35 min.
https://youtu.be/XxR9PDkYo7c
PHYS 1101: Lecture Seventeen, Part One
Welcome to lecture 17. This is the last lecture out of chapter six before exam two. It’s a very small section, so this will be a quick lecture for us. What we need to discuss is this last topic, which has to do with how fast work is done. That’s an important distinction in many cases. We call that term, a parameter, a variable that represents how fast work is done; we call that power. This is the power you’re accustomed to hearing about in the context, say, of your electric bill, or your energy consumption. Very specifically, power is the rate at which work is done.
This is equivalent to, you can think of this as, the rate that energy is consumed, because we know energy and work are synonymous. Work is a form of energy. A couple of reminders here for you; whenever we use the term rate in science, we mean per second. As an example, the rate as someone is walking along would be their meters per second; their distance per second. Power, as I have said, is energy, or work, per second. Per second we know translates to “every second this is how much energy is consumed.”
It makes mathematical sense to write that as a fraction, because then I know that if I have energy per second, all I have to do is multiply by, for example, two seconds. I know that because two seconds have gone by, I’ve used two times this amount of energy that I use every second. So that’s the logic of why per second we write as a mathematical fraction, it’s equivalent to that.
Let’s start out the lecture with a couple of warm up questions having to do with work. I’m going to put it in the context of an interesting problem that I’ve showed you before, but we haven’t talked about. We’re going to talk about the work that’s done on this small purple box. It’s an interesting combination in that this purple box ends up accelerating to the right, only because it’s attached or sitting against, that blue box.
The blue is also accelerating also to the right; obviously they must be if they’re attached and they’re moving together. The blue box accelerates because a force, p, is applied to the blue box. So if you do the force analysis for everything in direct contact with the blue box, you would have a normal force from the ground below, roughness at the surface, so friction force on the blue box. If you went around and said, “What else is touching the blue box?” there would be this something, perhaps there is this hand pulling. They tell you that p is applied to the blue box. And then as you went around the blue box you would realize there is a surface in contact here. This blue box feels forces from the purple box.
Our focus is going to be on the purple box. So remember you would follow that same reasoning to think through the forces on the purple box; you’ve got gravity. Then you would say, “What’s directly in contact with the purple box?” You hit a surface, of course you’ve got a normal surface. Then you potentially have friction. Okay, so thinking about all the forces on the purple box, you now want to consider what work those forces do.
To answer all of the questions that follow, I advise; number one, draw a free body diagram for this purple box, because it’s all about the work that gets done on the purple box. Only forces directly on the purple box can be adding energy to it directly or subtracting energy from it. I’m going to give you five points for each of these questions. Draw this free body dia gram. Imagine a short amount of time goes by, and the purple box does move to the right. You know to think about the work that gets done. You have to think about every force on the object. You have to think about the displacement, the real motion of the object during that time. And then consider, for example, the orientation of each force in that direction of motion.
Question 1: After moving one meter to the right, the work that force p does on the purple box is?
All I want to know here is can you work out is it positive, negative or zero? Is that force working to speed up the box, slow it down or having no effect?
Question 2: The work that the force of gravity on the small purple box does; positive, negative or zero?
All of these, I’ll give you plus five
Question 3: What’s the work that static friction does on the purple box?
Question 4: What’s the work that the normal force does on the small purple box?
As an aside, what object is causing the normal force on the box? That’s good to clear up in your mind.
Question 5: After moving one meter to the right, what’s the net work done on the purple box; positive, negative or zero?
Question 6: So considering forces on the purple box, which of those forces is directly responsible for the net work on the small- ?
Question 7. Here I have you switch your focus now to the blue box. We know that the purple box is in contact, it’s touching the blue box. Therefore, it is exerting, it’s the agent responsible for two forces on that bigger blue box. Those 2 forces are; a normal force, a surface pushing force, away from the surface, and a static fiction force.
