Understanding Work and Energy in Physics

Work is going to be the topic of this lesson in my brand new general physics playlist, covering a full year of algebra-based general physics. Now we're in a whole chapter on energy here, and I think you're going to find it a little bit refreshing in certain respects, because we're going to find out that energy and associated terms are all scalars, not vectors, which is great news when it comes to two-dimensional problems, as we'll see. My name is Chad, and welcome to Chad's Prep, where my goal is to take the stress out of learning science. Now if you're new to the channel, we've got comprehensive playlists for general chemistry, organic chemistry, general physics, and high school chemistry. And on chadsprep.com you'll find premium master courses for the same that include study guides and a ton of practice. You'll also find comprehensive prep courses for the DAT, the MCAT, and the OAT. All right, so we're going to start off talking about work in this whole chapter on energy. You should realize that work as well as potential energy and kinetic energy that we'll get to later in this chapter, these are all scalars. And that is fantastic news. So recall that being a scalar means they have magnitude. They do not have direction. And this is great because when we start adding energies together, we don't have to worry about breaking them into components because there's no direction associated with them. We're just going to add them as scalars, which is so much easier to do. So no adding vectors when it comes to energy terms. So work, potential energy, kinetic energy, plain old scalars. So you should also know that the SI unit for any of these, whether it be work, potential energy, or kinetic energy, is dual. So it symbolizes the capital J and it's a derived SI unit, obviously, it's not one of the base units. But you should be able to express it in the base units as well. And so you can figure out how to do that from any of your energy definitions or from the definition of work, as we'll see here. And since we haven't learned those energy definitions yet, or equations, then we'll start with the work equation. So work we commonly think of as force times displacement. So as long as we're dealing with a one-dimensional problem, it's usually how we look at it. Let's say everything's in the x dimension, then it's just simply force in the x dimension times the displacement in the x dimension. So what's force times displacement? Well, force has units of newtons, displacement has units of meters, and so a joule is the same thing as a newton meter. So, but notice we're still not quite in base units. The newton itself is a derived SI unit. So if you wanted to split this further and again express the joule in all base units, then we've got to take that newton and We can figure that out often most conveniently with force is equal to mass times acceleration. And so the Newton, the SI unit for force, must be a kilogram times a meter per second squared. And so a kilogram times a meter per second squared and then times another meter. Well, we're typically not going to write it like that, right? So we're just going to take this times that additional meter and just incorporate it in here with a squared. And so a joule. You'll often see it presented as a newton meter, but also as a kilogram meter squared per second squared expressed in all SI base units. Okay, so there's our SI unit here for any of our energies, whether it be work or kinetic energy or potential energy. Now, the term work itself, you know, we use it in the vernacular all day long, you know, I have to go to work and things of a sort, but it has a specific definition. And it's this force times this placement. It's not enough simply to apply a force. If you apply a force to something, but it doesn't move, you have done no work. So you've got to apply a force and have it move, and have it move in that same direction, so to speak. And so in one dimension, it's a rather simplistic definition. But once we start looking at this two-dimensionally, and you might be like, oh, Chad, you said that it's not a vector. You're right, it's not a vector. But it turns out it's a dot product. So we're not going to go into talking about dot products or anything, but we do have to account for the fact that direction... does matter even though work itself doesn't have direction because the quantities that make it up, if you look, force is a vector and displacement is a vector. And these are both vectors, and even though the dot product of the two is gonna end up being a scalar, so direction will come into play. And so it turns out when you're dealing with two dimensions, you can kind of think of it that if the force and the displacement point in the same direction, then it really is just force times displacement. However, if they don't point in exactly the same direction, Then you've got to factor in really only the component of the force that points in the same direction as the displacement. And the way we would accomplish that, as long as we're taking the angle with respect to the x-axis, let's say, is that f cosine theta is the component of the force in the x direction, and if the displacement's also in the x direction, we'd account for it like that. Well, in this case, all we really have to do is define theta here as the angle between the force and the displacement, and then we can write it this way every time. So work is going to equal... Force times the cosine of the angle between force and displacement and then times the displacement. And again, if force and displacement are in the same direction, Well then theta here is zero and the cosine of zero is one and that whole cosine just goes away. Notice also that this is