Sections | |||
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Intro to Conservation of Energy | 52 mins | 0 completed | Learn |

Energy with Non-Conservative Forces | 45 mins | 0 completed | Learn |

Escape Velocity | 12 mins | 0 completed | Learn |

Conservative Forces & Inclined Planes | 50 mins | 0 completed | Learn |

Motion Along Curved Paths | 107 mins | 0 completed | Learn |

Energy in Connected Objects (Systems) | 31 mins | 0 completed | Learn |

Solving Projectile Motion Using Energy | 33 mins | 0 completed | Learn |

Springs & Elastic Potential Energy | 63 mins | 0 completed | Learn |

Force & Potential Energy | 22 mins | 0 completed | Learn |

Concept #1: Conservative Forces are Path Independent

**Transcript**

Hey guys, so in this video I want to introduce two properties of conservative forces that you may need to know conceptually but that you're definitely going to have to know in terms of problem solving let's check it out. So first of all remember that there are two types of force, forces can be conservative and non-conservative the two types of conservative forces that we have are spring and weight or the force of gravity, now conservative forces have 2 key properties and these properties are also actually equivalent they're the same thing just looked at in different ways or one leads to the other, anyway the first one is that the work done by conservative forces does not depend on the path taken this is very important conceptually and for problem solving therefore we say that though the work done by conservative forces is path independent you might see that your professor talk about that or you might see that in the test so the work done by conservative forces does not depend the path its path independence, OK? That's why I've put in here that conservative forces are path independent, weight is a conservative forces we just talked about that so it's work depends only on the vertical displacements and not on the path taken so let me show an example real quick, here let's say that all of these block are exactly the same mass exactly the same thing and they start from the same heights from rest but they're going to fall in different ways they're going to take different paths to go from this point to this point the first block is going to fall this way, second block is going to fall this way down an inclined plane that has an angle of theta 1, this is a different inclined plane that has a different angle theta 2 and then this box going to go this way and get to the bottom, this one is going around a curved path like this and this one here is going to start here and I'm just going to draw a crazy path let's say it somehow moves this way and then it gets to the bottom, alright? In all of these cases because of all of these objects dropped the same exact height there will all have the same final speed, OK? And that's because the work done by the weight force which is what causes the object to speed up is path independent in fact I can show you using the conservation of energy equation, kinetic initial+ potential initial + work non-conservative= kinetic final + potential final, if you start from rest this is 0 you have a height so this is not 0, the object is falling at the absence of air resistance so there is no force no work done by U and no work done by friction so that's 0, kinetic energy at the end we have it because right before you hit the ground there is a final velocity there and the potential energy at the bottom is 0 so if we simplify this we get MGH=1/2MV squared and the masses cancel so you see that the final velocity = the square root 2GH final which is something we're going to see quite a bit this equation here so it's a very usual answer for a lot of problems where something's falling notice it depends on the height and not on the actual length of this arrow right so this this squiggly line here is longer than this line here the straight line but it doesn't matter all that matters is the heights, OK? So that's a really important point and we're going to see that a lot in problem solving.

The second one is that the net work done by conservative forces is 0 in a closed path sometimes this is also referred to as a closed loop, OK? The total work done by conservative forces in a closed path is 0 and that's probably the reason why they're called conservative because if you do something and then you move the object but bring it back to the beginning you didn't actually change the total makeup of the energy of the system, weight is a conservative force so I'll give you an example here and how the net work will be 0 if I started A and then I move this box all the way back to A so imagine this is you moving something sideways and then you pick it off up from the floor lift it to this height then you move it to the left and then you drop it back to the floor and it's back at the same point A, OK? When you move this thing sideways the work done by gravity or by the weight force is 0 because you're moving perpendicular to the weight, right? So, you're moving this way the weight force is this way because this makes an angle of 90 degrees the work is 0 when you move something up remember the work done by MG is the negative of the change in potential energy in other words the work done by MG is -MG delta H which means that if you because of this negative here if you going up your Delta H is positive, positive with a negative means that if you're going up your work by gravity is negative, right? Also, because another way to think about this if remember if you're going up gravity is pulling down against you, OK? Anyway, so when we go here the work done by MG is negative, when we go this way the work done by MG is also 0, again because it's perpendicular another way to look into it that the change in height is 0 and then for this piece here gravity does positive work because it works in the direction of motion, alright? So if I want to find the net work, the net work would be the work done by or the work that took place in each one of these intervals so I'm going to call this 1 2 3 4, there's no work at 1 and 3 and the work done at path or leg number 2 over here would be -MG delta H and for W 4 which is this part here it's positive MG delta and you can see hopefully that this is 0, the force cancel so the idea is whatever energy you have here if you do some crazy thing you come back here as long as the force that's moving this thing around is the weight force then you are a conservative force, OK? I'm sorry the net work would be 0 because it's a conserved force, cool. Friction however is not a conservative force so the net work will not be 0 because the net work is always 0 for that conservative forces so for example if you move objects from point A to B and back to A the net work will be work 1 plus work 2, 1 is when you move it that way 2 is when you move it back, the work done by friction remember is -FD and it's always opposite to motion, right? So work net let's move it over here, work net will be 1-F1D1+ -F2D2, these two negatives combine and the frictions will be the same friction as mu normal or mu MG and just to show that the same floor so the same mu, same mass same gravity so the frictions are the same, the distances are the same as well one is to the right and one is to left but its distance it doesn't matter the direction so what I have here is you actually have -FD-FD so -2FD instead of cancelling they combine and they're both negative this is clearly not 0 so in the first situation if you moved you would be back at the same energy no energy is wasted in this process but here you are wasting energy moving this way and you're wasting energy moving this way, alright? That's the idea, having said that I want to show you some practical applications of the first idea which is that the final velocity when we're calculating final velocity really anything with energy, what matters is not the path taken but the amount of height or the amount of vertical displacement so let's get started here.

