https://youtu.be/fZ4X0F07E3I
PHYS 1101: Lecture Fifteen, Part Six
Okay. Let’s move on now to a more complicated example where a person pulls a 75-kilogram toboggan for a distance of 35 meters. It’s along the snow, and a rope, again, is pulling it. This time the angle is 25 degrees, but now I can’t ignore friction. I do have a small frictional force. I know it’s small because the coefficient of kinetic friction is 0.005. They give me, again, the tension in the rope. It’s 94 Newtons. And I need to figure out the net work done on the toboggan.
Okay. For sure, my thoughts are immediately drawn to using energy conservation to solve it, first of all because it’s directly asking me for an amount of work done. Work is energy. It’s going to show up in the ideas behind this equation. Further, it’s the same kind of idea in that I know that this toboggan is going to speed up because it’s going to be accelerating from this rope, caused by this rope, and it’s moving a certain distance. So again, I have some notion of speed and position change.
Here’s where I need to start. I need to focus. What’s the object? All these forces, the motion — it’s all about the toboggan. Here’s my quick little sketch of it. I know that displacement vector has a magnitude of 35 meters. Let me just put a rope here and label it T. My tension force is up at an angle of 25 degrees. That’s my kind of physical sketch of the picture. I know the motion. I’m going to go ahead and draw it or envision that it’s moving to the right.
Okay. First step is preparing to use this equation. Let me think about the left side. To think about the net work that’s done, again, I need to picture all the forces and carefully go through and decide which force does end up doing work on the toboggan. So question 9 to you is, “What is the correct free-body diagram for this toboggan?” Because after I draw or pick this correct free-body diagram, what I need to work out for the left side is that the net work is the sum of the work that each of those forces does.
For this problem, I’m going to have potential work that gravity may do, potential work that the normal force may do, plus potential work that the rope does — the force from the rope — and I have friction now that I can’t ignore. So for each, we need to calculate that the work each does is the size of the force, cosine of the angle, times the distance that it moves. So to visualize this, I’ve got a picture of F, so I can picture the angle only after I have a clear picture of s. That displacement vector I drew above, but let me just draw it again here. I’m picturing the motion as a straight motion horizontally to the right.
Okay. Let’s go through each of those terms. You have a picture. You’ve decided on the direction of the force due to gravity, so you can do this calculation for the work that the force of gravity does. Question 10 is simply, is that quantity of the work that gets done positive, negative, or 0? Does gravity contribute to speeding up the toboggan, to slowing down the toboggan, or does it have no impact on the speed?
Question 11, follow the same exercise with the normal force.
Question 12, consider the tension force. Is the work that it does positive, negative, or 0? Said another way, more physically, “Does the tension force contribute to speeding up the toboggan, slowing it down, or has no effect?”
And then question 13, do this for the friction force.
Okay. After you’ve thought those through, I want to pull out and emphasize to you an important generalization that’s always, always true. Notice the important red box I’ve given this factoid to you in. You’ve seen it a couple of times now. Let me generalize. A normal force is always going to do 0 work, or perhaps said a better way is that a normal force never does work.
Here’s the normal force for this toboggan problem. The displacement for the toboggan is to the right. It’s along a surface. In fact, any object, if it moves along a surface, the normal force, by definition, is perpendicular to the surface, so it’s always going to be at 90 degrees to the motion. If it’s always at 90 degrees to the motion — let me draw it again here for you — we know it never can contribute to speeding up the object or slowing it down. This force simply contributes to balancing the weight of the object, and supporting the object. It has no impact — it never does — in the motion along the surface.
Therefore, one can generalize. You will always find that the angle is 90 degrees between the normal force and the motion for an object sliding along a surface, so it can never do any work. WFN‘s always 0. Let’s take another one of those contributions to our net work.