going to be important. What if the force and displacement are in opposite directions? All of a sudden then theta is 180 degrees and the cosine of 180 degrees is negative one and we'll be talking about negative work. And so if a force and displacement are in the same direction work is going to be positive but if force and displacement are in opposite directions work is going to be negative. Negative. So, and those are obviously the extremes, but anything between like 0 and 90 degrees, you're going to get a positive work, and anything between 90 and 180 degrees, you're going to get negative work. But the most common examples you'll see are when we've got force and displacement in the same direction or force and displacement in opposite directions. But you are now prepared for anything in between as well. Alright, so this is work, and we're ready to do some calculations here. So, uh, First question, how much work is performed by a person lifting a 10.0 kilogram object at constant velocity, big buzzword there, to a height of 2.0 meters? Okay, so we've got an object here, 10.0 kilograms, and we want to lift it to a height of two meters off the ground. Okay, well we're going to set up a free body diagram here and we know that this thing has a weight, mg. So and then somebody's lifting this thing so there's some sort of lifting force that's opposing that. I'm just going to call it F in this case. And we want to know how much work is done by that lifting force right there. That's what's going into our work equation. And so we've got work again equals force times the cosine of theta times displacement. And that's the force I need. Well the key is we're told that they're going to be lifting it at constant velocity. And if they're lifting at constant velocity, you know that the net force in this y direction must be zero. And so we could say that sum of the forces equals zero. It's an equilibrium according to Newton's first law. And so we'd have force minus mg equals zero. And so that lifting force must equal in magnitude the weight. Opposite direction obviously, but equal in magnitude to the weight. And you might have intuitively figured that out like, well if I want to lift that weight, then the force I lift up with has to equal that weight. Well, it does as long as we're not lifting it with any sort of net acceleration. In this case, we're told it's at constant velocity. So yeah, that force must equal the weight, but in opposite direction. All right, so now we know what that force is going to be, and we can substitute it right into this formula. Also, in this case, the displacement is going to be 2 meters in the upward direction. So the force is in the upward direction. They're in the same direction, and therefore the cosine of 0 is 1, and that cosine just goes away. And so here we're going to have work equals So mass times gravity, well we're told it's 10.0 kilograms times 9.8 meters per second squared, and then times the displacement of 2.0 meters. So, and we could let our calculator do some work, we probably do this in our head truth be told. So 10 times 9.8 is 98, 98 times 2 is 196, that's 196 joules. Now in this case we're gonna be limited to two sig figs here. by that 2 meters. Now that's 3 sig figs, and obviously to make it 2 sig figs it's going to round up to 200. So personally I hate putting bars over zeros, truth be told I've done it before, but my personal favorite then is just to make this 2.0 times 10 to the second joules as a way of writing 200 with just 2 sig figs. Alright, the next question says, how much work must be done to raise the 24 kilogram mass 2.0 meters at constant velocity with and without the pulley system in the diagram. So, if you recall, we have the same pulley system we studied in the last chapter, and it gives us a mechanical advantage so that we can raise this 24 kilogram mass here with less force than it would normally be required to raise it. So what we're gonna find, there's no such thing as a free lunch, and even though it's gonna require less force, it is not gonna require any less energy. There's no such thing as a free lunch here, so as we'll see. So let's do it without the... Without the pulley system first and figure out how much energy this is going to be so and in this case We'll figure out how much work it is to raise it two meters and again. We know it's got a weight downward It's got a force upward that we're going to apply and then from there work is going to equal force times cosine theta times the displacement And in this case we can see that that force is going to have to equal the weight again this is being raised at constant velocity and so That mass was again 24 kilograms, gravity is 9.8 meters per second squared, and then the displacement is going to be 2 meters. We'll let a calculator do some work for us here. All right, so 24 times 9.8 times 2. We're going to get 470.4, which two sig figs all round down to 470. Joules. Okay. Now again, there's no such thing as a free lunch. This should come out to exactly the same thing. Now if you recall though, in this case the tension on these cables gets to pull on each of these pulley wheels that's attached to the 24-kilo-year mass on both sides. We got a tension force pulling there, the tension pulling there, the tension pulling there, and there, and there, and there. And so the tension gets a chance to pull up on these pulleys that are attached to the mass in six different places. And so with this 24 kilogram mass being essentially just about 240 newtons, over here we have to lift with 240 newtons. 24 kilograms times 9.8 is almost 240 newtons. But here we're only going to have to pull up with one-sixth of that. Well, I say pull up, we have to pull on the cable with roughly one-sixth of that force. Instead of roughly 240 newtons, it's going to be closer to 40 newtons. 