So, I have a mass of 2 kilograms a block of 2 kilograms is released from the top of a smooth inclined plane, smooth means that friction will be 0, that is 10 meters long and makes 37 with the horizontal let me draw this, so 2-kilogram block, the angle is 37, the plane has a length of 10 meters and I want to know what is the speed of this thing at the bottom here? So, what is the final speed? The initial speed will be 0 because it's released from rest, let me put that in here and we're going to use the conservation of energy because I have a problem where an object is changing speeds while also changing heights so the energy question is a good one for that, KI+UI+work non-conservative=Kf+Uf, there's no kinetic energy at the beginning because the speed is 0, there is potential energy because you have a height the work done by non-conservative forces is the work done by you plus the work done by friction you didn't do any work you released this block and you're just watching and this is a smooth plane so there is no friction so this doesn't exist, there is kinetic energy at the end and because we have a speed and we're looking for that speed and the potential which is 0 because when you're at the bottom you are at the height of 0 so this is MGH initial 1/2MV final squared, I can cancel the masses and the final velocity will be the square root of 2GH which you should be familiar with, right? This is going to show up a lot I mentioned, the thing here is we don't have the heights, right? And by the way this shows us real quick that it depends only on the height, right? But the problem is we don't really have the height we have the length of the plane and we have the angle well in a triangle whenever you know two things from a triangle you can find the other two what I mean by two things is you have the let's say the angles here you have adjacent opposite hypotenuse as long as you know two of these four you can find the others by using either the Pythagoras theorem or SOH CAH TOA but I'm just going to remind you of this shortcut here if you know the length here and you know the angle the height is simply L sine of theta, OK? So, if you didn't already know this one this is a good one to remember because it's going to show up a lot in these plane problems they put this a lot to make sure that you know how to convert from when you're given L and theta instead of H which is what you need here, OK? So, let's put it in the initial height is the length of this thing it's all the way at the top so it's 10 meters sine of 37 so the initial height is 6, OK? So, let's plug that in here and then we're done, square root of 2 gravity is 10 and the initial height is 6 so the answer here will be 11 meters per second, OK? That's it for this one very straightforward, here I have something that rather a practice problem for you guys, it's a little different instead of a block coming down it's going up the hill I'm going to draw this for you, just because it's a little bit sometimes confusing what exactly this means but a block reaches the bottom of an inclined plane with 20, right? So, when I said a block reaches the bottom of an inclined plane with 20 meters per second it could mean that you're reaching the bottom going up or it could be that you're reaching the bottom on your way that way, right? In this case I'm asking how far up the plane will block slide which implies the block got to the bottom and now it's going to go up, OK? So, the block's here and I want to know how far up the plane does it go? and it gets here with an initial velocity if 20 I won't put anything else I just wanted to help you a little bit here let's give that a shot.

Practice: A 3 kg block is initially moving on a flat surface when it reaches the bottom of an inclined plane with 20 m/s. If the plane is smooth and makes 53° with the horizontal, how far up the plane will the block slide?