Question 14 has you calculating another contribution to our net work. This is the work that the frictional force does on this object. You have already answered or determined whether you expect this to be a positive quantity, negative, or . . . I guess it’s not 0, because I don’t even give you that as an option.
Let me do a quick sketch here to emphasize that exercise that you likely have already gone through. For the work due to friction, I have to calculate the size of the force times the cosine of the angle between the force and the displacement times that magnitude of the displacement.
Let me draw these forces. My motion is to the right. Friction is in a direction that always opposes motion. Notice that word “always”, because I’m about to make another generalization for you. If friction opposes the motion, then my angle, as I swing from this one vector to the next, is 180 degrees. That’s always going to be the case for friction, because it always is going to directly oppose the motion. So when I go to calculate it, my cosine of the angle is always going to be minus 1.
That means the work is going to be the magnitude of the frictional force times the cosine of 180 times s, and this whole term is -1. Plug it into your calculator. You’ll see. The cosine of 180 is always -1. That means the work that friction does is going to be the magnitude of the frictional force times the distance, but it’s negative. It’s -1 times that.
Okay. You’re going to find this is always true for the work that friction does. It’s always the entire frictional force times the distance, and it’s always negative, minus, that. Friction, that force, all of that force does work to slow the object down. That’s what the negative sign physically captures. So there’s my pink, important box for you. The work that friction does is always negative. It’s always the whole magnitude of the frictional force times the distance.
Okay. Question 15, try out calculating the work that the tension does for this particular problem. You can enter that in WebAssign to three digits. So before I work out these numbers for this problem, I’m going to now focus my attention in thinking about the right side of my basic equation. I’m now thinking about this KEf minus KE0.
In order to say anything about that, again, ultimately I’d be asking, “Do I know the mass?” which I do in this case. “Do I know the final speed? Do I know the initial speed?” And if you read through the problem again, you’ll see that actually you’re not told a final speed. You don’t know that the toboggan starts from rest. In fact, you don’t have any information other than the mass to talk about specifics for the right side of the equation.
So before I go much further, I’ve just thought through some specifics with respect to the right side. Before I get too far, let me go through the list of knowns and decide what it is I’m trying to solve for, what I want. It’s useful to do this fairly early on in the problem, before you head down a path where you’re spending time calculating something that you don’t need, knowns, though. What do I know about this problem? I was told a kinetic friction coefficient value of 0.005, the mass of this object — 75 kilograms — and it was moved a distance of 35 meters. The tension in the rope was 94 Newtons.
What do I want? I’ve scrolled back up to the problem here so we can read it carefully. I’m asked for the net work that’s done on the toboggan. What variable do I use to represent that? Well, it’s Wnet. It is the left side of this equation. We want to solve for the total amount of energy that’s either added to or taken away from this object.
Our basic equation tells us we could get that one of two ways. If we happen to know the initial speed and the final speed, we could put in real numbers for these one-half mvf2 squared minus one-half mv02. We could calculate the right side, in which case that number would be the net work because the two sides are equal. Or I know that the net work can be calculated by carefully looking at all of the forces that are on this object and adding up the work contribution that they all may do. And because I don’t have any speed information, that’s the approach I’m going to have to take to calculating the net work.
So for this problem, in fact, I don’t have to worry too much about the right side. I just need to try to directly calculate the net work based on the force information and the distance information that I have. So I’m going to write here that I want the variable I’ve called Wnet, and that I’ve written above as being the work due to the tension plus the work due to the kinetic friction. The other forces on the object are perpendicular to the motion and don’t do any work.
Okay. Well, this is what I want. But I don’t have direct numbers for either of these yet, so I’ve got to go through, break them down and calculate them. So I’m always going to come back to my basic equation. So let me kind of, as I move to the right here, say . . . I don’t know. Let’s first focus on the kinetic friction. What do I know about that? The kinetic friction, the work that it does, I’ve pointed out, is minus the size of the kinetic friction force times the distance.
