hello class and welcome to unit 3 energy now energy is a really exciting unit we cover all sorts of things in it and it really does a great job of explaining so many things with really really simple results ok and so I think you're going to enjoy this unit because it's going to let us manipulate some really really neat relationships with some very simple equations ok it does a similar thing where with forces we were able to summarize some of our kinematics and and connect them to the real world now with energy we can summarize some of the things that we had in forces and make them easier to use all right we're starting off with five point one work and you know what work is well you're doing your homework or you're doing whatever kind of work you might think of but in physics work means a very specific thing so our mechanical work is very specifically a force applied over a distance so if I push something with five Newton's of force over ten meters that's me doing work on it okay we have an equation for this it is w1 work is equal to F Delta D cosine theta so we have our force our displacement and our angle theta and we have units for this of joules which is J joules and jewels are named after a famous physicist of course so we have force Delta D and we have our cosine theta now theta hopefully I think we've worked with this angle before but in this case theta means so theta is the angle between the force F and the movement Delta D so I can be applying a force and it might not be the same direction as the movement itself okay and you can see the picture to the right here we have a man pulling his luggage and you can see he's applying a force up on this angle whereas the the luggage itself is just moving horizontally like this so you can see that there's some angle theta between those two okay now we do have a special case that's this next line here the special case is when theta is equal to zero when theta is equal to zero that means that the force and the displacement are in the same direction which means that there's no angle in between them okay so in this case our equation simplifies w is just equal to F Delta D because the cosine of zero is equal to one so that cosine theta just disappears okay so we're going to be doing some some examples here and I'm promising you that work and energy they're going to be really really useful and interesting for now though we're just going to calculate them and you have to trust me that it will be useful so first off how much mechanical work does a person do on a shopping cart if they apply a force of 25 Newtons in the forward direction and displace the cart three and a half meters in the same direction so we have work is equal to F Delta D remember this is our special case where theta is equal to zero so I don't need to do cosine theta so F Delta D we have our force is 25 Newtons and Delta D is three and a half meters and this gives me a value of 88 joules perfect now I'll just point out I could have left in the cosine theta part of the equation which would have been cosine of zero degrees and that cosine zero just becomes one so we get the same answer okay if you prefer to just always use the full equation F Delta D cos theta that's fine I usually do that myself okay the next problem here says a curler a force of 15 Newtons on a curling stone and accelerates it from rest to a speed of 8 meters per second in three and a half seconds assuming that the ice surface is level and frictionless how much mechanical work does the curler do on the stone so we have our same equation again they're in the same direction there's no angle so we have W because F Delta D okay that's our work now we have the force 15 Newtons but Delta D we're gonna have to work for right we we don't have Delta D but we have let's see VI was equal to zero our final velocity is equal to eight meters per second and our delta T is equal to three and a half seconds so we can use an equation to relate all these that equation is Delta D is equal to V I plus VF over two times delta T okay so we'll just plug in our values here we have zero plus 8 over 2 times 3.5 and this gives us a value of 4 times 3.5 which is equal to 14 point zero meters so there's our Delta D and now we can go back and and do W equals F Delta D which is 15 times 14 that's our Delta D and it gives us two point one zero times 10 to the 2 Newtons or sorry not Newtons attend to the two jewels there we go that's our work so those are two examples where there is no angle between them this next example says work done when force and displacement are in different directions so we want to calculate the mechanical work done by a custodian on a vacuum a vacuum cleaner if the custodian exerts an applied force of 50 Newton's on the vacuum hose and the hose makes a 30 degree angle with the floor there's a picture to the right here to help you see what's what's going on the vacuum cleaner moves three meters to the right on a level flat surface so you can see our displacement is to the right our force is up at an angle of 30 degrees so we have work is equal to F Delta D cosine theta which is equal to 50 that's our force Delta D is 3 + cosine of 30 degrees we put in all those numbers and we get a final value of one point three zero time is 10 to the Joule 10 jul 10 to the 2 joules sorry and notice I'm using scientific notation for my answers I could have just said 130 joules as well but this is this is a good way of writing our answers ok second page we've got a few more special cases right remember special cases are where something strange happens or something different so the first one here says Ron beer where's his backpack as he walks forward in a straight hallway there's a picture of him there he walks at a constant velocity of 0.8 m/s for a distance of 12 meters how much mechanical worked as Ron beer do on his backpack well notice that the backpack has a force upwards and a force downwards okay there's no horizontal force because he's walking at a constant speed so he's not accelerating but his movement is horizontal so the forces are at 90 degrees - his movement okay so here I could say that this angle here is 90 degrees so he's applying this force upwards at an angle of 90 degrees so let's just put that in our equation and see what happens I've got work is equal to F Delta D cosine theta and so we have our applied force well we don't even know what the applied force is it's gonna turn out we don't need to but let's just say here it's the applied force times Delta D which is a distance of 12 meters times the cosine of 90 degrees that was our angle here okay and here's where things fall apart because you'll know the cosine of 90 degrees that's equal to zero so this whole thing that means that he's done