Understanding Work, Energy, and Power Concepts

Hello everybody, my name is Iman. Welcome back to my YouTube channel. Today we're continuing our MCAT physics lecture on work and energy. Last time we talked about energy and particularly focused on kinetic and potential energy. We also talked about total conservation of mechanical energy and we discussed the first law of thermodynamics. Now what we want to move into is discussing work. Now, please note that we will dive into work and its components heavily also in the next chapter on thermodynamics. But we're going to introduce it here. Work is not energy, but a measure of energy transfer. In fact, there are only two forms of energy transfer, work and heat. Now, in physics, work is the energy transfer. to or from an object through the application of force along a displacement. And so in its simplest form, it is often represented as the product of force and displacement. So we're going to be really focusing in on work as a way of energy transfer. We'll talk about heat a lot in the next chapter. Like we said, Energy is transferred through the process of work when something exerts forces on or against something else. So work is force dot product displacement. In other ways, we can write force times displacement times cosine theta. All right, so W is work. F is the magnitude of the applied force. And D is the magnitude of the displacement through which the force is applied. Theta is the... angle between the applied force vector and the displacement vector. You'll notice that work is a dot product, right? So it's a function of the cosine of the angle between the vectors. This also means that only forces or components of forces parallel or anti-parallel to the displacement vector will do work. And so that means transfer energy. We've already said that the SI unit for work is joules. And while this suggests that work and energy are the same thing, remember that they're not. Work is the process by which a quantity of energy is moved from one system into another. Now, as described, work is a process of energy transfer. In systems of gases, then, we therefore approach work as a combination of pressure and volume changes. All right, so that's going to be really important. For a gas system contained in a cylinder with a movable piston, We can analyze the relationship between pressure, volume, and work. When the gas expands, it's going to push up against the piston, and that's going to exert a force that causes the piston to move up and the volume of the system to increase. Now, when the system is compressed, the piston is going to push down on it. the gas and that's going to exert a force that decreases the volume of the system. We say that work has been done when the volume of the system has changed due to an applied force. Gas expansions and compression processes can actually also be represented in graphical form with volume on the x-axis and pressure on the y-axis. Such graphs are termed as PV graphs, and we can see these right here, right? Volume on the x-axis, pressure on the y-axis. Fantastic. And we're going to see later on in the next chapter how work is demonstrated and we'll see it here too, but how work is equal to pressure. Times change in volume. All right, so that's another mathematical expression for work. All right, now let's discuss a couple of different pressure-volume graphs and how we can determine work from these. All right, now when a gas expands, we say that the work was done by the gas. All right, and the work is positive. Okay, let me repeat that. When a gas expands, we say that the work was done by the gas and the work is positive. All right, so let's write this down. When gas expands, work done. Sorry, work. Work is done by gas, all right? And then work is positive, okay? Now, what if a gas is compressed, all right? What if a gas is compressed? I'm gonna shorthand it, all right? Then we say that work was done on the gas, all right? Work done on the gas and work is negative. Now there are infinite number of paths between initial and final state here. Different paths require different amounts of work. And you can calculate the work done on or by a system by finding the area under the pressure volume curve. We're going to go through different scenarios A, B, C, and D. In scenario A, Note that if the volume stays constant, if there's no change in volume as pressure changes, then the delta V change in volume is going to be zero. And considering that our formula for work is pressure times change in volume, if change in volume is zero, then no work is done because it's P times zero, and that equals zero. Okay, so if the volume stays constant as pressure changes, there is no work done because look at this, there is no area to calculate. This is called an isovolumetric or isochoric process. All right, and we can see that demonstrated in a pressure volume graph. in graph A here, all right? What about graph B here, all right? On the other hand, if pressure remains constant as volume changes, all right, then the area under the curve is a rectangle of length P, all right, of length P. That's length P and width delta V. All right, so for processes in which pressure remains constant, this is called isobaric processes, as shown in plot B, and you can calculate the work done as pressure times delta V. So phenomenal. Now, what about what about graph C? All right. Graph C shows a process in which neither pressure nor volume are held constant. OK, the total area under the curve, then you're going to have to break it down into region one. All right. Which is a triangle and region two, which is a square. All right, a rectangle. Region 1 is a triangle, all right, where area can be determined like this, one half delta V times delta P, all right, and you can figure that out from the graph. And then you can calculate for region 2, which is a rectangle, so it's going to be pressure 2 times change in volume here. Alright, and then the work, the total work done is just the sum of region 1 plus the sum of region 2. Alright, so that's for graph C. And then graph D, this one is much more complicated and sort of outside the... scope of MCAT, it shows a closed cycle in which after certain interchanges of work and heat, you return to an initial state. Now, because work is positive when the gas expands and negative when the gas is compressed. The work done is the area enclosed by the curve. But calculating the work done in this situation would require calculus, and we don't have to worry about that here in this MCAT prep. You'll learn about it if you take a look at the previous video. Take physical chemistry, and I actually cover it in my physical chemistry