Okay. I have the distance, but I don’t know big the kinetic friction force is. So now I’ve got to step into it a little deeper and say, “What’s the kinetic friction force? How big is that?” Well, I know kinetic friction, the magnitude of that force, is the coefficient times the normal force. Now I peel that layer away, and I say, “Well, what’s the normal force? How big is that?”
Here’s a sketch of the forces on my toboggan. Let me label them for us. The downward force has to be gravity. Here’s my kinetic friction, here’s my normal force, and here’s my tension. I know that this tension — it’s at an angle — I can replace with its two components. So I’m going to do that, because I need to work out the magnitude of the normal force, which means I have to balance the forces vertically.
So let me write below here I need to figure out FN. And to do that, I’m going to solve Newton’s Second Law for the vertical acceleration, because I know my acceleration’s not up or down. It’s straight to the right. That’s equivalent to saying that these forces balance, which is what I said visually. I have FN plus Ty minus mg has to be equal to 0. That means FN has to be equal to mg minus the vertical component of the tension.
Okay. I’ve got numbers for m and g, but I’ve got to do a quick trig exercise to get a value for Ty. Here’s my quick sketch to do my trigonometry. Ty is the side opposite to the angle, so Ty is the hypotenuse times the sine of 25 degrees. The tension magnitude was 94 Newtons times the sine of 25 degrees. And that gives me 39.73 Newtons. Okay. Now I have Ty. I had to take several steps there breaking it down to calculate what I needed, but I’m getting there.
So now that I have Ty, I’m going to step back out and say now I can calculate my normal force. It’s mg minus Ty. That is my 75 kilograms times my 9.8 minus this 39.73 Newtons. And the number I get for that is 695.3 Newtons.
Okay. Now that I have my normal force, remember we needed the normal force to calculate the friction force. I can now back out and do this calculation, which means I’m here, calculating that the friction force is the coefficient times the normal force, which in my case is 0.005. Now, times 695.3 Newtons, and that gives me a value of 3.476 Newtons. Notice how small it is. The friction force . . . there’s not going to be much friction on snow.
Now that I have the friction force, let’s remind ourselves why we needed that. We needed that to then calculate the work that friction does. It’s minus that force times the distance, so let’s back out and do that calculation. The work that kinetic friction does is –fk times the distance, so we now have the numbers we need to do that — -3.476 Newtons times my 35 meters. And that gives me a value of -122 joules. A Newton times a meter is a unit of energy called the “joule”. That’s the work that friction does.
Okay. Let’s go back and see how far we are. We’ve calculated the work that friction does, and that we needed to do to calculate one piece of the total work that gets done, of the net work. So we have this now. Now we need to go get the tension — the work that the tension does.
So now I’m going to go a little to the left here and write, “What’s the work that the tension does?” Well, the work that tension does is equal to the size of the tension force times the cosine of the angle between — 25 degrees — times s, the distance. That’s the work. When you plug in the numbers for that, given my tension was 94 Newtons, cosine of 25 times 35 meters, you end up with . . . let’s see here. I ended up with 2,860, again, Newton times a meter — joules, a unit of energy.
Okay. The tension does positive work. It does a lot of positive work. This force has a large contribution to speeding up the object. It’s adding kinetic energy. The friction opposes the motion, and it does a small amount of negative work that’s slowing the object down. It’s removing kinetic energy. But by far more positive energy is added compared to what’s taken away, so in the end, the toboggan definitely speeds up. But it’s not going quite as fast as it would if I had no friction.
Okay. Let’s say that. Let’s put these pieces together and say that in the equation now. I’m ready to calculate the net work. It’s the work that the tension does plus the work of friction. That is plus 2860 joules minus 122 joules. Most of it’s positive. Small amount of negative. I end up with a final answer . . . I’m sorry about that. I looked at the wrong number here. The work that the tension force does is 2,980. When I subtract 122, I end up with my net work is 2,860 joules. That’s my final answer.