https://youtu.be/OI3vCAIY-kw
PHYS 1101: Lecture Fifteen, Part Seven
Okay, next example. “A man pushes horizontally on a 60-kilogram crate, causing it to move across a rough floor at a steady speed. The coefficient of kinetic friction is 0.45. How much work has the man done after pushing the box 2 meters?”
I see words in here directly asking me about how much work something has done. In this case, it’s the man. My speed doesn’t change, but I have speed information, and I have, again, position information. The position has changed by 2 meters. So I immediately am drawn to energy conservation. Think about energy arguments for this object. What’s the object? It’s going to be the crate. That’s what these forces are acting on, causing it to move. I’m going to make it move to the right a distance, again, 2 meters.
What forces do I have? Let me sketch them real quick on this. I’ve got force due to gravity. I’m going to have a normal force. In this case, the man pushes horizontally. I’m going to do Fm for that force. The motion is to the right. The floor is rough so I know I’ve got friction opposing that motion. The crate is moving along at a constant speed so that tells me no acceleration. So all of my forces have to balance.
Just do a quick, logical check here. It looks good. Everything up balances down, and right left balances. So the basic equation that defines this energy constraint is that this net work is equal to the change in kinetic.
Okay, let’s think about the left side. I need to consider the work that gets done by all forces on the box as it moves to the right. Question 16 for you. The work done by gravity on the load, this is on the box, on the crate. A, B, or C?
So let’s work on that left side a bit. The net work is the work done by all of these forces. Let me rewrite this here. I want to be consistent. The work is done by a certain force, and that’s the force caused by the man. So Fman. And I think that matches my label. I guess I did Fm above.
Okay. I’ve emphasized normal forces never do work. In this case, gravity is perpendicular to s. This is a 90-degree angle. Or if you go this way, it’s a 270-degree angle. Whichever angle you use, you end up with 0 work. The cosine of this angle is 0.
What work does the force from the man do? And then the work of friction? I could flesh that out a bit more, but before I get too far, I’m going to think about the right side of my equation. Let me just note here that I at least have simplified that the left side drops down to just these two terms. Well, what about the right side? KEf minus KE0. Remember each of these is one-half m times vf2. Or times the speed squared, so it’s vf2. And the last one contains v02. Let me point out that my speed doesn’t change. vf equals v0 so the kinetic energy doesn’t change. Right? If you plugged in numbers for these, you’d get the same thing.
This would be like 5 minus 5. It’s the same thing. This has to go to 0. So my right side is 0. That will always be the case. If an object is moving along at constant speed, the kinetic energy doesn’t change. The object isn’t speeding up or slowing down. That means that the net work has to be 0, but individual forces can still be doing work. Some must be doing positive work though, while some is doing negative work. And the negative and the positive contributions have to be the same for it to balance, or add up, to 0. And that’s what’s going to happen here.
My visual, remember, said that the size of these two forces were equal. The force that the man exerts is parallel to s, so the work is going to be this Fm times s, and that’s positive. Force is in the same direction as motion. Friction is opposite. Remember friction always does minus friction times the distance amount of work. And so this one is doing positive work. Friction is doing the same amount of negative work. Those two combine and balance each other to give us 0.
Okay. So let’s put it together now, but first let’s jot down a quick list of knowns, and the thing that we want. We know a coefficient of kinetic friction. It’s pretty large in this case. It’s a rough floor. We know the mass of the crate is 60 kilograms, this object we’re focused on. And the distance is our 2 meters. What do we want? Okay, again, it has to be variable that’s going to show up in our equations. Our equation, again, is work net is equal to KEf minus KE0. And we’ve concluded that the right side is 0, so I have that the net work equals 0.
This net work breaks up into work that the force of the man exerts plus the work of friction. Those have to add up to 0. Well, what do I want? It asked me for the work that the man has done after pushing the box. You have to identify the force that this agent causes to the object because when I do my calculation for work, I’m always focused on the work that a particular force does. So the man is responsible for this force Fm. Therefore I have to go after the work that this force does.
So what do I want? Specifically, I’ve got to solve for the variable, work, of the man. The work that the man exerts, does on the object. Okay, so I’ve got to calculate. Well, let me just rearrange this. I want to solve for this as a variable, which simply means adding the friction to the other side. So all I need is a number for this, and then I take the negative of that, and I’ve got my answer.
So what is this, the work that friction does? Well, it’s the size of the force times the distance, and it’s negative, minus 1 times that. Friction always does negative work. Again, I know the distance, how big is the frictional force? Well, the frictional force is the coefficient times the normal force.
What’s the normal force? Well, that is given by the second law for the vertical part to the acceleration, which is 0. ay is the sum of the forces in y over m. There’s no acceleration up down. This leads to the fact that the vertical forces have to balance. If you go back to the picture, we only had two vertical forces plus FN minus mg. In this simple case, FN is equal to mg for this object.
Okay? Let’s go back. I’m going to plug that in. fk, then, is the coefficient times the normal force. Take that back a step then. I can plug that in for the work that that friction force does, and that’s equal to –fk times the distance, which is minus μk mg times s. A common mistake I’ve seen is for the work that gets done, there tends to be a focus on the force alone, and often people forget you have to multiply by the distance. So if you get a wrong answer, check to see if you’ve made that mistake.
And we’re ready to plug in numbers now. -0.45 times 60 kilograms times 9.8 meters per second squared times 2 meters, and we end up with a value of -529 joules. That’s the work that friction does. If we now take that number back a level, if we go back, we realize that the thing that we want to solve for is minus that quantity. So we’re going to substitute that back into this equation.
We’re going to substitute in minus the work that friction did. It’s minus 529 joules. Those two negatives cancel. So the sign works out to tell me that the man does positive work on the crate. His work is adding kinetic energy, speeding it up, and he does 529 joules of work. At the same time the man does this positive work, the friction force is doing an equal amount of negative work, which is why the crate doesn’t speed up. At best, the man is just able to keep it moving at a constant speed.