Example #1: Energy Problems in Inclines

**Transcript**

Alright guys we're now going to solve a few more inclined plane problems using energy, let's check it out so here I have a 2-kilogram block at the bottom of an incline and we're going to push it up with a force of 20 that is directed up the plane, OK? So, the force is 20 up the plane so looks like that, this makes an angle of 37 it's a smooth plane so the friction is 0 and I want to know after I've done this for 5 meters after I have to push this up the plane in other words after I pushed it this way so that the object is over here somewhere so after I've done this for a distance of 5 meters what kind of final velocity does it have? We're going to use the conservation of energy equation because my speeds changing my height's changing and there's work done by you so all these 3 guys are in the energy equation so we can solve this very straightforward like this, the initial kinetic energy is 0 because this starts from rest it says that right there it's originally at rest, the potential energy is 0 as well because we're at the bottom, the work by non-conservative forces is the work done by you plus the work done by friction but there is no work done with friction however there is work done by you, OK? The kinetic final is 1/2MV final squared, there is kinetic energy at the end, I'm looking for that velocity and there is potential energy so I'm going to replace that with MGH final, OK? Work done by you, well the work done by any force is F D cosine of theta it says here that you push it up the plane so you push it this way which is the direction that the object is going to move so it moves in that direction the angle between these two is 0 degrees because they're perpendicular.....Parallel to each other the cosine of 0 is 1 so that the work done by F becomes just simply F D, OK? let me go ahead and move that FD in here so F D let me plug in numbers, the force is 20 and the distance is 5, on the other side we have 1/2 the block is 2, the final velocity is what we're looking for, the block is 2 gravity is 10 and the final height is..... Notice that we don't have a final height once again so we have to use our little equation here, if I know the length or the distance same thing and I know the angle theta I can find the height by using in this case we're calling it D so D sine of theta and 3 sine of 37....I'm sorry 5 sine of 37 which is 3, Cool? So, height final is 3 meters that's what goes there I have everything except V final so I can now solve this, OK? This is 100, these 2 cancel V final squared + 60 you move this over I have V= the square root of 100 - 60 the square root of 40 and that gives us 6.32 meters per second. That's it for this, let's try the next one.

Example #2: Energy Problems in Inclines

**Transcript**

Alright, so now we have a block that's sliding down an inclined plane but the plane now has friction so we have the height's the speed's changing and then there's the work done by friction all these three elements are part of the energy equation so we're going to use the energy equation to solve this, OK?

I have a 3 kilogram object here I am going to release it from rest, it is placed on an inclined plane and you can assume that it's placed with an initial velocity at 0 and it's going to slide down, 8 meters above the ground means this distance here the height is 8 meters and the angle is 53, in previous problems I gave you the length and you had find the height, now I'm giving you the height, the coefficient of friction right there between the block and the plane is 0.5, so as always let's start with the energy question, KI+UI+work non-conservative = kinetic final + potential final, no kinetic energy at the beginning because there's no speed the potential exists because you have a height, the work done by non-conservative forces is the work done by you plus the work and by friction in this case you're just watching this block go down but there is work done by friction, I also have a let me expand this out here this is MGH initial and then on this side here there is kinetic energy because we get to the bottom with some final velocity but there is no potential energy because when you're at the bottom your height is 0, alright? And we're looking for V final so we have to know all these other numbers here, just to plug in something real quick MGHi, the work done by friction is negative FD and 1/2MV final squared, alright? Let's start plugging in numbers, I got a 3, a 10 the initial height is an 8 that was given to us. What about friction? We have to calculate friction so I want to remind you that friction is always mu normal but on an inclined plane we have on an inclined plane if you remember normal is perpendicular to the surface so normal does not equal MG normal equals MGy, OK? So, normal equals MGy and MGy I hope you remember as long as the angle is down here which is the correct place for the angle the more usual place for the angle...MGy would be MG cosine of theta, remember X goes with.... IÕm sorry X goes with cosine but on the inclined plane it's backwards, OK? So, this means that friction will be mu normal and normal is all this so Mu MG cosine of theta and if we plug this in mu is 0.5 mass is 3 gravity we're going to round it to 10 and the cosine of 53 if you do all this I have it here you get a 9, OK? 9 Newtons so we can plug this in here it's a negative 9 because of this negative here the distance is.... Now what is the distance here? Well usually are in previous problems I give you the distance and you get the height now I'm giving you the height which was good here because we needed the height but it's bad here because we now need the distance and in a lot of problems if I give the height you still have to find the distance or if I give you the distance you still have to find the height, we can use that little equation and D sine H equals D sine of theta but in this case I don't want H I want D, so D= H/ sine of theta let's start with numbers there already, 8/sine of 53 And that's actually 10 meters so 10 goes right here, 1/2 (3) V final squared, notice how we have all numbers now it's just a matter of doing the algebra here so this is 240-90=1.5V final squared and if you calculate this you get that V final is 10 meters per second, alright? That's it for this one let's go into the next one.

Example #3: Energy Problems in Inclines

**Transcript**

Hi guys so in this problem we're trying to calculate how fast do I have to throw or launch this block from the bottom of the plane so that it get all the way to the top of the plane, OK? So, let me draw that.