Draw the blue box, draw forces on it, and think through the origin, or appreciate the existence of both of these forces. These two forces are the action-reaction pairs; two forces that are on this purple box. At that contact point between these two objects, that’s where you have, or you think about the action-reaction pairs between two objects. Ask yourself this; is the work that’s done by the normal force on the blue box from the purple box, is that positive, negative or zero?
So you’re looking at one of the forces on the blue box. It’s the normal force caused, not by the ground, but by the small purple box. Draw that force. Consider the direction of motion for the blue and work out the overall sign of the work that’s done by that force.
Let’s start our new material. We just need to go over the concept of power. Let me introduce it by having you compare two scenarios. Let’s imagine I’m here. Here’s Angela, and Angela’s holding up this box. The box is the object. I’ve drawn it in red here. What are the forces on the box? Well, I have a force due to gravity straight down. Then the overall effect of my contact on the box has to be consistent with the overall motion for this box. What do I want that motion to be?
Let’s presume that I’m carrying this box at just a steady pace up this straight hill. Our velocity vector then is nice and constant, so we know I don’t have any acceleration. So when I go to draw forces on the box, I can come to the conclusion that whenever effect I’m having, I’m the only thing in contact with the box, it must all end up leading to an overall upward force that exactly balances gravity, because that’s the scenario that I need to be consistent with my acceleration picture. There can’t be a net force on this object. So here’s the force due to Angela, and here’s the weight. They’re equal and opposite.
Do we agree that that would be the same scenario, and the same force condition for the picture on the right? On the right, I want everything to be the same. I’m going up at the same speed. The only difference is that the slope is steeper now, but again, the forces have to balance. The only difference between the two scenarios is simply this slope of this hill. What’s different? Your intuition would tell you that as I carried the box up the hill I had a very different sensation in one case versus the other. Yet everything we’ve learned so far in this class, in terms of the work, would tell us that the amount of work that I do is the same.
If I go through my general equation and I appreciate that the variable that I want is the work that I do; that then I mathematically, have to go after, or represent by calculating the work that the force that Angela exerts does. I know that that force will do the work that’s the size of my force times the cosine of the angle, times the displacement. I can try to calculate it directly based on this direct calculation, or I could flesh it out by looking at what it must be, starting with our main equation. That’s the direction to take because that’s easier for me to calculate. Use our starting equation, okay?
On the left side, let’s start thinking through what some of these terms are. I just think about the left side first. I know I have to include the work of every force other than gravity. That means I only have work due to the force of Angela. That in fact is the variable that I want to solve for. So let’s think about the right side of our starting equation; v final equals my initial speed. That’s going to tell me that the initial kinetic energy equals the final kinetic energy. If these two are equal, they have the same number, one’s negative, so they’re going to cancel. That tells me that KE final minus KE initial goes to zero, equal to zero.
What about the potential energy terms on the right side? I have h final minus h initial. That’s the same for both scenarios. Let’s just call that the overall height. Let’s leave it h final. I know the h initial is zero. Let’s put these pieces into our main equation and see what we have.
On the left side I know I’ve simplified it to work due to the force of Angela. On the right, I know I’ve worked out that the kinetic energy terms combine to be zero, but that I do have the two potential energy terms. For example, PE final is mgh final. PE initial is mgh initial. That’s where I defined my zero, my lowest altitude. That term goes to zero. But I do have this term. So the work that Angela does looks like it simplifies to simply PE final, and that is equal to mgh final. That’s what I worked out here. Okay that’s the same for both scenarios. So what is different?
We certainly have the sensation that the time is different, that it takes to deliver this energy, this work that I’m doing, the work that this force does, it has to be delivered over a very different time interval here, compared to here. That’s the essence of this term power. It’s the rate they we’re going to be doing that work; delivering that energy for that particular force.
So that’s quiz question 8 for you. What is the difference? Of those two scenarios; which occurs over a longer time interval?