39 point something, if you might recall from the last chapter. So... the force is less, that kind of seems like we're getting a free lunch, doesn't it? So, and we set up our equation, we're like work equals force cosine theta times displacement. So, oh, and by the way, I didn't mention it, but it goes without saying here, cosine theta disappeared because the force and the displacement were in the same direction, and cosine of zero was one. So in this case, we're not applying the force directly to it, we're actually applying the force to the cable. So, and we're pulling it down, which is going to pull the mass up and we still want this mass to rise two meters. That's going to be crucial here but we're going to actually be pulling down on the cable and it turns out it wouldn't have to be perfectly down we could pull it off at an angle as long as we're putting a tension in this cable that's the key. And so in this case that force instead of being 24 kilograms times 9.8 meters per second squared it's going to be just 1 sixth of that so in this case instead so it's gonna be 1 sixth of again 24 kilograms. times 9.8 meters per second squared and all of a sudden it's looking like well the force is only one-sixth of the force Chad and we still only want it to rise two meters so if we plug two meters into that isn't the work going to come out to just one-sixth of the value? It might look like that but again the key is this we want this mass to rise two meters but that's not the displacement we're worried about here so when we're actually doing the work the displacement is how far down we've got to pull this cable. When we pull this cable, the whole pulley system is cinching together. And with it going around the six wheels here, it turns out that when you pull it downward, so the cables are actually only going to raise this one-sixth of that distance. And so if I want this to rise two meters, I actually have to pull this cable down a total of 12 meters. And that's where we're not getting that free lunch. And so even though the overall force, so is one sixth of the force, the displacement we've got to generate is going to be six times greater. And so all said and done, when you calculate out that work, it's still going to come out to 470. joules rounded to two sig figs. Cool. So mechanical advantage is going to decrease the amount of force we need to apply, but it is not going to reduce the amount of work that needs to be done whatsoever. All right, so last two problems. are going to be related. First one is a 10.0 kilogram object on a horizontal surface, coefficient of kinetic friction is given, is pulled horizontally. So with a force of 50.0 Newtons yielding a horizontal displacement of 2.0 meters, how much work was performed by the pulling force and then how much work was performed by friction? So two questions here, work of the pulling force and work by friction. All right, so we still know that work... F cosine theta times displacement. And if we look at the pulling force, the pulling force is pulling to the right, and there's going to be a displacement to the right. That should be positive work. So on the other hand, though, the displacement's going to be the right. Friction always opposes the motion, and so friction's going to pull back the opposite direction, and so the work for friction should be negative instead. Alright, and so from this case if we look at the pulling force, well the pulling force is in the horizontal direction And so is the displacement and so we're just going to have 50.0 Newtons cosine of 0 is 1 so that falls out and then times that 2.0 meters, and we're gonna get a work of a hundred Joules and again if I want to write that in two sig figs, I'll make that 1.0 times 10 to the second Joules All right, so that was the work done by the pulling force now. What about the work done by friction? Well for the work done by friction We got the same setup for solving for work, but we got to figure out what that force is. So this placement is still going to be 2.0 meters. Well, in this case, if we set up a free body diagram, so let's start with the force of friction actually. Let's just go there. So this force is going to be the force of friction, and we know that the force of friction is kinetic friction mu times the normal force and then times cosine theta times d. Alright, so we know coefficient of kinetic friction is probably 0.2, the normal force we haven't figured out yet. But we know that the weight points down, and the normal force is going to counteract that. And we set up some of the forces there. We can see that some of the forces in the y direction add up to 0. And so in this case, the normal force minus mg equals 0. And that normal force equals mg, pretty common. for the normal force to equal the weight in magnitude. Alright, so we'll substitute that in. Weight, there's our coefficient of kinetic friction. So the mass in this case was given as 10 kilograms. Gravity is 9.8 meters per second squared. So, and then cosine theta here is going to be cosine of 180 degrees. Again, the displacement's to the right as I've drawn it. Whereas friction is force pointed in the exact opposite direction. So and then finally times the displacement of 2 meters. And we'll let our calculator do some work for us here. So in this case times 0.2 times 10 times 9.8 times cosine of 180 is just going to make it negative and then times 2. And we're going to get negative 39.2. Joules. In this case actually if I round that to two sig figs I'll have to round that down to 39 joules. Cool and again it's negative just because the force the displacement are in opposite directions not in the same direction. So we'll