Guys, let's check out this problem here. We've got a 20 kilogram box that's moving along the floor, and we've got a downward force on it. So, let me go ahead and sketch this out. We have a 20 kilogram box that's on the floor, and we're pushing down on this box. I'm going to call this f down, and that's 30 newtons. We know that the box is going to keep its velocity constant, v equals 2. We want to figure out how hard we have to push the box horizontally so that we can keep this box at a constant speed. So, basically, there's another force right here, which I'll just call regular f, and that's basically what we're trying to figure out. We have the coefficient of friction. So what we want to do first is draw a proper free-body diagram. Let's go ahead and do that. So, basically, our free-body diagram is going to look like this. We have a downwards mg, and we look for any applied forces. We know there's 2. We have one that acts downwards. That's f down. We know what that is. And then we have our horizontal force, which is our f that we're trying to look for. Now remember there are 2 other forces. We have a normal force because it's on the floor. And then because these two surfaces are in contact and they're rough, we have some coefficient of friction. There's going to be some friction. Now, if this box is moving to the right, then remember, kinetic friction always has to oppose that motion. It always is in the opposite direction of velocity. So your fk points to the left like this. That's your free-body diagram.
Now, we want to figure out this force here. So what we want to do is write our F = ma. But first, I'm going to pick a direction as positive. So I'm just usually going to choose up and to the right to be positive. That's what we'll do here. So your sum of all forces in the x-axis equals mass times acceleration. We're going to start with the x-axis because that's where that force pops up. Alright. So we got our forces. When we expand our sum forces, we got f is positive and then we've got fk is to the left. What about this acceleration here? What about the right side of this equation? Well, remember, what we're trying to find is how hard we need to push this so that the box is moving at a constant 2 meters per second. If the velocity is constant, then that means the acceleration is equal to 0. So really, this is an equilibrium problem. So we've got 0 here. So basically, our applied force, our mystery f, has to balance out with the kinetic friction. So, basically, when you move this to the other side, it's equal to fk. So that means your force is equal to μkN. Remember that is the equation for kinetic friction. We have μk = 0.3. What about this normal force here? You might be tempted to write mg in place of the normal force, but I want to warn you against that because you should never assume that normal is equal to mg. When you look at your free-body diagram, you have to look at all the forces that are acting in the vertical axis and then basically use F = ma to solve for that normal. So, we're going to go over to the y-axis here and solve for normal. This is the sum of all forces in the y-axis equals may. Similar to the x-axis, an acceleration is going to be 0 in the y-axis because that would mean, basically, the block isn't going to go flying into the air or go crashing into the ground. That doesn't make any sense. So now we expand our forces. We've got normal, then you've got mg, and then you've got this f down. This basically this additional force that's pushing the block down, and that's equal to 0. So, when you solve for this, you're going to get n is equal to when you move both of these over to the other side, you're going to get 20 times 9.8 plus 30. If you go ahead and work this out in your calculator, you're going to get 226. So this is the number that you plug back into this equation here. So, basically, your force is equal to 0.3 times 226. So if you go ahead and work this out, what you're going to get is 67.8, and that's your answer. So you get 67.8 Newtons is how hard you need to push this thing.