one playlist. But you don't have to worry about it so much for the MCAT. The whole point is to show you that work can be calculated as a function of pressure. pressure and volume, and that you can work from pressure volume graphs to calculate that based off of that equation. And we cover different graphs and different scenarios to do just that. Now, let's discuss a few more things here quickly before we move into our last topic. Those are going to be power and the work energy theorem. Power refers to the rate at which energy is transferred from one system. to another. It's going to be calculated from this expression here. Power, all right, where power, P is power, all right, it's equal to W over T. W is work, all right, which is equal to change in energy, all right, and that's all over T, which is time over which work is done. So power is equal to work over time. It's also equal to change in energy over time, all right. The S, Unit for power is watt. That's watt or W, which is equal to joules per second. All right. Now, in the later chapters, we're going to identify additional ways to calculate power and electric currents. But for now, note that many of the devices we use every day, toaster ovens, light bulbs, phones, cars, they're all quantified by the rate at which those appliances transform electrical potential energy into other forms like thermal, light, sound, and kinetic energy. Power is calculated in many different situations, especially those involving circuits, resistors, and capacitors. And we're going to talk about those later down the line, I think in like chapter six and seven. For now, this is all you need to know. This expression, power is equal to work over time or it's equal to change in energy over time. With that, we can discuss work energy theorem here. The work energy theorem is a powerful expression of the relationship between work and energy. All right? Between work and energy. And it's calculated. by this expression work net is equal to change in kinetic energy. So that's equal to kinetic final minus kinetic initial. Now in mechanical application, it is going to offer a direct relationship between the work done by all the forces acting on an object and the change in kinetic energy of that object. the network done by forces acting on an object is going to result in an equal change in the object's kinetic energy. And we can use this expression to calculate that. All right. Phenomenal. So that's power and that's work energy theorem. All right. Now, We can look at how this, we're applying here a net force, all right, and it's moving at some initial velocity, all right, and we have a displacement and our final velocity. This can be under the guise of positive work, all right, and if we apply force net in an opposite direction while it moves in an initial velocity with some sort of displacement and final velocity, then we're going to see that this is going to be negative work, all right? In essence, the network done by forces acting on an object will result in an equal change in the object's kinetic energy. And we can calculate that by kinetic final minus kinetic initial. This is just 1 half mv final squared minus 1 half mv initial squared. And that's how you can calculate and see what gives you positive work, what gives you negative work, all right? Fantastic. If the V final is greater than the V initial, you're going to see that you're going to have positive work. All right. If the opposite is true, if you have if if V final is less than V initial, you're going to get negative work values. All right. You're going to calculate a negative work, a number. All right. And so that. is work energy theorem. Our last topic for this chapter is, oh, actually, let's do a practice problem. Look at that, we have a practice problem to do. A lead ball of mass 0.125 kilograms is thrown straight up in the air with an initial velocity of 30 meters per second. Assuming no air resistance, Find the work done by the force of gravity by the time the ball is at its maximum height. All right. Now, one thing to note, this answer could be calculated using kinematics and determining the maximum height of the ball. But you can also actually use the work energy theorem to calculate, and it's much simpler that way. So what you can do is you can set up that your work net is equal to your kinetic final minus your kinetic. initial all right your kinetic final when it reaches that maximum height we talked about that when it reaches its maximum height the velocity is equal to zero so what that means is that your your velocity final is equal to zero so if we write this out 1 half MV final squared minus 1 half MV initial square if you know your MV if your V final is zero this whole term is equal to zero So all you have now is 0 minus 1 half m v initial squared Alright, we know we know what V initial is. It's 30 meters per second. We know what the mass is so we can simply plug things in 0 minus 1 half mass is 0.125 kilograms times 30 meters per second squared all right and this if you plug this into a calculator is going to be equal to minus 56.25 joules Fantastic. So that's how you can approach a problem like this. All right. Now we can actually move into our last and final topic for this chapter, which is mechanical advantage. All right. Mechanical advantage. Now we're going to talk about here. All right. What the difference? The difference for a given quantity of work, all right? So for a given quantity of work, any device that allows for work to be accomplished through a smaller applied force. is going to be said to provide mechanical advantage. All right. And that is the topic of this section, mechanical advantage. So one example of mechanical advantage. is using inclines. Sloping inclines like hillsides and ramps make it easier to lift objects because they're going to distribute the required work over a larger distance, thereby decreasing the required force. And so it is one example of mechanical advantage. In addition to the inclined plane, five other devices are considered the classic simple machines designed to provide mechanical advantage. What are those devices? Well, in addition to inclined planes, we have wedge, which is two merged inclined planes. We have wheel and axle. We have lever. pulley, and screw. Screw is just a rotated inclined plane. Of these, inclined plane, lever, and pulley are the most frequently tested on in the MCAT. All right, we're going to talk about pulleys here in this chapter a lot, but to kind of set this up, let's define mechanical advantage. All right, mechanical advantage is the ratio of magnitudes of the force exerted. on an object by a simple machine to the