find out that when work is positive it is kind of adding energy to whatever your object is when force when work is negative it's taking away energy from it as we'll see in the next section. Alright, if we take a look at this one and see what's different now. So instead of pulling perfectly horizontal with that 50 newton force, it's going to be pulled up at an angle of 30 degrees. That's going to make this a little harder as we'll see. So let's read the whole thing though. A 10.0 kilogram object on a horizontal surface, mu k equals 0.2, is pulled with a force of 50.0 newtons at an angle of 30 degrees above the horizontal, yielding a horizontal displacement of 2 meters. So displacement's still horizontal though, 2 meters. How much work was performed by the pulling force and how much work was performed by friction. Okay, so we'll start with the work done by the pulling force here. And again, work is equal to F cosine theta times displacement. And once again, so our force here of 50 newtons, but we don't have cosine of zero now. It's going to be cosine of 30 degrees and then times the displacement of 2.0 meters. Cool, now one way to look at this, so one we just plugged it into the formula, so what you might think of is that this F cosine theta is really just the component of the force that's in the same direction as the displacement. So in this case the displacement's in the x direction, so I need the component of the 50 Newtons that's also in that same direction. Well if you look it would just be 50 Newtons times cosine 30, that's kind of where that's coming from, so at least one way to look at it. So that's my personal way of just conceptually understanding it. So but from here it's plug and chug. Let our calculator do the work for us. So 50 times the cosine of 30 times 2. It's gonna get us 86.6 newtons which rounding to 2 sig figs. Ahah, newtons. Joules, we're doing energy here. So 87 joules rounded to 2 sig figs. Now notice this is smaller than the hundred joules we had right back here done by the pulling force. So in this case, because not all 50 newtons is pointed in the direction of the displacement like it was back here, now only a smaller component of it. Same displacement of 2 meters, but a lower force in the direction of that displacement, leading to less work done by that pulling force. Now we've also got to get the work done by friction here. Once again, this is the force of friction. So in that force of fric... Friction is equal to mu kinetic times your normal force times cosine theta and times your displacement. And once again, we know the coefficient of kinetic friction is 0.2. We know the angle theta is going to be 180 degrees, that displacement is horizontally to the right. Friction always opposes the motion, it's to the left. And the displacement is still 2 meters, but we still need that normal force. And so we'll set up the free body diagram. We still have the object's weight. Pointing down we still have a normal force Pointing up and you might be like it worked out the same way Chad so normal force is still gonna equal the weight Except we got to count for one other thing right so that 50 Newton's not only had an X component that in this case was 50 times cosine 30, but it also has a y component that's equal to 50 newtons times sine of 30 degrees. And so now when we go to set up the sum of the forces in the y direction equaling zero, there are three parts to that, not just two. And so in this case, we're going to have that normal force pointing up positive. We're also going to have this 50 sine 30 pointing up, so that's positive. And then minus mg equaling zero. And so we solve for that normal force. It's not simply going to equal mg. It turns out that normal force is going to equal mg minus 50 sine 30. Well, sine of 30 conveniently is...... is one half and 50 times a half is 25. So it's gonna be mg minus 25 newtons on the other side there. If we look at this, so the mass again is 10 kilograms, gravity is 9.8 meters per second squared, 10 times 9.8 is 9.8. 98, and 98 minus 25 is going to get us a normal force of just 73 newtons. And so if we look, that's less than what we had back over here. Here, the normal force was the weight of 98 newtons. And so in this case, the normal force was perfectly having to balance all the weight. Whatever the weight pulled down with, the normal force had to balance that back. Maybe you look at it according to Newton's third law, if you will. For every force, there's an equal and opposite force. Whereas here, the normal force doesn't have to balance all of the mg by itself. itself, so to speak, because it's getting a little help from this 25 Newtons of the component of the force that's pulling upward. And so the normal force only has to balance the remaining 73 Newtons instead. So there's less of a contact force and there's going to be less friction as a result as well. Alright, so that coefficient of kinetic friction is still 0.2. So our normal force is now just 73 newtons. Times, again, cosine of 180 degrees. They're in opposite directions. This will make work negative. And times that displacement of 2.0 meters. And we'll let our calculator do some more work. All right, so 0.2 times 73 times cosine of 180 and times 2. It's going to get us negative 29.2 joules, which rounded to 2 sig figs will round down to negative 29 joules. Let me make sure I got that right. Yep, negative 29 joules. Cool, and so we see not only did the pulling force do less work here than back on this first problem, but we see friction did less work as well because there was less normal force, less contact force, and therefore less of a frictional force there as well. Cool, if you have found this lesson helpful, then consider giving it a like. Happy studying.

