Ch 07: Work & EnergyWorksheetSee all chapters
All Chapters
Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Concept #1: Work Done by Gravity

Transcript

Hey guys, so now we're going to talk about the work done by gravity as things are going up and down how much work does gravity do, I think this is pretty straightforward but I want to give you a clean explanation of it so that if you see this in a test you can just knock it out of the park, alright? Let's check it out so the work done by gravity, remember positive work is when a force is helping motions to make you move faster something like that, negative work is when it's hurting motion friction is always pulling against you so it's hurting you motion does negative work and forces that don't affect motion at all do no work or they do zero work, OK? Here's our work equation how we calculate work, now when an object is falling gravity does positive or negative work on it? What do you think based on this definition here helps motion, hurts motion do you think when an object is falling gravity does positive or negative work? Gravity does positive work on it, right? Because it's helping motion its what causes falling in the first place so if you have a mass and it's falling, MG pulls it down so gravity is helping motion they go in the same direction that's what you get, now if an object is going up gravity is pulling against it and gravity does negative work on it it's that simple, OK? Let's talk about one more thing, what about if you're lifting something, well the force to lift something is MG this is a little bit counterintuitive and I'll show you why this might be confusing, so if you're lifting something first of all every object near the earth is going to be pulled down by an MG, OK? To lift it your force to lift is MG, so at first you must look at this and say well if it's 100 down and I pull it with 100 up wouldn't it cancel? And it does you're right it cancels if this is a 100 then your force needs to be a 100.00001 but that's kind of silly so it has to be slightly greater than MG so we just basically set it equals MG, OK? But you understand it needs to be a little more but mathematically kind of rounds to MG so whenever you're lifting something the force the bare minimum force you need to lift it is MG, OK? Now in physics, lifting this is a very specific thing for physics lifting in physics if that's all you get implies that you're taking it from an initial velocity of 0 to a final velocity of 0 so there's a book here I'm going to lift it, not moving.... not moving, right? So, it's either from initial velocity of 0 to a final velocity of 0 and another thing you might hear is that it's you're lifting something with a constant speed, right? So that's the same idea here and the reason why you'll be at constant speed is again if you get the object moving and then these are the same thing so the object accelerates a little bit to start moving but then moves at a constant speed, OK? So, lifting or lifting with a constant speed this could be another one means that the force you need to do that is MG the idea is that you need to barely overcome the force of weight that's pull you down, so what is the work to lift something, right? Well if the force to lift is MG then the work to lift is the force to lift which is MG times the distance, now let's say you're lifting something a height of H, right? So, the distance would be H x the cosine of theta. Now if I pull something up and as a result the object moves up the angle between these two is 0 so this is going to be 0 right here and you know the cosine of 0 is 1 so I'm going to say the work to lift is MGH, now instead of calling this MGH I'm actually going to call this MG delta H because MGH means that implies that you're starting from the ground or maybe you're starting from the ground maybe you're starting from 2 to 3, so your delta H from 2 to 3 is 1 so MG delta H. I hope that you remember MGH, MGH was your potential energy gravitational potential energy is MGH, so MG Delta H is basically your change in gravitational potential energy, OK? change in gravitational potential energy, so long story short work lift is this that's how you're going to calculate it but sort of conceptually it's the change in potential energy and this should make sense if you have a box on the floor it has no potential energy when you lift it gains a 100 Joules of potential energy and that's because you did a 100 Joules of work so you pumped 100 Joules of potential energy into this thing, if you lift the you're doing positive work if you bring it down you're doing negative work because at that point if you move something down you're stealing potential energy from the object, right? You can think about it that way. Gravity does negative work if you lift something because as you're going up gravity is pulling you down so it's going opposite to the displacement so the reason why that's going to happen is because you're going to have MG that's the force of gravity so the work done by MG is MG if you go up that way height but gravity is pulling you down, gravity is pulling you down which means the angle here will be 180, Ok? so this will be the cosine of 180. And this is going to be -1 so I can say that it's -MG Delta H so instead of being the change in potential energy it's going to be the negative change in potential energy which means instead of being (final-initial) it's (initial-final), OK? The important one to remember is this one and this is more of a conceptual one, so notice how these things are completely opposite, OK? So, you can just remember that the work to lift is MG delta H and the work done by gravity as you're moving something up and down is negative MG delta H, OK? If you're moving up then your delta H is positive and if you're moving down then your delta is negative, right? (Final-initial), OK? So, this is pretty straightforward we're going to some examples here and I think you guys are going to get it, alright? That's it MG delta H which is should make sense because it's the gravitational potential energy, you lift something you give it potential energy and that's coming from the work that you did, OK? So, you lift a 3-kilogram box straight up from the floor and place it on a shelf 2 meters tall shelf above the floor so that's let's say there's a 3 kilograms book here or box rather it's on the floor and let's say over here there's a shelf, alright? You're going to move this guy and put it over here. I want to know how much work did you do? And this is very straightforward, OK? The work done by you is the same thing as the work done to lift and the work done to lift is MG delta H, OK? If this is 2 meters high I can see that this is H=0, this is H is 2 so your delta H=+2, OK? Very straightforward, mass is 3, Gravity is 10 this +2 so this is sixty Joules.