force actually applied on the simple machine. And so we can write this as F out over F in. All right. What you notice is that it's a ratio, so it's going to be dimensionless. Now, reducing the force needed to accomplish a given amount of of work does have an associated cost with it. However, all right, the distance through which the smaller forces must be applied in order to do the work is going to be thereby increased. All right, that's just just a direct consequence of mechanical advantage. All right? So reducing the force needed to accomplish a given amount of work is going to cause the distance through which the smaller force must be applied in order to do the work to be increased. All right? Inclined planes, levers, and pulleys, they do not magically just change the amount of work necessary to move an object from one place to another. And that is why applying a lesser force over a greater distance is going to achieve the same results, is going to achieve the same changes in position accomplished by the same amount of work. So we're going to increase the distance through which those smaller forces have to be applied in order to do the same amount of work. Fantastic. Now, let's quickly see what that looks like through an example problem before we talk about pulleys. All right. A block weighing 100 newtons is pushed up a frictionless incline over a distance of 20 meters to a height of 10 meters, like we see in this. image right here. What we want to do is we want to find A, the minimum force required to push the block, B, the work done by the force, and C, the work required or the force required and the work done by the force if the block were simply lifted vertically 10 meters. Fantastic. Let's start with A. All right, we want to find the minimum force required to push the block. All right, the first thing we want to do is find the minimum force required to push the block and so we have to draw a free body diagram of the situation all right so here's our block it looks like this all right that's our block all right we're going to draw our free body diagram we have our force of gravity pushing it downwards all right we have our normal force perpendicular to the surface All right, and then we have the force pushing it upwards this way All right, the minimum force needed is going to be a force that will push the block with no acceleration parallel to the surface of the incline. What that means is the magnitude of the applied force, F, is going to be equal to that of the parallel component of gravity. All right, so gravity is going to have A component, all right, it's going to be the parallel component of gravity. So what we're doing here, all right, is we're essentially defining this to be our plane of reference. This is our x-plane, this is our y-plane. Force of gravity is out of this plane, so it's going to have... what we call a X component and a Y component to this gravity all right this force that's going to push the block all right is gonna be equal to that of the parallel component of gravity. This means that the magnitude of the applied force is equal to that of the parallel component of gravity. All right? And so that means F is going to equal mg sine. theta all right because we're moving we're moving our force of gravity let me draw this better actually all right force of gravity is going to have a opposite component this way and an opposite component this way it's going to have a minus x component all right and a minus y component that we're breaking the force of gravity into so that it is in our frame of reference of what we defined x and y to be now if this if we're moving this block this way due to this force you All right, it has to be parallel to that component of gravity. So F equals mg sine theta. mg represents the weight of the object, which is 100 newtons, as given by the problem here. So F is equal to 100 newtons. sine theta. All right. Using trigonometry, sine theta is just the ratio of the length of the opposite side to the hypotenuse. And so that's going to be the opposite side, 110 meters, I'm sorry, over the hypotenuse, which is 20 meters. All right. And so force is going to be equal to 100 newtons times 10 over. 20. All right. And so this is going to equal 50 newtons. All right. So for part A, the minimum force required to push the block is going to be 50 newtons. All right. We drew our free body diagram. All right. We determined we have a force of gravity downwards. We have our force of pushing this block this way, and we have our normal force. We've defined our plane of x and y to be in the direction, our y direction to be in the direction of the normal force and our x direction to be of the force applied to this block. That means our force of gravity is has a both x and y component based off of our x y plane. We break it into its x and y components. If the block is moving in the direction of the force applied. All right. This means the magnitude of the applied force is going to be equal to that of the parallel component of the gravity. So we write that F is equal to Fg parallel, which is just mg sine theta. We know what mg is. That's the weight, which we're given to be 100 newtons. And sine theta is equal to. The ratio of the length of the opposite side, opposite side over hypotenuse. And we're given both of those. That's 10 meters over 20 meters. Right. And we plug this into our formula to get a force of 50 newtons. All right. The minimum force required to push the block is 50 newtons. Now, part B asks us. Well, what is the work done by the force? Well, we know that work is equal to force times displacement cosine theta. All right, here theta represents the angle between the force and the displacement. vectors, not the angle of the inclined plane. All right. Now, because the force and displacement vectors are parallel, that means that theta is equal to zero. All right. And so now we have force times displacement cosine zero, which is just equal to one. And so here work is equal to force times displacement. All right. We know what those are. Force we just calculated in part A is 50 newtons. displacement is 20 meters, and that is going to give us 1,000 joules. All right, so the work done by the force is 1,000 joules. Now, part C asks us, well, what is the force required and the work done by the force if the block were simply lifted vertically 10 meters? Well, to raise the block vertically, an upward force equal to the object's weight would have to be generated, where the object's weight here is 100 newtons. So then the work done by the lifting force is going to be work is equal to force times distance cosine theta. For the block to be raised vertically, the