Now if you're new to the channel, we've got comprehensive playlists for general chemistry, organic chemistry, general physics, and high school chemistry. And on chadsprep.com you'll find premium master courses for the same that include study guides and a ton of practice. You'll also find comprehensive prep courses for the DAT, the MCAT, and the OAT. All right, so we're going to start off talking about work in this whole chapter on energy.

You should realize that work as well as potential energy and kinetic energy that we'll get to later in this chapter, these are all scalars. And that is fantastic news. So recall that being a scalar means they have magnitude. They do not have direction. And this is great because when we start adding energies together, we don't have to worry about breaking them into components because there's no direction associated with them.

We're just going to add them as scalars, which is so much easier to do. So no adding vectors when it comes to energy terms. So work, potential energy, kinetic energy, plain old scalars. So you should also know that the SI unit for any of these, whether it be work, potential energy, or kinetic energy, is dual. So it symbolizes the capital J and it's a derived SI unit, obviously, it's not one of the base units.

But you should be able to express it in the base units as well. And so you can figure out how to do that from any of your energy definitions or from the definition of work, as we'll see here. And since we haven't learned those energy definitions yet, or equations, then we'll start with the work equation.

So work we commonly think of as force times displacement. So as long as we're dealing with a one-dimensional problem, it's usually how we look at it. Let's say everything's in the x dimension, then it's just simply force in the x dimension times the displacement in the x dimension.

So what's force times displacement? Well, force has units of newtons, displacement has units of meters, and so a joule is the same thing as a newton meter. So, but notice we're still not quite in base units. The newton itself is a derived SI unit. So if you wanted to split this further and again express the joule in all base units, then we've got to take that newton and We can figure that out often most conveniently with force is equal to mass times acceleration.

And so the Newton, the SI unit for force, must be a kilogram times a meter per second squared. And so a kilogram times a meter per second squared and then times another meter. Well, we're typically not going to write it like that, right? So we're just going to take this times that additional meter and just incorporate it in here with a squared.

And so a joule. You'll often see it presented as a newton meter, but also as a kilogram meter squared per second squared expressed in all SI base units. Okay, so there's our SI unit here for any of our energies, whether it be work or kinetic energy or potential energy.

Now, the term work itself, you know, we use it in the vernacular all day long, you know, I have to go to work and things of a sort, but it has a specific definition. And it's this force times this placement. It's not enough simply to apply a force.

If you apply a force to something, but it doesn't move, you have done no work. So you've got to apply a force and have it move, and have it move in that same direction, so to speak. And so in one dimension, it's a rather simplistic definition.

But once we start looking at this two-dimensionally, and you might be like, oh, Chad, you said that it's not a vector. You're right, it's not a vector. But it turns out it's a dot product.

So we're not going to go into talking about dot products or anything, but we do have to account for the fact that direction... does matter even though work itself doesn't have direction because the quantities that make it up, if you look, force is a vector and displacement is a vector. And these are both vectors, and even though the dot product of the two is gonna end up being a scalar, so direction will come into play. And so it turns out when you're dealing with two dimensions, you can kind of think of it that if the force and the displacement point in the same direction, then it really is just force times displacement. However, if they don't point in exactly the same direction, Then you've got to factor in really only the component of the force that points in the same direction as the displacement.

And the way we would accomplish that, as long as we're taking the angle with respect to the x-axis, let's say, is that f cosine theta is the component of the force in the x direction, and if the displacement's also in the x direction, we'd account for it like that. Well, in this case, all we really have to do is define theta here as the angle between the force and the displacement, and then we can write it this way every time. So work is going to equal...

Force times the cosine of the angle between force and displacement and then times the displacement. And again, if force and displacement are in the same direction, Well then theta here is zero and the cosine of zero is one and that whole cosine just goes away. Notice also that this is going to be important.

What if the force and displacement are in opposite directions? All of a sudden then theta is 180 degrees and the cosine of 180 degrees is negative one and we'll be talking about negative work. And so if a force and displacement are in the same direction work is going to be positive but if force and displacement are in opposite directions work is going to be negative.

Negative. So, and those are obviously the extremes, but anything between like 0 and 90 degrees, you're going to get a positive work, and anything between 90 and 180 degrees, you're going to get negative work. But the most common examples you'll see are when we've got force and displacement in the same direction or force and displacement in opposite directions. But you are now prepared for anything in between as well.