The question we’re after here with this notion of rate that work is done is how quickly it’s done. So we’re going to define a parameter that captures just that. We’re going to define the power to be the average rate at which the work is done by a particular force. We mathematically represent that as, we’re going to use p for power, and that’s going to be equal to; we can always estimate the average power as being the total work that got done divided by the time. This will give us energy per second when we multiply this fraction out. That’ll tell me, for every second that went by, if I did this small incremental amount of work every second, if 20 seconds goes by, I’ll end up with 20 times that amount. This is the right fraction we need to work that out.
Question 9 then; after answering question eight, and you’ve got through which scenario takes a longer amount of time, answer question nine. Think it through, which scenario corresponds to a larger power requirement? Answer that from two perspectives; from the perspective of directly thinking about calculating this p or estimating what it would be, it’s the work done, divided by the time. The work that gets done we know is the same for the two scenarios, but the time is different. So thinking about that fraction, which is going to give us a smaller power or a larger power?
Then check your intuition. If you think of you being the person caring the box up; is that the scenario that’s more tiring for the person responsible for delivering that power? Can you point out the units you’ll be dealing with for power? Power was work divided by time. The units on the left have to be consistent with the units on the right. Power has to have units of work, which is joules. That’s energy divided by time, which our SI unit for that is seconds. So it’s joules per second. A joule per second is then called a watt, a term you’re familiar with probably. You have 100 watt light bulbs in your light sockets. Joule per second is a watt. I’m going to try to write as a funny capital W so you don’t get it confused with the w we mean for work. Watt is the unit for power.
Let me point out one mathematical consequence of this definition of power. One way to use it is that you directly calculate the power as work over time; remembering that this represents energy per second. It’s delivered to, or added to or taken away from this object. Think about rewriting that equation by multiplying both sides by t. Just move t to the left side.
Now I have an equation that tells me that the work that’s done is equal to power times the time. We write that so I can read it from left to right, and remind you that this is the total energy added or taken away from this object by one particular force. One way to calculate that was directly. We can calculate the work that gets done by that force from start to finish.
Or if we were given the average power delivered by that force we could just calculate that power times the time. That’s what this says. Let me point out what the units would be in that case. On the right hand side I would have units of watts multiplied by time. That gives me units of total energy or energy which is in joules. Just want to point out that if you have a look at your power bill, you’ll see a reflection of this. You’re charged every month for the total amount of energy that you use and that number is reported in 100 kilowatts times hours. This number tells you how many kilowatt hours you’ve used; it’s the product of a power and the units of kilowatts, give you an idea that you have 100 watt light bulbs all around your house. That ends up adding up in a hurry if you have all the lights on in your house. So every hour that goes by, it ends up being convenient to measure the power that you’re using in kilowatts. That’s times ten to the third. So that’s 100,000 watts per hour. So here’s the multiplication of time and the product of those two things tells your total energy consumption. Our problem solving steps are pretty straightforward for using this concept.
We’re going to need are all we’ve learned so far about calculating the work that a specific force does, so we may need the work equals force times the cosine of the angle, times the distance. We can think of our main starting equation, when we’re trying to work with a power problem, is that the power equals the work over time. You’ll see in one of the examples that I do, that for objects that are moving along at a constant velocity, another way of calculating this work over time, you’ll see it’s equivalent to it being one of the forces on the object, times that speed to get the power. You’ll see that in a minute.
So in addition to being able to calculate work, here are some extra steps to think about. As always, it’s always one object you’re focused on. What’s the power that’s added to or taken away to this object? Sketch that scene. What’s the scope of the problem, the start to the finish that you’re calculating this average power consumption over? Draw this displacement vectors. You have a picture of the direction of the motion.
As always, power is going to be associated with one specific force on that object. It’s going to be the power that that force delivers. So draw your free body diagram so you can clearly see that force in the context of the other forces on the object. Make yourself a clear list of the knowns and the variables that you need. What’s the variable that’s going to show up in your equation that we use mathematically to represent this physical quantity that it’s asking you about? For example, if it asks you to calculate the power delivered by Angela as she carries a box up the hill, the variable you’re after is p. Specifically, it’s the power due to force Angela.