upward force has to be equal to the object's weight, which is 100 newtons. All right. with a displacement of right 10 meters that we're simply lifting it vertically up times cosine theta which is zero all right and that's simply going to be one here and so it's a hundred times 10 meters which is equal to 1 000 joules the same amount of work is required in both cases but twice the force is needed to raise the boat to raise the block vertically compared with pushing it up the incline All right, so that's how we do that practice problem. Now, we also want to focus, right, that's inclines. We also want to talk about pulleys. Pulleys utilize the same paradigm to provide mechanical advantage as the inclined. plane there's a reduction of necessary force at the cost of increased distance to achieve a given value of work or energy transference in practical terms pulleys allow heavy objects to be lifted using a much reduced force all right Again, mechanical advantage here makes it easier to accomplish a given amount of work. Why? Because the input force necessary to accomplish the work is going to be reduced. The distance through which the reduced input force must be applied, however, is increased by the same factor. Now here when we talk about pulleys, we can talk about efficiency, which is work outputted versus work inputted. And this can be defined through this equation, load times load distance over effort times effort distance. The load is the output force of a simple machine, which acts over a given load distance to determine the work output of said simple machine. The effort is the input force of a simple machine which acts over a given effort distance to determine the work input of the simple machine. Efficiency is just the ratio of the machine's work output to work input when non-conservative forces are taken into. account. Efficiency is often expressed as a percentage by multiplying the efficiency ratio by 100 percent. All right. And the efficiency of a machine gives a measure of the amount of useful work that's going to be generated by the machine for a given amount of work put into the system. All right. Now, to really see this in action, we're going to do a quick practice problem. All right. We're going to do a quick practice problem. problem. Everything that I just stated are written in bullet points here so that we can keep it in mind. Let's do a practice problem with this system of pulleys. All right, what this says here is the pulley system in this figure has an efficiency of 80%. A person is lifting a mass of 200 kilograms with the pulley. Find A, the distance through which the effort must move to raise the load. distance of four meters be the effort required to lift the load and see the work done by the person lifting the load through a height of four meters all right so let's start with a the dip we want to find the distance through which the effort must move to raise the Load a distance of four meters. Now, for the load to move through a vertical distance of four meters, all six of the supporting ropes must shorten four meters also, all right? This may also be accomplished by pulling six times four, all right, which is equal to 24 meters of the rope through the setup. So, the effort must move through a distance of... 24 meters, right? Because look, this is a system of six pulleys. Increasing the number of pulleys decreases the tension in each segment of the rope, and that leads to increase in the mechanical advantage of the setup. If we are trying to move a total of distance of four meters, right? The distance through which the effort must move to raise the load of four meters, it has to move four meters. in all six of the supporting ropes. They all have to shorten by four meters also. And so that means the effort must move through a total of four meters for all six of the ropes. So that's a total of 24 meters. All right. Now, B says, well, what is the effort required to lift the load? Well, to calculate the effort required for part B, the equation for efficiency should be used. The load is the weight of the object being lifted. lifted. It's going to be equal to the mass of the object times the acceleration due to gravity g. And so the effort distance calculated in part a was 24 meters. And so then we can calculate the efficiency here. That's load times load distance. All right. Over effort times effort distance, effort times effort distance. Right. All right. So this is going to be equal to what we know, the efficiency is 80 percent. That's zero point eight. That's going to be equal to 200 kilograms. That's the mass times nine point eight meters per second squared. That is the acceleration due to gravity. Right. Because the load is the weight of the object being lifted. All right. And it's equal to the mass of the object times the acceleration due to gravity. And the distance here is 4 meters over the total effort, which we're trying to find. to the total distance of all the ropes trap of all of all um throughout the whole distance which we calculated in part a to be 24 meters what we're trying to figure out then is we're trying to isolate this effort term all right we're trying to isolate this effort term, which requires us to shift things around. All right. We can approximate this to about 10. So 200 times 10, that's 2000. 2000 times 4. All right. 2000 times 4 and we're going to move 0.8 and 24 meters over to the other side here. All right. So it's over 0.8 times 24. This gives us about 408 newtons. All right. So the effort required to lift the load is 408 newtons. And we calculate that using the efficiency equation, moving things around to solve for effort. All right. Load is just the mass times the acceleration due to gravity. And the load distance is just the total distance that the the distance that was traveled, which we calculated from A to be 24 meters. All right. So this is the total 4 meters. The effort distance is the 24 meters. And then we just shifted things around to solve for effort. Then the last problem for C says, well, what is the work done by the person lifting the load through a height of 4 meters? Well, we can actually calculate work in as the effort times the effort distance. All right. And that's going to be simply 408 newtons times 24. All right. All right, that's going to give us about 9,800 joules. All right, and that's how we would go about solving this problem. All right, that's all I have for you. In the next video, we'll work through more practice problems for this chapter. Let me know if you have any questions, comments, concerns down below. Other than that, good luck, happy studying, and have a beautiful, beautiful day, future doctors.