Alright, so this is work, and we're ready to do some calculations here. So, uh, First question, how much work is performed by a person lifting a 10.0 kilogram object at constant velocity, big buzzword there, to a height of 2.0 meters? Okay, so we've got an object here, 10.0 kilograms, and we want to lift it to a height of two meters off the ground. Okay, well we're going to set up a free body diagram here and we know that this thing has a weight, mg. So and then somebody's lifting this thing so there's some sort of lifting force that's opposing that.

I'm just going to call it F in this case. And we want to know how much work is done by that lifting force right there. That's what's going into our work equation. And so we've got work again equals force times the cosine of theta times displacement. And that's the force I need.

Well the key is we're told that they're going to be lifting it at constant velocity. And if they're lifting at constant velocity, you know that the net force in this y direction must be zero. And so we could say that sum of the forces equals zero.

It's an equilibrium according to Newton's first law. And so we'd have force minus mg equals zero. And so that lifting force must equal in magnitude the weight. Opposite direction obviously, but equal in magnitude to the weight. And you might have intuitively figured that out like, well if I want to lift that weight, then the force I lift up with has to equal that weight.

Well, it does as long as we're not lifting it with any sort of net acceleration. In this case, we're told it's at constant velocity. So yeah, that force must equal the weight, but in opposite direction. All right, so now we know what that force is going to be, and we can substitute it right into this formula.

Also, in this case, the displacement is going to be 2 meters in the upward direction. So the force is in the upward direction. They're in the same direction, and therefore the cosine of 0 is 1, and that cosine just goes away. And so here we're going to have work equals So mass times gravity, well we're told it's 10.0 kilograms times 9.8 meters per second squared, and then times the displacement of 2.0 meters. So, and we could let our calculator do some work, we probably do this in our head truth be told.

So 10 times 9.8 is 98, 98 times 2 is 196, that's 196 joules. Now in this case we're gonna be limited to two sig figs here. by that 2 meters. Now that's 3 sig figs, and obviously to make it 2 sig figs it's going to round up to 200. So personally I hate putting bars over zeros, truth be told I've done it before, but my personal favorite then is just to make this 2.0 times 10 to the second joules as a way of writing 200 with just 2 sig figs. Alright, the next question says, how much work must be done to raise the 24 kilogram mass 2.0 meters at constant velocity with and without the pulley system in the diagram.

So, if you recall, we have the same pulley system we studied in the last chapter, and it gives us a mechanical advantage so that we can raise this 24 kilogram mass here with less force than it would normally be required to raise it. So what we're gonna find, there's no such thing as a free lunch, and even though it's gonna require less force, it is not gonna require any less energy. There's no such thing as a free lunch here, so as we'll see. So let's do it without the... Without the pulley system first and figure out how much energy this is going to be so and in this case We'll figure out how much work it is to raise it two meters and again.

We know it's got a weight downward It's got a force upward that we're going to apply and then from there work is going to equal force times cosine theta times the displacement And in this case we can see that that force is going to have to equal the weight again this is being raised at constant velocity and so That mass was again 24 kilograms, gravity is 9.8 meters per second squared, and then the displacement is going to be 2 meters. We'll let a calculator do some work for us here. All right, so 24 times 9.8 times 2. We're going to get 470.4, which two sig figs all round down to 470. Joules. Okay. Now again, there's no such thing as a free lunch.

This should come out to exactly the same thing. Now if you recall though, in this case the tension on these cables gets to pull on each of these pulley wheels that's attached to the 24-kilo-year mass on both sides. We got a tension force pulling there, the tension pulling there, the tension pulling there, and there, and there, and there. And so the tension gets a chance to pull up on these pulleys that are attached to the mass in six different places.

And so with this 24 kilogram mass being essentially just about 240 newtons, over here we have to lift with 240 newtons. 24 kilograms times 9.8 is almost 240 newtons. But here we're only going to have to pull up with one-sixth of that. Well, I say pull up, we have to pull on the cable with roughly one-sixth of that force. Instead of roughly 240 newtons, it's going to be closer to 40 newtons.

39 point something, if you might recall from the last chapter. So... the force is less, that kind of seems like we're getting a free lunch, doesn't it? So, and we set up our equation, we're like work equals force cosine theta times displacement. So, oh, and by the way, I didn't mention it, but it goes without saying here, cosine theta disappeared because the force and the displacement were in the same direction, and cosine of zero was one.