I have a 3-kilogram block down here and we're going to launch it up in such a way that it gets to the top all the way over here so what we're looking for is the launch speed or the initial velocity and the angle here is 53 and there is a coefficient of friction between the block and the plane which is 0.4, OK? Let's jump right into the energy equation and see what happen, K initial + U initial work non-conservatives = K Final + U final. Is there a kinetic energy at the beginning? There is kinetic energy at the beginning because you launch this thing with a certain initial speed so I'm going to write that in here, there is no potential energy because in the beginning your height is 0, the work done by non-conservative forces is the work done by you plus the work done by friction now in this case there is no work done by you but there is work done by friction, I already discussed this remember that when you throw something you do work in getting it from 0 to whatever velocity but once it leaves your hand which is really when the problem starts you don't do any work, right? So, you did work to get this thing moving but right after it started moving to here you didn't do any work and by the way I forgot to mention this is 5 meters long I just saw it there, OK? So, there is no work done by you but there is work done by friction and the work done by friction is -FD but I'll get back to that in a little bit. Is there a kinetic energy at the end? The answer is no and this is kind of tricky but I want to kind of show you why this has to be 0 with numbers let's say you toss this I'm just going to make up some numbers here, let's say you toss this with a 100 meters per second from the bottom and it gets to the top with 10 meters per second, well we want the minimum speed that you need at the bottom, if you got to the top with 10 it's because you could actually have thrown this a little bit slower so let's say maybe if you throw it with 90 or maybe not 90, if you throw with 95 and again I'm just making up numbers so it doesn't matter it gets to the top with 2 meters per second which means you could have tossed a little bit slower and maybe 94 is the number that gives you a 0, right? That's what you want this is the slowest You can go if you do 93 it just means you're never going to get to the top, OK? So, these are just some made up examples made up numbers but anyway I hope you get it that this has to be 0, I do have potential energy because we have a height, OK? And out of all of this what we're looking for is V initial right here so let's start plugging numbers as much as we can, 1/2 the mass 3 the initial velocity is what I'm looking for, the work done by friction I'm going to calculate this here on the side so remember friction is mu normal and on an inclined plane normal is up this way, which will then equal to MGy so it's mu and instead of normal I'm going to write MGy but remember further that MGy in an inclined plane is MG cosine of theta, Y usually goes with sine at least the way I do it and here it's going to go with cosine because remember that's how it works with incline planes you can think of it as it being backwards, cool? So that's my friction I can put that in here let's do that real quick and then we'll plug in a bunch of numbers so negative friction, friction is mu MG cosine of theta D If we plug in all of these numbers I have it here if we plug in all these numbers we get a 7.2 but let me......Actually this is -7.22 that's this piece here and then the distance is 5, right? So, I calculated this earlier and I got that the work done by friction is -7.22 x 5 this is -7.22 And this is a 5 right there, OK? By parts, so the mass is 3 gravity is 10 and the final height is we don't have the final height but we can use our equation to get that so distance, angle and height is distance sine of theta and in this case the distance is 5 meters sine of 53 which is 4 meters, OK? Let me just make sure that's what I had here, 4 meters that's right so that's the number that's going to go here I'm a little bit messy here with some stuff at the bottom so I'm going to move here I have 1.5 V initial squared equals this whole number here is 120 and it's going to go to the right side so you can become positive 120 plus.... I'm sorry actually this is 120 and then this other number will be 36.1, OK? And when you calculate V initial you're going to get the square root of some number and the final answer I'm going to skip here to the answer is going to be 10.2 meters per second, OK? So, if you toss with 10.2 it will get to the top with exactly 0 so we just kind of barely stop at the top and that's your answer, alright? That's it for this one.

Practice: When a 4-kg block is released from rest from the top of an inclined plane, it reaches the bottom with 4 m/s. The incline is 5 m long and makes 37° with the horizontal. Calculate the magnitude of the frictional force acting on the block.

Practice: When a block of unknown mass is released from the top of an inclined plane of length L meters, it slides down. The incline makes an angle of Θ degrees with the horizontal, and the coefficient of kinetic friction between the block and the plane is µ. Derive an expression for the speed of the block at the bottom of the plane.

Practice: A block of unknown mass is released from a distance D1 from the bottom of an inclined plane, then slides on a horizontal surface, and up a second inclined plane, as shown. Both planes make an angle of Θ degrees with the horizontal. The horizontal surface is smooth, but the coefficient of friction between the block and the two inclines is µ. Derive an expression for the maximum distance D2 that the block will reach on the second incline.

0 of 8 completed

Concept #1: Conservative Forces are Path Independent

Practice #1: Conservation of Energy in Inclines

Example #1: Energy Problems in Inclines

Example #2: Energy Problems in Inclines

Example #3: Energy Problems in Inclines

Practice #2: Energy Problems in Inclines

Practice #3: Energy Problems in Inclines

Practice #4: Energy Problems in Inclines

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