Hello everybody, my name is Iman. Welcome back to my YouTube channel. Today we're continuing our MCAT physics lecture on work and energy.

Last time we talked about energy and particularly focused on kinetic and potential energy. We also talked about total conservation of mechanical energy and we discussed the first law of thermodynamics. Now what we want to move into is discussing work. Now, please note that we will dive into work and its components heavily also in the next chapter on thermodynamics.

But we're going to introduce it here. Work is not energy, but a measure of energy transfer. In fact, there are only two forms of energy transfer, work and heat. Now, in physics, work is the energy transfer.

to or from an object through the application of force along a displacement. And so in its simplest form, it is often represented as the product of force and displacement. So we're going to be really focusing in on work as a way of energy transfer. We'll talk about heat a lot in the next chapter.

Like we said, Energy is transferred through the process of work when something exerts forces on or against something else. So work is force dot product displacement. In other ways, we can write force times displacement times cosine theta.

All right, so W is work. F is the magnitude of the applied force. And D is the magnitude of the displacement through which the force is applied.

Theta is the... angle between the applied force vector and the displacement vector. You'll notice that work is a dot product, right? So it's a function of the cosine of the angle between the vectors.

This also means that only forces or components of forces parallel or anti-parallel to the displacement vector will do work. And so that means transfer energy. We've already said that the SI unit for work is joules. And while this suggests that work and energy are the same thing, remember that they're not. Work is the process by which a quantity of energy is moved from one system into another.

Now, as described, work is a process of energy transfer. In systems of gases, then, we therefore approach work as a combination of pressure and volume changes. All right, so that's going to be really important. For a gas system contained in a cylinder with a movable piston, We can analyze the relationship between pressure, volume, and work. When the gas expands, it's going to push up against the piston, and that's going to exert a force that causes the piston to move up and the volume of the system to increase.

Now, when the system is compressed, the piston is going to push down on it. the gas and that's going to exert a force that decreases the volume of the system. We say that work has been done when the volume of the system has changed due to an applied force.

Gas expansions and compression processes can actually also be represented in graphical form with volume on the x-axis and pressure on the y-axis. Such graphs are termed as PV graphs, and we can see these right here, right? Volume on the x-axis, pressure on the y-axis.

Fantastic. And we're going to see later on in the next chapter how work is demonstrated and we'll see it here too, but how work is equal to pressure. Times change in volume. All right, so that's another mathematical expression for work.

All right, now let's discuss a couple of different pressure-volume graphs and how we can determine work from these. All right, now when a gas expands, we say that the work was done by the gas. All right, and the work is positive.

Okay, let me repeat that. When a gas expands, we say that the work was done by the gas and the work is positive. All right, so let's write this down. When gas expands, work done.

Sorry, work. Work is done by gas, all right? And then work is positive, okay?

Now, what if a gas is compressed, all right? What if a gas is compressed? I'm gonna shorthand it, all right?

Then we say that work was done on the gas, all right? Work done on the gas and work is negative. Now there are infinite number of paths between initial and final state here. Different paths require different amounts of work.

And you can calculate the work done on or by a system by finding the area under the pressure volume curve. We're going to go through different scenarios A, B, C, and D. In scenario A, Note that if the volume stays constant, if there's no change in volume as pressure changes, then the delta V change in volume is going to be zero.

And considering that our formula for work is pressure times change in volume, if change in volume is zero, then no work is done because it's P times zero, and that equals zero. Okay, so if the volume stays constant as pressure changes, there is no work done because look at this, there is no area to calculate. This is called an isovolumetric or isochoric process.

All right, and we can see that demonstrated in a pressure volume graph. in graph A here, all right? What about graph B here, all right? On the other hand, if pressure remains constant as volume changes, all right, then the area under the curve is a rectangle of length P, all right, of length P.

That's length P and width delta V. All right, so for processes in which pressure remains constant, this is called isobaric processes, as shown in plot B, and you can calculate the work done as pressure times delta V. So phenomenal. Now, what about what about graph C? All right. Graph C shows a process in which neither pressure nor volume are held constant.

OK, the total area under the curve, then you're going to have to break it down into region one. All right. Which is a triangle and region two, which is a square. All right, a rectangle.

Region 1 is a triangle, all right, where area can be determined like this, one half delta V times delta P, all right, and you can figure that out from the graph. And then you can calculate for region 2, which is a rectangle, so it's going to be pressure 2 times change in volume here. Alright, and then the work, the total work done is just the sum of region 1 plus the sum of region 2. Alright, so that's for graph C. And then graph D, this one is much more complicated and sort of outside the...

scope of MCAT, it shows a closed cycle in which after certain interchanges of work and heat, you return to an initial state. Now, because work is positive when the gas expands and negative when the gas is compressed. The work done is the area enclosed by the curve. But calculating the work done in this situation would require calculus, and we don't have to worry about that here in this MCAT prep.

You'll learn about it if you take a look at the previous video. Take physical chemistry, and I actually cover it in my physical chemistry one playlist. But you don't have to worry about it so much for the MCAT. The whole point is to show you that work can be calculated as a function of pressure. pressure and volume, and that you can work from pressure volume graphs to calculate that based off of that equation.