That’s step number four. To calculate a specific power, you’re going to have to identify a specific force. My example, I would focus in on force Angela. That would be the force that I would go over here to plug into the right side of my equation, what I would use to calculate the work done. Then we have our normal steps, which are; work with the starting equation p equals w over t. Focus in on this variable you want to solve for. Keep customizing that equation, substituting into it until you see your variable appear and then you can solve for it.
Here’s our first quick example. Let’s say we’re asked; what’s the power needed to lift a 49 kilogram person a vertical distance of five meters in 20 seconds?
Here’s my sketch of the scene. Here’s the person; pretty small person, 49 kilograms being lifted. My displacement vector then is up. Here are the forces on this person.
Often in these problems they don’t directly tell you if the person is accelerating or not. But, when they say something like what power is needed, perhaps that gives you the feeling that you’re looking for a minimum power requirement. What’s needed to lift the person? If I want to get the person lifted more quickly, if I want to accelerate this person up, this force is going to be bigger; it’s going to have higher power requirements. I’m just looking for the minimum power – no acceleration. That’s going to be the smallest lift force needed, just enough to balance the force due to gravity.
What are my knowns for this problem? I know the magnitude of the displacement is five meters. The variable t will represent the time from start to finish of this motion over which this work gets done. I also know the mass of this object: 49 kilograms.
Okay, what’s the force of interest? What power is needed to lift the person? I’ve just labeled it here, force lift. I know there’s some upward force that balances the force of gravity. If I’m calculating the power to lift it, it’s got to be force lift that I focus on. This then is the force that I’m going to use. I go to calculate the right side of my starting equation. What variable do I want? I want p. I want to calculate that directly. So my starting equation is p equals work divided by time. I have the value for time. I don’t have a number for this work yet. Here’s the variable that I want, p, so I have to go calculate this variable w. I need to further realize that what I’m focusing on is, very explicitly, the work due to force lift. What is the work that’s done? Work of force lift.
Work is always the size of the force, times the cosine of the angle, times the distance. Here are the two vectors that I need to look at to consider calculating this work; the direction of the force lift, and the direction of the displacement. These are aligned. They’re parallel, pointing in the same direction. I have zero degrees between the two. There’s no angle between them, and so the cosine of zero is one. The work that gets done is force lift times the distance. I’m ready to now plug that into my main equation. The power, the variable that I want, is equal to force lift times the distance. That’s plugging in for the numerator, work. Then I have to divide by the time. And I’m done now. Now, I do have numbers. Here’s my starting equation again. I’m ready to substitute in. I know that that’s work lift divided by time. I just calculated that. It’s the force lift times the distance divided by t.
I’m getting closer. I don’t directly know the value of force lift yet. So let’s go do that one. Every time I’m faced with needing a value for a force, I often consider whether Newton’s second law will help me out. I know my acceleration is zero in this problem. So force lift is up. Minus mg is down. These two balance. I know force lift is mg, I can substitute that in. Let me show an arrow here indicating that I’m going to do that. So I’m back to my main equation and I’m going to make that substitution; mgs divided by t. The power that I’m after then, my solution to this problem, is this mass, 49 kilograms times 9.8, times five meters. Then I have to divide by 20 seconds. I end up with a number that’s about 120 joules per second, which is equal to 120 watts. It’s a little more than a 100 watt light bulb.
Let’s do a little more complicated problem now. Here I have a motorcycle. I’m told the mass. It’s traveling now at a steady speed of 20 meters per second. The force of air resistance is 200 Newtons. While I’m reading that here, I’m going to jot down here air force resistance is 200 Newtons. That’s going to oppose the motion. Find the power necessary to sustain this speed if, a, the road is level, b, the road a slope upward; 37 degrees with respect to the horizontal.
