block moving. So therefore, it's not actually moving just yet, which means the kind of friction that you're going up against is going to be static friction. And the moment where you actually get the block moving as we've seen in the previous videos is that's equal to the \( f_s_{\text{max}} \) threshold. Then when you finally actually get it moving, when it's moving with some constant speed, now you're going up against kinetic friction because the velocity is not 0. So that's really the difference between these two diagrams here. In one you're going up against \( f_s_{\text{max}} \) and then the other one you're going up against kinetic friction. Alright. So how do we then calculate these forces? Basically, just use our \( f = ma \). So we have our \( f = ma\) here. Now we just pick a direction of positive. It'll be to the right for both of these diagrams here. So you've got \( f - f_s_{\text{max}} \). And then what's the acceleration? Well, the moment right when I get the block to move, the acceleration is still equal to 0. So we're gonna use \( a = 0 \) for this even though we're actually getting the block to move. We want to figure out the force that you need right before that happens when \( f = f_s_{\text{max}} \). So you have \( f = f_s_{\text{max}} \). Remember that has an equation that's mu static times the normal force. And so the normal force if you're just looking at a block sliding horizontally just gonna be equal to mg So basically \( f = \mu_{\text{static}} \times mg \) and so this is gonna be \( 0.7 \times 5.1 \times 9.8 \), and you're gonna plug this in. You're gonna get 35 newtons. So basically, assuming that this is the coefficient of static friction, you have to push this block with at least 35 Newtons to get it moving. So then we can figure out the other force here by using basically the exact same method. So now we're gonna do the sum of all forces equals mass times acceleration. Here, we know the acceleration has to be 0 because the velocity is gonna be So your forces are \( f \), except now it's not \( f_s_{\text{max}} \), you're just using \( f_k \). So those have to cancel. So your \( f = f_k \) which is mu k times normal. So your force is equal to \( \mu_k \times mg \), right, just like we had before. And so this is gonna be \( 0.5 \times 5.1 \times 9.8 \). So now if you plug this in, you're gonna get 25 Newtons. So let's talk about that. So we got these two different numbers here, which means that our answer is actually answer choice C. It takes 35 newtons to get this block to start moving. Once it actually is moving with some velocity, then it only takes 25 newtons. You don't have to push as hard to keep it moving at constant speed. So because the mu static is always greater than or equal to mu kinetic, what that means is that it always is harder to get something moving than it is to keep it moving. This is something you've definitely experienced before in everyday life. You push something really, really heavy and you have to push really hard at first. But once you actually get it to move, basically, the kinetic friction coefficient decreases, and so it's easier to keep it moving. You don't have to push as hard. Alright. So that's it for this one, guys. Let me know if you have any questions.