And we cover different graphs and different scenarios to do just that. Now, let's discuss a few more things here quickly before we move into our last topic. Those are going to be power and the work energy theorem.

Power refers to the rate at which energy is transferred from one system. to another. It's going to be calculated from this expression here.

Power, all right, where power, P is power, all right, it's equal to W over T. W is work, all right, which is equal to change in energy, all right, and that's all over T, which is time over which work is done. So power is equal to work over time.

It's also equal to change in energy over time, all right. The S, Unit for power is watt. That's watt or W, which is equal to joules per second.

All right. Now, in the later chapters, we're going to identify additional ways to calculate power and electric currents. But for now, note that many of the devices we use every day, toaster ovens, light bulbs, phones, cars, they're all quantified by the rate at which those appliances transform electrical potential energy into other forms like thermal, light, sound, and kinetic energy.

Power is calculated in many different situations, especially those involving circuits, resistors, and capacitors. And we're going to talk about those later down the line, I think in like chapter six and seven. For now, this is all you need to know. This expression, power is equal to work over time or it's equal to change in energy over time. With that, we can discuss work energy theorem here.

The work energy theorem is a powerful expression of the relationship between work and energy. All right? Between work and energy.

And it's calculated. by this expression work net is equal to change in kinetic energy. So that's equal to kinetic final minus kinetic initial.

Now in mechanical application, it is going to offer a direct relationship between the work done by all the forces acting on an object and the change in kinetic energy of that object. the network done by forces acting on an object is going to result in an equal change in the object's kinetic energy. And we can use this expression to calculate that.

All right. Phenomenal. So that's power and that's work energy theorem. All right. Now, We can look at how this, we're applying here a net force, all right, and it's moving at some initial velocity, all right, and we have a displacement and our final velocity.

This can be under the guise of positive work, all right, and if we apply force net in an opposite direction while it moves in an initial velocity with some sort of displacement and final velocity, then we're going to see that this is going to be negative work, all right? In essence, the network done by forces acting on an object will result in an equal change in the object's kinetic energy. And we can calculate that by kinetic final minus kinetic initial. This is just 1 half mv final squared minus 1 half mv initial squared. And that's how you can calculate and see what gives you positive work, what gives you negative work, all right?

Fantastic. If the V final is greater than the V initial, you're going to see that you're going to have positive work. All right.

If the opposite is true, if you have if if V final is less than V initial, you're going to get negative work values. All right. You're going to calculate a negative work, a number. All right. And so that.

is work energy theorem. Our last topic for this chapter is, oh, actually, let's do a practice problem. Look at that, we have a practice problem to do.

A lead ball of mass 0.125 kilograms is thrown straight up in the air with an initial velocity of 30 meters per second. Assuming no air resistance, Find the work done by the force of gravity by the time the ball is at its maximum height. All right.

Now, one thing to note, this answer could be calculated using kinematics and determining the maximum height of the ball. But you can also actually use the work energy theorem to calculate, and it's much simpler that way. So what you can do is you can set up that your work net is equal to your kinetic final minus your kinetic. initial all right your kinetic final when it reaches that maximum height we talked about that when it reaches its maximum height the velocity is equal to zero so what that means is that your your velocity final is equal to zero so if we write this out 1 half MV final squared minus 1 half MV initial square if you know your MV if your V final is zero this whole term is equal to zero So all you have now is 0 minus 1 half m v initial squared Alright, we know we know what V initial is. It's 30 meters per second.

We know what the mass is so we can simply plug things in 0 minus 1 half mass is 0.125 kilograms times 30 meters per second squared all right and this if you plug this into a calculator is going to be equal to minus 56.25 joules Fantastic. So that's how you can approach a problem like this. All right. Now we can actually move into our last and final topic for this chapter, which is mechanical advantage. All right.

Mechanical advantage. Now we're going to talk about here. All right.

What the difference? The difference for a given quantity of work, all right? So for a given quantity of work, any device that allows for work to be accomplished through a smaller applied force. is going to be said to provide mechanical advantage. All right.

And that is the topic of this section, mechanical advantage. So one example of mechanical advantage. is using inclines.

Sloping inclines like hillsides and ramps make it easier to lift objects because they're going to distribute the required work over a larger distance, thereby decreasing the required force. And so it is one example of mechanical advantage. In addition to the inclined plane, five other devices are considered the classic simple machines designed to provide mechanical advantage.

What are those devices? Well, in addition to inclined planes, we have wedge, which is two merged inclined planes. We have wheel and axle.

We have lever. pulley, and screw. Screw is just a rotated inclined plane. Of these, inclined plane, lever, and pulley are the most frequently tested on in the MCAT. All right, we're going to talk about pulleys here in this chapter a lot, but to kind of set this up, let's define mechanical advantage.

All right, mechanical advantage is the ratio of magnitudes of the force exerted. on an object by a simple machine to the force actually applied on the simple machine. And so we can write this as F out over F in. All right. What you notice is that it's a ratio, so it's going to be dimensionless.