Let’s do a first. Here’s a quick sketch to let settle in my mind what motion I’m describing. I’m considering: motorcycle, it’s moving to the right, constant steady speed, straight line, 20 meters per second. I’m going to have some displacement. But here it’s not obvious, they’re not asking us for a specific time interval or a specific distance. If you’re faced with that, and yet we’ve been working with equations where we’re picturing a specific scope, sometime it’s helpful to consider a particular time interval. Let’s consider one second. In one second that’s gone by, if I’m going 20 meters every second, then I know my displacement is 20 meters. What we’ll conclude, and what you’ll see, is that every second I go 20 meters, and a force is going to end up doing so much work every second. You’ll see then that that just means that my power consumption is just some steady amount.
Okay, forces on the object. I know I have the weight and the normal force on the ground, the balance. If I’m going at a steady speed, my right-left force is balanced too. This force f, resistance, let’s just call it force res, has to oppose the motion. That’s going to be the motorcycle fighting the airflow as it moves forward. These two forces have to balance.
So let’s do our list of knowns. We’re going to consider a distance of 20 meters, a time of one second. My speed is 20 meters per second. Let’s see how we need that, if we need it. I know the mass of this object is 250. I know the value for the resistance force is 200 Newtons.
What variable do I want? I want power, but every time I think about power I’ve got to identify the specific force that’s responsible for this particular power I’m focused on. I need the power necessary to sustain this speed. I have got to pick one of these forces. What force is responsible for sustaining this speed? It can’t be either of the vertical forces, they don’t do any work. No contribution in the direction of motion. This force is what’s holding me back. This is the force that’s directly responsible for this forward motion. This is the force that is overcoming, or balancing the resistance force. I don’t know the details of that, and I don’t need to worry about what’s causing this force.
I’m going to call it force of propulsion. It turns out that it’s the static friction that the road is putting on these tires. I think your book has a nice discussion of propulsion forces if you’d like to read more about that. Without knowing the details, just from Newton’s second law I can infer that the net effect of the motor turning the wheels and then moving against the ground has to have the effect of a forward force that balances this air resistance if I’m going to move forward and still maintain this constant speed. So, I want the power generated by force propulsion. If you clearly identify the force of interest, take your time to pick that out, then you won’t be confused when you go to plugging into our starting equation; when you go to calculate power as being the work divided by t. To calculate work, you have to pick a specific force to do that. Take the time. Slow down. What’s the force you have to focus on? Let’s work with this equation.
We know the time, but we don’t know that work. Let’s go calculate that. The work that this particular force does is the size of that force cosine of the angle times the displacement. This force, it’s in the same direction as s, so might cosine, all the forces in that has the component in that direction. Cosine of zero degrees is one. I’ll just note it here so you see it explicitly. That’s my cosine term and then times the distance.
So, work in my equation is going to be a work of force propulsion. And for that I can substitute in force of propulsion times the distance. Let’s make that substitution. That’s my numerator in my main equation I’ve substituted for. Now I have to divide by t.
Okay, I need to go over to the side and point out; let’s just rewrite this equation a little bit. This power I’m trying to calculate is equal to the force of propulsion times, then I’m just going to write this fractions: s divided by t. Our problem, remember we said let’s consider one second going by. In that one second I went 20 meters, so I would be dividing by one second. That’s 20 meters per second. This fraction that we have here is nothing more than the speed of this object, 20 meters per second.
So, if you have an object that’s moving along at a steady speed, nice straight line; the force is doing just steady work as this object moves along. I can replace that, the power that I want calculate, as being the force of interest, times the speed. That’s that other equation that I pointed out you may want to use.
Let’s write that the power is equal to the force of interest, force propulsion, times the speed. What’s force propulsion? To figure out the size of the force, I’m inclined to consider the second law. In this case, the propulsion force is a horizontal force. I’m in equilibrium, horizontal and vertical. Acceleration is zero. So I know that force propulsion is just equal to the force of the air resistance. I had a number for that. It’s 200 Newton’s.