So in this case, we're not applying the force directly to it, we're actually applying the force to the cable. So, and we're pulling it down, which is going to pull the mass up and we still want this mass to rise two meters. That's going to be crucial here but we're going to actually be pulling down on the cable and it turns out it wouldn't have to be perfectly down we could pull it off at an angle as long as we're putting a tension in this cable that's the key.

And so in this case that force instead of being 24 kilograms times 9.8 meters per second squared it's going to be just 1 sixth of that so in this case instead so it's gonna be 1 sixth of again 24 kilograms. times 9.8 meters per second squared and all of a sudden it's looking like well the force is only one-sixth of the force Chad and we still only want it to rise two meters so if we plug two meters into that isn't the work going to come out to just one-sixth of the value? It might look like that but again the key is this we want this mass to rise two meters but that's not the displacement we're worried about here so when we're actually doing the work the displacement is how far down we've got to pull this cable.

When we pull this cable, the whole pulley system is cinching together. And with it going around the six wheels here, it turns out that when you pull it downward, so the cables are actually only going to raise this one-sixth of that distance. And so if I want this to rise two meters, I actually have to pull this cable down a total of 12 meters.

And that's where we're not getting that free lunch. And so even though the overall force, so is one sixth of the force, the displacement we've got to generate is going to be six times greater. And so all said and done, when you calculate out that work, it's still going to come out to 470. joules rounded to two sig figs. Cool. So mechanical advantage is going to decrease the amount of force we need to apply, but it is not going to reduce the amount of work that needs to be done whatsoever.

All right, so last two problems. are going to be related. First one is a 10.0 kilogram object on a horizontal surface, coefficient of kinetic friction is given, is pulled horizontally.

So with a force of 50.0 Newtons yielding a horizontal displacement of 2.0 meters, how much work was performed by the pulling force and then how much work was performed by friction? So two questions here, work of the pulling force and work by friction. All right, so we still know that work... F cosine theta times displacement. And if we look at the pulling force, the pulling force is pulling to the right, and there's going to be a displacement to the right.

That should be positive work. So on the other hand, though, the displacement's going to be the right. Friction always opposes the motion, and so friction's going to pull back the opposite direction, and so the work for friction should be negative instead. Alright, and so from this case if we look at the pulling force, well the pulling force is in the horizontal direction And so is the displacement and so we're just going to have 50.0 Newtons cosine of 0 is 1 so that falls out and then times that 2.0 meters, and we're gonna get a work of a hundred Joules and again if I want to write that in two sig figs, I'll make that 1.0 times 10 to the second Joules All right, so that was the work done by the pulling force now.

What about the work done by friction? Well for the work done by friction We got the same setup for solving for work, but we got to figure out what that force is. So this placement is still going to be 2.0 meters.

Well, in this case, if we set up a free body diagram, so let's start with the force of friction actually. Let's just go there. So this force is going to be the force of friction, and we know that the force of friction is kinetic friction mu times the normal force and then times cosine theta times d. Alright, so we know coefficient of kinetic friction is probably 0.2, the normal force we haven't figured out yet.

But we know that the weight points down, and the normal force is going to counteract that. And we set up some of the forces there. We can see that some of the forces in the y direction add up to 0. And so in this case, the normal force minus mg equals 0. And that normal force equals mg, pretty common. for the normal force to equal the weight in magnitude.

Alright, so we'll substitute that in. Weight, there's our coefficient of kinetic friction. So the mass in this case was given as 10 kilograms.

Gravity is 9.8 meters per second squared. So, and then cosine theta here is going to be cosine of 180 degrees. Again, the displacement's to the right as I've drawn it. Whereas friction is force pointed in the exact opposite direction.

So and then finally times the displacement of 2 meters. And we'll let our calculator do some work for us here. So in this case times 0.2 times 10 times 9.8 times cosine of 180 is just going to make it negative and then times 2. And we're going to get negative 39.2.

Joules. In this case actually if I round that to two sig figs I'll have to round that down to 39 joules. Cool and again it's negative just because the force the displacement are in opposite directions not in the same direction.

So we'll find out that when work is positive it is kind of adding energy to whatever your object is when force when work is negative it's taking away energy from it as we'll see in the next section. Alright, if we take a look at this one and see what's different now. So instead of pulling perfectly horizontal with that 50 newton force, it's going to be pulled up at an angle of 30 degrees.

