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Net Work & Constant Speed

Patrick Ford
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Hey guys, I have a very important problem here that's going to teach you a very important conceptual idea about the work energy theorem. That's going to allow you to sort of take a shortcut in some of your problems that you might see on homework or tests. So let's go and take a look here. We have five kg boxes on a flat surface, but it's rough, which means there's some kinetic friction. So I have this box like this, we have the masses five. And I'm going to be pushing on this box with a constant force. This is some applied force like this. Now, what happens is I'm moving at a constant eight m per second. So there has to be another forces canceling my push. And that's the force of friction. So, I've got my F. K. Here. I also have the MG and the normal force just to label all my forces. But I want to do in this. Problem is I want to figure out the network that's on the box. So remember that. To figure out network, you're just gonna have you're gonna do one of two things. You can add together all of the works done if you know all the forces right? By using some of all works. Or if you don't know all the forces you can use F net D. Cosign Theta now are applied force here is actually unknown. We've seen something like this before. So we're gonna have that the network is equal to the Net Force times D. Cosign Theta here. So all we have to do is just figure out the net force to figure out the work that's um that's the network. So we're gonna go over here and figure out the network of the net force by using F. Equals M. A. Right? So you're some of all forces equals mass times acceleration. So your forces really related to the mass times acceleration. But we actually know in this problem that we're moving in a constant eight m per second. So we know that we're going to be pushing this box, friction is gonna be opposing me but I'm going to be going at a constant speed. So because of this, I know that my acceleration is actually equal to zero. I don't have to work out. My forces actually can't because I don't know what the applied forces. Right? So basically what happens is that my net force because this is equal to zero is equal to zero. The forces have to cancel because I'm not actually accelerating. So what this means here is if I have a net force of zero, I'm just gonna plug this into my work net equation. You can probably expect what's going to happen once I multiply zero into deco Cynthia to the whole thing is going to cancel and I get a network that's equal to zero jewels. That's the answer. Here is another way you can think about this, remember that the network is equal to the change in the kinetic energy. That's the work energy theorem. So if I had zero network that's done on an object and that means that the change in the kinetic energy is also equal to zero. This brings me to a really important conceptual idea. So the work energy theorem is very useful conceptually because if you ever see that an object has constant speed in a problem and you're asked to calculate things like kinetic energies and works, you just know that they change. The kinetic energy is going to be zero and therefore the network is going to be zero. Remember that your kinetic energy is equal to one half mv squared. So if you're v never changes, your kinetic energy will never change. And that just means that there was zero work done on you net. It could just mean that there's no forces acting on you or just could mean that all the all the works are going to cancel out. So that's it for this one. Guys, let me know if you have any questions.