For part B. what is the work done by gravity? The work done by gravity or the work done by weight same thing those are used interchangeably technically not the same thing but they're used interchangeably, the work done so it's going to be instead of MGH if you remember it's -MG delta H, now the variables are exactly the same the only thing that happens different here is that you have a negative so it's just going to be -60, alright? Check it out, -3, 10 this is +2 I'm still going up the reason why it's negative it's not because +2 becomes negative but because there's a negative in front of the equation so it's -60 Joules, itÕs says the box then falls to the floor how much work does gravity do then? So, the box is falling here's a 3-kilogram box it's falling so I can say the initial heights is zero...And I'm sorry the initial height is 2 and the final height is 0 so what is my change in height? My change in height is -2, Why? Because I dropped 2, OK? So, if I go here the work done by gravity is always negative MGH or MG delta H, -3 gravity we're using 10 my delta H is -2 right here, OK? Just to be clear this -2 comes from here and if you do this the negative cancel you get a positive 60 joules and this should make sense because gravity does positive work on the way down so you gave this object's 60 Joules of potential energy and then meanwhile was doing negative work against you and when it drops gravity gives it back the 60 Joules of potential energy that it kind of stole on the way up, OK? So, gravity does negative work going down and positive work... I'm sorry negative work going up and positive work going down, OK? That's the gist of it we're going to do a few more problems I want you to try this one it's a little tricky but I think you might be able to get it, letÕs give this one a shot.

Practice: You push on 3 kg box against a wall for a distance of 2 m with a 100-N force that makes 53° with the horizontal, as shown. The box-wall coefficient of friction is 0.3. Calculate the work done by:

(a) you,
(b) friction,
(c) gravity.

Practice: A 70 kg person hikes from the bottom to the top of a 1,000 m hill with varying speeds. The path you take is very irregular, with varying inclinations. How much total work does gravity do on you during the entire hike?

Concept #2: Work on Inclined Planes

Transcript

Hey guys in this video I want to talk about work problems, work done by forces. In situations where the object is moving along an inclined plane. And the reason why this requires itÕs own video is because these questions can be pretty tricky and I want you to be ready for some of these things. LetÕs check it out. So first remember theta in the work equation is the angle between the force and the displacement. Right force and displacement they form two vectors, two arrows. ThereÕs a very specific definition as to what theta is. And a lot of questions will give you the wrong angle to try to trick you into using the wrong angle because they want to make sure they know what youÕre doing. So I think itÕs helpful to have sort of a healthy amount of paranoia every time you see a theta to make sure that itÕs the right one. Okay. One quick thing here before we start. When you have an angle, when you have an incline plane like this, the planeÕs length which is this, I can call this L or d, the distance going up or down, is related to itÕs height and to itÕs angle by this equation h equals L sin of theta. I need you to remember this as well as obviously the equation for potential energy m g h and the work done by force which is this one. Okay. IÕm gonna do an example that I want you guys to try one practice problem. So letÕs check this out. I have a 100 kilogram crate sliding at a constant 7, so let me write some of this stuff down. Mass is 100, velocity is 7, itÕs a constant velocity which tells me the acceleration is zero. From the top of a 12 meter long ramp, that makes 37 with the horizontal. So let me draw a little ramp here. And this ramp makes 37 with the horizontal down here. It slides 12 meters so IÕm gonna say that the entire amounts of, the length of your displacement or your distance is 12 and you start up here and youÕre going to slide down. So first we wanna know is what is the crateÕs initial potential energy. So potential initial is potential initial gravitational because thereÕs no spring so the potential here has to be gravitational only, and itÕs just if you remember m g h initial, potential energy m g h. The mass is 100, gravity weÕre going to use 10 just to make it simpler and my original height is I donÕt have it. Right, I donÕt have the original height, but I can find the original height and thatÕs because h is L or d sin of theta so itÕs 12 sin of 37 and 12 sin of 37, I have it here, is 7.2. So thatÕs my height initial and thatÕs whatÕs gonna go right here, 7.2. When I multiply this whole thing I get 7200 Jules as my initial potential energy. Cool. ThatÕs it. The next question says use the work equation, which is this equation right here to calculate the work done by m g x, m g y and friction. So the work done by m g x. Remember the work done by any force, since I want to use this equation, is that force m g x distance cosine of theta. But before I do anything IÕm gonna have to calculate m g x. So letÕs do that real quick. M g x, I hope you remember is m g sin of theta so mass, gravity weÕre gonna use 10 and sin of 37. When you do this real quick you get a 600. M g y is m g cosine of 37 and thatÕs 800. Okay these are forces, this is a height but this is a force so these are Newtons. So m g x is this number right here, okay. So 600 goes right here. What is the distance. Well the distance you moved was 12 and what is the angle. The angle is the tricky one, right. So some people might have thought that that was 37. ItÕs not. ItÕs the whole purpose for the existence of this question is to make the point itÕs not that angle. Okay. Check it out. You are moving this way, and m g x is this way. Those two are parallel to each other which means they make an angle of zero degrees with each other. Okay. So this is the cosine of zero. And when you multiply all of this you get that the work done by m g x is 7200 Jules. I got a positive which should make sense because even though IÕm not going directly down, I am losing height and if you remember, the work done by m g is m g delta h. So if I had an initial energy of 7200 the work done by m g should be 7200 if I fell the whole way down. So that kind of matches up. What about the work done by m g y. So m g y, m g y is 800, the distance is 12 cosine of what? Well youÕre moving this way delta x, m g y is always into the plane so this makes an angle of 90 which means this is zero. And you might remember the work done by m g y is always zero because itÕs always perpendicular to the plane, itÕs always into the plane. The same thing is the work done by normal. So it shouldnÕt be a surprise that we got a zero here. The last point I wanna make is that the work done by mg can be thought of as the work done by m g x plus the work done by m g y, but the work done by m g y will always be zero so the work done by m g is the same as the work done by m g x. Okay. And if you remember, the work done by m g is also m g negative, m g drop in height. In this problem, my change in height was negative so these two negatives cancel and my total work done by m g turned out to be positive which is consistent with the idea that if youÕre going down, m g does positive work. Okay. So thatÕs the basic idea here. Explored a few things. Key points, so we got to use our h equals L sin equation, cool, but a key idea here is that this angle here is zero. Okay, not 37. ThatÕs the angle for the plane. Another point thatÕs important to make is that I do use this angle in these equations here, right. When youÕre decomposing m g x and m g y you have to use the angle at the bottom of the incline, but for work thereÕs a very specific definition which is the angle between the direction or the displacement in the force and in this case m g x and delta x went down. IÕve said that a few times already in this video alone, hopefully it clicks. Let me know if you guys have any questions, and I want you to do this next practice problem. LetÕs give this a shot.