Okay, ready to go back now. My power is equal to the force of the propulsion times the speed, that’s 200 Newton’s times 20 meter per second. That’s 4000 watts.
Let’s do part b. Let’s see how much this changes now when the motorcycle is fighting the same headwind, but now is headed up a hill at an angle of 37 degrees. Using a tilted axis is going to make the most sense again so that most of our forces are already aligned to our axis. We will have the force of gravity though off at an angle. So I’ll have to break it up into the x component along the slope, and the vertical component perpendicular. You did this before, and you are probably remembering that this is also my 37 degrees. So wy has a magnitude of mg times the cosine of 37 degrees, and wx is mg sine of 37 degrees. Let me explicitly draw these forces and get the arrows in the right direction here.
Okay, I’ve made my list of knowns. What do I know? It’s the same resistance force. All right, let’s make our list of knowns. I have the same magnitude of the resistance force of 200 Newton’s. I have same mass. The speed is again 20 meters per second. What do I want? I need the power again, but very explicitly the power that what force is responsible? That’s going to be the force of propulsion. This also may ultimately the force that’s overcoming air resistance. In this case it has got to overcome a component of gravity as well. This is ultimately what the engine is indirectly responsible for. Quite directly on the motorcycle it’s a static friction force on the tires that’s providing this forward force. That’s, of course, indirectly going to be related to engine being responsible for turning the wheels. But rigorously for physics, it’s the static friction force. It’s the only force that can be on the motorcycle in this direction that’s moving it forward.
Our starting equation; p equals, we’re going to go with the force times the speed. Have your eyes been drawn to this as you’re starting an equation. If ever you have a scenario with something moving along a straight line, steady speed; have a number for the speed, what’s the force? Specifically what is the force we’re interested in and what its value or what represents it? It’s the force of the lift, of the propulsion.
My thoughts again are to Newton’s second law to see if I can figure out the size of this force, the requirements on it. I’m in equilibrium again, the masses drop out. So I’m really left with asking how these forces balance. That’s with going to be the requirement on the force of propulsion. It’s the only positive force I have up the hill. I’ve two negative forces; the air resistance and the part of gravity that’s along the slope. Notice that it’s down the hill, same direction as the resistance, because the object is moving up to hill. If I do a little algebra and rearrange for that then the propulsion force now doesn’t simply have to balance with the air resistance only, but it also has to be large enough to overcome wx.
I have a value for force resistance, wx. I’ve done my trigonometry so I know what to substitute in for that. I’m going to do that here. It’s the mg sine of 37 degrees. Again, I’m just going to substitute in the positive value, because up here in my vector equation when I started, I explicitly took the direction into account and so chose the proper sign here. I just want this to represent the positive magnitude of it. So that’s what I plug in for that.
All right, this is what my force of propulsion has to be now. I’m going to back that out and plug that back into my starting equation now. Force of propulsion, I have concluded is equal to force of the resistance plus mg sine of 37. Pulling that out then into my final equation, or plugging it in, my power is going to be the force of propulsion, times the speed. That’s going to be f resistance, plus mg sine of 37 degrees, times the speed.
Putting in my numbers for that, I have 200 Newtons as my air resistance. It is the same as it was before, but now my power requirements have just gone up because I have to include this gravitational component. I have 250 kilograms. It’s not going to be small either because that motorcycle and person is heavy; 250 kilograms times, 9.8 times, the sine of 37 degrees. And then, I need to multiply by my 20 meters per second. I have Newtons being added to kilogram-meter per second squared, that’s a Newton as well.
In fact I end up with 200 Newtons, plus 1,470 Newtons, that I then have to multiply by the 20 meters per second. Before my propulsion had to be this large, it’s got to be more than seven times greater, eight times that now. Multiply these two together, I end up with 33,000 watts. That’s 33 kW. Much larger than when the motorcycle is on just a straight, flat road in. That brings us to the end of lecture 17.