Now, reducing the force needed to accomplish a given amount of of work does have an associated cost with it. However, all right, the distance through which the smaller forces must be applied in order to do the work is going to be thereby increased. All right, that's just just a direct consequence of mechanical advantage. All right? So reducing the force needed to accomplish a given amount of work is going to cause the distance through which the smaller force must be applied in order to do the work to be increased.

All right? Inclined planes, levers, and pulleys, they do not magically just change the amount of work necessary to move an object from one place to another. And that is why applying a lesser force over a greater distance is going to achieve the same results, is going to achieve the same changes in position accomplished by the same amount of work.

So we're going to increase the distance through which those smaller forces have to be applied in order to do the same amount of work. Fantastic. Now, let's quickly see what that looks like through an example problem before we talk about pulleys.

All right. A block weighing 100 newtons is pushed up a frictionless incline over a distance of 20 meters to a height of 10 meters, like we see in this. image right here. What we want to do is we want to find A, the minimum force required to push the block, B, the work done by the force, and C, the work required or the force required and the work done by the force if the block were simply lifted vertically 10 meters.

Fantastic. Let's start with A. All right, we want to find the minimum force required to push the block.

All right, the first thing we want to do is find the minimum force required to push the block and so we have to draw a free body diagram of the situation all right so here's our block it looks like this all right that's our block all right we're going to draw our free body diagram we have our force of gravity pushing it downwards all right we have our normal force perpendicular to the surface All right, and then we have the force pushing it upwards this way All right, the minimum force needed is going to be a force that will push the block with no acceleration parallel to the surface of the incline. What that means is the magnitude of the applied force, F, is going to be equal to that of the parallel component of gravity. All right, so gravity is going to have A component, all right, it's going to be the parallel component of gravity.

So what we're doing here, all right, is we're essentially defining this to be our plane of reference. This is our x-plane, this is our y-plane. Force of gravity is out of this plane, so it's going to have...

what we call a X component and a Y component to this gravity all right this force that's going to push the block all right is gonna be equal to that of the parallel component of gravity. This means that the magnitude of the applied force is equal to that of the parallel component of gravity. All right? And so that means F is going to equal mg sine.

theta all right because we're moving we're moving our force of gravity let me draw this better actually all right force of gravity is going to have a opposite component this way and an opposite component this way it's going to have a minus x component all right and a minus y component that we're breaking the force of gravity into so that it is in our frame of reference of what we defined x and y to be now if this if we're moving this block this way due to this force you All right, it has to be parallel to that component of gravity. So F equals mg sine theta. mg represents the weight of the object, which is 100 newtons, as given by the problem here. So F is equal to 100 newtons. sine theta.

All right. Using trigonometry, sine theta is just the ratio of the length of the opposite side to the hypotenuse. And so that's going to be the opposite side, 110 meters, I'm sorry, over the hypotenuse, which is 20 meters.

All right. And so force is going to be equal to 100 newtons times 10 over. 20. All right.

And so this is going to equal 50 newtons. All right. So for part A, the minimum force required to push the block is going to be 50 newtons. All right.

We drew our free body diagram. All right. We determined we have a force of gravity downwards.

We have our force of pushing this block this way, and we have our normal force. We've defined our plane of x and y to be in the direction, our y direction to be in the direction of the normal force and our x direction to be of the force applied to this block. That means our force of gravity is has a both x and y component based off of our x y plane.

We break it into its x and y components. If the block is moving in the direction of the force applied. All right. This means the magnitude of the applied force is going to be equal to that of the parallel component of the gravity. So we write that F is equal to Fg parallel, which is just mg sine theta.

We know what mg is. That's the weight, which we're given to be 100 newtons. And sine theta is equal to.

The ratio of the length of the opposite side, opposite side over hypotenuse. And we're given both of those. That's 10 meters over 20 meters.

Right. And we plug this into our formula to get a force of 50 newtons. All right.

The minimum force required to push the block is 50 newtons. Now, part B asks us. Well, what is the work done by the force? Well, we know that work is equal to force times displacement cosine theta. All right, here theta represents the angle between the force and the displacement.

vectors, not the angle of the inclined plane. All right. Now, because the force and displacement vectors are parallel, that means that theta is equal to zero.

All right. And so now we have force times displacement cosine zero, which is just equal to one. And so here work is equal to force times displacement. All right.

We know what those are. Force we just calculated in part A is 50 newtons. displacement is 20 meters, and that is going to give us 1,000 joules.

All right, so the work done by the force is 1,000 joules. Now, part C asks us, well, what is the force required and the work done by the force if the block were simply lifted vertically 10 meters? Well, to raise the block vertically, an upward force equal to the object's weight would have to be generated, where the object's weight here is 100 newtons.