Practice: You push a 2-kg box from the bottom of an inclined plane with 50 N for 10 m. The incline makes 37° with the horizontal, and the box-incline coefficient of friction is 0.6. Find the work done by:

(a) you,
(b) friction, and
(c) gravity.

Example #1: Work on Inclined Planes

Transcript

Alright so here's an extra example, this is a little bit weird so I'm going to do this for you guys I have a 5-kilogram box that's pulled at 3 meters per second so mass=5, velocity=3 when I say velocity is 3 I'm implying that the acceleration is 0 which is called equilibrium which also means that if you remember it also means that forces will cancel equilibrium which means the forces cancel, OK? So here is you're being pulled up on a conveyor belt, right? So, this belt is moving there's an object here being moved by the belts, so 5 kilogram objects being pulled up at the velocity of 3 meters/second so there's a force that pulls you up and I'm going to call this F belt, there's a force that pulls you up why is it that there has to be a force? Because otherwise you would be falling due to MGX over here, OK? You're moving with a velocity of 3 for a distance of 10 and the angle here is 37 degrees, OK? A little different how much work is done by gravity? So the work done by gravity is negative MG delta H, mass is negative mass is 5, gravity is 10 and I need the delta H, I don't have delta H but I can find it again, H is L sine of theta, this is a very popular equation for these inclined plane problems to see if you are familiar with how to go from one to the other, the length here is 10 sine of 37 which is simply 6, this 6 goes here and this is -300 joules and it should make sense that it's negative work because as you grow up gravity is doing negative work against you, OK? That's where the number comes from.

Question B is sort of more of a conceptual question which force is pulling the object up? As you go in a conveyor belt like this it pulls you up which force does that, I hope you know this I hope you realize this but if you don't I'll tell you and then hopefully remember F belt is really static friction, right? If I have an object in my arm and I pull my arm and the object comes with it, it's because friction of my arm is dragging along it's causing some traction, OK? So, conveyor belts work because of friction if you were ice the object would just stay there, OK? That's it and I want to know how much work does this force do? In other word, what is the work... By the way this is static friction I mentioned static I said static and then I put that one there, what is the work done by this force? What is this work done by the static friction? The work done by kinetic friction is -Fd but that's not the case for static friction, to find the work done by static friction you're just going to use the regular work equations with no shortcuts because in this case it's actually going in the positive direction so you couldn't say negative because the angle is not negative the angle is 0 being not 180 it's 0 and I'm going to show you, the work done by a force is that force distance cosine of theta, friction's pulling me up and the displacement in that direction so the angle is 0, OK? But how much is friction? How much friction do I get? The key point in this problem to figure out how much friction you have is the fact that this moves at a constant speed 0 acceleration which means forces cancel, which forces here would cancel? Friction which is the force of the belt cancels out MGX, so I can say that friction static=MGX which in turn equals MG sine of theta, let's calculate real quick, MG sine 37 this is 30.... 30 Newtons, so this thirty goes here, what is the distance? You go 10 up the plane and we already mentioned the angle is 0, The angle is 0 that comes from here, so this work is 300 Joules and this should make sense as well, the static friction on the conveyor belt is (This is part C by the way how much work you do?) the static friction on the conveyor belt is pulling you up so there's 300 which is the same thing that gravity did, gravity did negative 300 to lift something you do a positive work and gravity does the negative of that, cool? That's it for this one let me know if you guys have any questions.