zero work zero joules of work and that is true that's what happens that the the work is only done by force in the same direction as his movement okay needs to be in the same direction so here his force is straight up his movement is horizontal they're perpendicular to each other which means none of the four cities doing is in the direction of his movement so he's doing no work by walking this way okay so we'll look at another example here where something different happens it says how much mechanical work is done on a stationary car if a student pushing with a 300 Newton force fails to displace the car so the student is pushing with 300 Newton's so we've got work is equal to F Delta D cosine theta well in this case he's you know let's say that theta is zero degrees but the problem is there's no displacement the car doesn't move he's pushing but the car doesn't move so here Delta D equals zero therefore our work is equal to zero there's no there's no work done so again both of these are examples where no work actually gets done and you see I've sort of done the equation so you can see it but as soon as you see a problem where it says the force is perpendicular to the movement like with Ron beer you know that no work is gonna be done same with if the if the object doesn't move at all you know that no work is done and you don't even have to do the equations okay this last special case here it's as a shopper pushes a shopping cart on a horizontal surface with the horizontal applied force of 41 Newtons for 11 meters the cart experience is a force of friction of 35 Newtons and we want to calculate the total mechanical work done on the shopping cart now this I guess isn't exactly a special case it's just a bit different because now we have two forces and we want to find the total amount of work so we can look at this two different ways either we could just add find the net force on this object we could do the net force on the object times its Delta D cos theta that would give us a result I'm gonna look at it a bit differently first I'm going to look at the work done by the shopper so I'm gonna say the shopper here let's find out how much work the shopper is doing so her work is F Delta D cosine theta which is equal to so the force is 41 Newtons that she's applying times 11 meters cosine and let's say that she's pushing it straight forward so that our angle is zero degrees here so this gives us that the shopper is doing this much work 41 times 11 so we get 451 joules okay 451 joules and that's how much work the shop is doing that's great now we can look at the the cart the friction now how much work does friction do because friction is the other the other force on this thing so friction likewise is doing work E is equal to F Delta D cosine theta okay and we'll plug in our numbers here so it says it's doing 35 Newton's of force over the same distance 11 cosine and now my angle well if I draw a little Freebody diagram to the left here this was my applied force here's my friction which we can say it's at a hundred and eighty degrees to the movement so cosine 180 degrees and you might know that cosine 180 gives us negative one so our our answer ends up being negative so we get negative 385 joules there we go and so if I want my total well I can do the work from my shopper plus the work from my friction this gives me 451 joules - 385 joules and that gives me a total of 66 joules done so that was my total amount of work so I could call that W net my network excellent so you can see that we solved it this way as I pointed out we could also have just found the net force first and multiply that by the distance it would have gotten us the same answer here okay great so there's one more topic and that is graphing work done so when we graph the work being done here well there's three pictures this first picture shows a constant amount of force being applied over a distance so you see we've got force on the y axis distance on the on the x axis and you can see that over this whole duration we've got a constant amount of force up here being applied and if we want to find our work we'll work is F Delta D or F Delta D cos theta but let's say that in these problems the angle is all zero degrees so W is equal to F Delta D well you can see we have F here Delta D here and so to find work it's just the area of this box so in this case we can say that the work is equal to the area under the curve so you can take the area of that ranked rectangle so we could say base times height something like this work is equal to base times height okay and we have the second picture the same thing but we have now positive work just like in the first one and negative work that's where our force is in the opposite direction of movement and you can see it works the same way we have negative force times our displacement and it'll give us a negative amount of work done so it's the same story here so we can say when F is negative the area is negative great now we've got one last situation here which is where the force is changing and this is kind of cool because we don't deal with changing force or changing acceleration much in this class but this picture shows us that we can actually do it fairly easily the same way where if the force let's say is increasing from this point zero over the amount of distance until we have a much larger force by the end well still we just want to take the area under the curve that's going to be our work this is our work here and to get the area well it's a triangle so we can just either take the so we can say work is equal to 1/2 times base times height here because it's a triangle but another way of looking at that is we can find our average force right and then we can just multiply that by the whole displacement you can see the area of that rectangle is the same value as the area of that triangle so we could also say W is equal to the average force F average Delta D and both of these are true and the second one is actually very useful that even if I had a graph of work being done they looked crazy very squiggly like anything like this doesn't matter sorry if this is my force being done over some distance well I can still find the work it's going to be the area underneath the curve here and there's positive and negative sections and another way to do that is I could find the average force over that whole duration and it's still going to be F average Delta D it doesn't matter what the force is doing over that whole duration our work is still going to be that product okay that's the whole lesson try out the homework I hope that actually applying it will will make a lot more sense but this idea of work we're going to be using a lot so get nice and comfortable with it and enjoy thank you