So then the work done by the lifting force is going to be work is equal to force times distance cosine theta. For the block to be raised vertically, the upward force has to be equal to the object's weight, which is 100 newtons. All right.

with a displacement of right 10 meters that we're simply lifting it vertically up times cosine theta which is zero all right and that's simply going to be one here and so it's a hundred times 10 meters which is equal to 1 000 joules the same amount of work is required in both cases but twice the force is needed to raise the boat to raise the block vertically compared with pushing it up the incline All right, so that's how we do that practice problem. Now, we also want to focus, right, that's inclines. We also want to talk about pulleys. Pulleys utilize the same paradigm to provide mechanical advantage as the inclined. plane there's a reduction of necessary force at the cost of increased distance to achieve a given value of work or energy transference in practical terms pulleys allow heavy objects to be lifted using a much reduced force all right Again, mechanical advantage here makes it easier to accomplish a given amount of work.

Why? Because the input force necessary to accomplish the work is going to be reduced. The distance through which the reduced input force must be applied, however, is increased by the same factor.

Now here when we talk about pulleys, we can talk about efficiency, which is work outputted versus work inputted. And this can be defined through this equation, load times load distance over effort times effort distance. The load is the output force of a simple machine, which acts over a given load distance to determine the work output of said simple machine.

The effort is the input force of a simple machine which acts over a given effort distance to determine the work input of the simple machine. Efficiency is just the ratio of the machine's work output to work input when non-conservative forces are taken into. account.

Efficiency is often expressed as a percentage by multiplying the efficiency ratio by 100 percent. All right. And the efficiency of a machine gives a measure of the amount of useful work that's going to be generated by the machine for a given amount of work put into the system.

All right. Now, to really see this in action, we're going to do a quick practice problem. All right. We're going to do a quick practice problem. problem.

Everything that I just stated are written in bullet points here so that we can keep it in mind. Let's do a practice problem with this system of pulleys. All right, what this says here is the pulley system in this figure has an efficiency of 80%. A person is lifting a mass of 200 kilograms with the pulley.

Find A, the distance through which the effort must move to raise the load. distance of four meters be the effort required to lift the load and see the work done by the person lifting the load through a height of four meters all right so let's start with a the dip we want to find the distance through which the effort must move to raise the Load a distance of four meters. Now, for the load to move through a vertical distance of four meters, all six of the supporting ropes must shorten four meters also, all right? This may also be accomplished by pulling six times four, all right, which is equal to 24 meters of the rope through the setup.

So, the effort must move through a distance of... 24 meters, right? Because look, this is a system of six pulleys. Increasing the number of pulleys decreases the tension in each segment of the rope, and that leads to increase in the mechanical advantage of the setup. If we are trying to move a total of distance of four meters, right?

The distance through which the effort must move to raise the load of four meters, it has to move four meters. in all six of the supporting ropes. They all have to shorten by four meters also.

And so that means the effort must move through a total of four meters for all six of the ropes. So that's a total of 24 meters. All right. Now, B says, well, what is the effort required to lift the load? Well, to calculate the effort required for part B, the equation for efficiency should be used.

The load is the weight of the object being lifted. lifted. It's going to be equal to the mass of the object times the acceleration due to gravity g.

And so the effort distance calculated in part a was 24 meters. And so then we can calculate the efficiency here. That's load times load distance.

All right. Over effort times effort distance, effort times effort distance. Right. All right.

So this is going to be equal to what we know, the efficiency is 80 percent. That's zero point eight. That's going to be equal to 200 kilograms. That's the mass times nine point eight meters per second squared.

That is the acceleration due to gravity. Right. Because the load is the weight of the object being lifted. All right. And it's equal to the mass of the object times the acceleration due to gravity.

And the distance here is 4 meters over the total effort, which we're trying to find. to the total distance of all the ropes trap of all of all um throughout the whole distance which we calculated in part a to be 24 meters what we're trying to figure out then is we're trying to isolate this effort term all right we're trying to isolate this effort term, which requires us to shift things around. All right.

We can approximate this to about 10. So 200 times 10, that's 2000. 2000 times 4. All right. 2000 times 4 and we're going to move 0.8 and 24 meters over to the other side here. All right.

So it's over 0.8 times 24. This gives us about 408 newtons. All right. So the effort required to lift the load is 408 newtons.

And we calculate that using the efficiency equation, moving things around to solve for effort. All right. Load is just the mass times the acceleration due to gravity. And the load distance is just the total distance that the the distance that was traveled, which we calculated from A to be 24 meters.

All right. So this is the total 4 meters. The effort distance is the 24 meters.

And then we just shifted things around to solve for effort. Then the last problem for C says, well, what is the work done by the person lifting the load through a height of 4 meters? Well, we can actually calculate work in as the effort times the effort distance.

All right. And that's going to be simply 408 newtons times 24. All right. All right, that's going to give us about 9,800 joules. All right, and that's how we would go about solving this problem.

All right, that's all I have for you. In the next video, we'll work through more practice problems for this chapter. Let me know if you have any questions, comments, concerns down below. Other than that, good luck, happy studying, and have a beautiful, beautiful day, future doctors.

Transcript for:Understanding Work, Energy, and Power Concepts

Transcript for:
Understanding Work, Energy, and Power Concepts