That's going to make this a little harder as we'll see. So let's read the whole thing though. A 10.0 kilogram object on a horizontal surface, mu k equals 0.2, is pulled with a force of 50.0 newtons at an angle of 30 degrees above the horizontal, yielding a horizontal displacement of 2 meters.

So displacement's still horizontal though, 2 meters. How much work was performed by the pulling force and how much work was performed by friction. Okay, so we'll start with the work done by the pulling force here.

And again, work is equal to F cosine theta times displacement. And once again, so our force here of 50 newtons, but we don't have cosine of zero now. It's going to be cosine of 30 degrees and then times the displacement of 2.0 meters.

Cool, now one way to look at this, so one we just plugged it into the formula, so what you might think of is that this F cosine theta is really just the component of the force that's in the same direction as the displacement. So in this case the displacement's in the x direction, so I need the component of the 50 Newtons that's also in that same direction. Well if you look it would just be 50 Newtons times cosine 30, that's kind of where that's coming from, so at least one way to look at it.

So that's my personal way of just conceptually understanding it. So but from here it's plug and chug. Let our calculator do the work for us. So 50 times the cosine of 30 times 2. It's gonna get us 86.6 newtons which rounding to 2 sig figs. Ahah, newtons.

Joules, we're doing energy here. So 87 joules rounded to 2 sig figs. Now notice this is smaller than the hundred joules we had right back here done by the pulling force. So in this case, because not all 50 newtons is pointed in the direction of the displacement like it was back here, now only a smaller component of it. Same displacement of 2 meters, but a lower force in the direction of that displacement, leading to less work done by that pulling force.

Now we've also got to get the work done by friction here. Once again, this is the force of friction. So in that force of fric...

Friction is equal to mu kinetic times your normal force times cosine theta and times your displacement. And once again, we know the coefficient of kinetic friction is 0.2. We know the angle theta is going to be 180 degrees, that displacement is horizontally to the right.

Friction always opposes the motion, it's to the left. And the displacement is still 2 meters, but we still need that normal force. And so we'll set up the free body diagram. We still have the object's weight. Pointing down we still have a normal force Pointing up and you might be like it worked out the same way Chad so normal force is still gonna equal the weight Except we got to count for one other thing right so that 50 Newton's not only had an X component that in this case was 50 times cosine 30, but it also has a y component that's equal to 50 newtons times sine of 30 degrees.

And so now when we go to set up the sum of the forces in the y direction equaling zero, there are three parts to that, not just two. And so in this case, we're going to have that normal force pointing up positive. We're also going to have this 50 sine 30 pointing up, so that's positive.

And then minus mg equaling zero. And so we solve for that normal force. It's not simply going to equal mg. It turns out that normal force is going to equal mg minus 50 sine 30. Well, sine of 30 conveniently is...... is one half and 50 times a half is 25. So it's gonna be mg minus 25 newtons on the other side there.

If we look at this, so the mass again is 10 kilograms, gravity is 9.8 meters per second squared, 10 times 9.8 is 9.8. 98, and 98 minus 25 is going to get us a normal force of just 73 newtons. And so if we look, that's less than what we had back over here. Here, the normal force was the weight of 98 newtons. And so in this case, the normal force was perfectly having to balance all the weight.

Whatever the weight pulled down with, the normal force had to balance that back. Maybe you look at it according to Newton's third law, if you will. For every force, there's an equal and opposite force.

Whereas here, the normal force doesn't have to balance all of the mg by itself. itself, so to speak, because it's getting a little help from this 25 Newtons of the component of the force that's pulling upward. And so the normal force only has to balance the remaining 73 Newtons instead.

So there's less of a contact force and there's going to be less friction as a result as well. Alright, so that coefficient of kinetic friction is still 0.2. So our normal force is now just 73 newtons. Times, again, cosine of 180 degrees.

They're in opposite directions. This will make work negative. And times that displacement of 2.0 meters. And we'll let our calculator do some more work. All right, so 0.2 times 73 times cosine of 180 and times 2. It's going to get us negative 29.2 joules, which rounded to 2 sig figs will round down to negative 29 joules.

Let me make sure I got that right. Yep, negative 29 joules. Cool, and so we see not only did the pulling force do less work here than back on this first problem, but we see friction did less work as well because there was less normal force, less contact force, and therefore less of a frictional force there as well.

Cool, if you have found this lesson helpful, then consider giving it a like. Happy studying.

Transcript for:Understanding Work and Energy in Physics

Transcript for:
Understanding Work and Energy in Physics