Ch 09: Work & EnergyWorksheetSee all chapters
All Chapters
Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
Ch 03: Vectors
Ch 04: 2D Kinematics
Ch 05: Projectile Motion
Ch 06: Intro to Forces (Dynamics)
Ch 07: Friction, Inclines, Systems
Ch 08: Centripetal Forces & Gravitation
Ch 09: Work & Energy
Ch 10: Conservation of Energy
Ch 11: Momentum & Impulse
Ch 12: Rotational Kinematics
Ch 13: Rotational Inertia & Energy
Ch 14: Torque & Rotational Dynamics
Ch 15: Rotational Equilibrium
Ch 16: Angular Momentum
Ch 17: Periodic Motion
Ch 19: Waves & Sound
Ch 20: Fluid Mechanics
Ch 21: Heat and Temperature
Ch 22: Kinetic Theory of Ideal Gasses
Ch 23: The First Law of Thermodynamics
Ch 24: The Second Law of Thermodynamics
Ch 25: Electric Force & Field; Gauss' Law
Ch 26: Electric Potential
Ch 27: Capacitors & Dielectrics
Ch 28: Resistors & DC Circuits
Ch 29: Magnetic Fields and Forces
Ch 30: Sources of Magnetic Field
Ch 31: Induction and Inductance
Ch 32: Alternating Current
Ch 33: Electromagnetic Waves
Ch 34: Geometric Optics
Ch 35: Wave Optics
Ch 37: Special Relativity
Ch 38: Particle-Wave Duality
Ch 39: Atomic Structure
Ch 40: Nuclear Physics
Ch 41: Quantum Mechanics
Intro to Energy
Intro to Calculating Work
Work By Gravity & Inclined Planes
Work By Variable Forces (Springs)
Net Work & Kinetic Energy
More Work-Energy Problems

Concept #1: Calculating Net Work


Hey guys, so now that we have seen different ways that we can calculate work in different situations let's talk about how to calculate the total work on an object which is also called the net work of an object or on an object, so the net or total work done on an object is simply the addition of all the works done to the object or on the object, in other words the work done by all forces that act on the object, OK? So work net or net work is simply the sum of all works, so if you have an object and there's three forces let's say acting on it, there are 3 forces acting on the object and they all do work on the object you'd have something like this, Work 1 + Work 2 + Work 3 and if you can find all these numbers or you have all these numbers you just add them up and if you had more obviously you would just keep adding this, OK? But there's another way sometimes you're not going to know all the forces or you're not going all the works but you will know the net force, remember the net force is the sum of all forces this is a vector addition so instead of knowing a bunch of the forces maybe you just know the net force and that works as well, remember work is F D cosine of theta so the work done by the net force is the net work, OK? So, I can do that instead I can just find the net force and multiply it by D cosine of theta, theta is still the angle between your displacement and your net force, OK? So depending on what you have you're going to use one or the other, one point here that's important to make is that work and energy are scalars they're not vectors they don't have direction so unlike forces where we would treat X and Y forces separately because they are vectors with work we combine X and Y to form net work let me give an example real quick if you have a force 3 going this way and a force 4 going this way the net force is not 7 (3+4=7) but instead it's a 5 because it's a vector, now if a force going this way does a work of let's say 10 and a force going this way gives you a total work of 20 then the net work is going to be 30, notice I didn't draw an arrow because work is not a vector, right? When I drew work here is the work done by this force here, OK? So you just add the numbers because they're scalars so let's try this out here's an example, you pull this box with a constant force for a distance of 5 meters up the plane so sort of 5 meters this way, smooth so there's no friction and the magnitude of the work done by each force is shown below so this is the arrow indicates the force not the work, work doesn't have to actually but I'm telling that the work done by normal 0 the work done by F is 75 and the work done by MG is 60, this is the magnitude of the works, OK? And I want to know what is the net work done to each so there's two ways you can calculate net work you can add up all individual works or you can find the net force, here I'm given the work so that's what I'm supposed to do so net work or work net is going to be the sum of all the works which is all these guys added together, the only thing you have to be careful here is that I'm giving the magnitude meaning I'm giving you these works as positives and some of them might be negative and here you have to remember when is a work positive or when does a force do positive work and when does a force do negative work, positive work is in the direction of motion and negative work is against motion, this guy here is 0 because it's perpendicular to motion and so you should remember that the work done by normal is always 0 because it's always perpendicular to motion, this guy goes in the direction of motion so it's +75 and MG is going against the motion because if you remember MGy is perpendicular so it doesn't do anything so the work done by MG is the same as the work done by MGx against the motion. Another thing another way you can remember this is that you were effectively up even though you're going at an incline you're still going up so when you go up gravity does negative work against you, OK? So, when you combine these it's not 75+60 but it's 75+-60 or 75-60 and the answer is 15 Joules, so you do have to realize that this number is a negative very important, OK? And then I'm asking for the net force acting on the box, how do you find net force? Well sort of the straight forward way to find net force is you use this equation the sum of all forces but I don't know any forces right so I'm kind of stuck here and I have really no way of doing this I don't even know the mass of this box so I couldn't even begin to calculate it, so another way that you can find net force now that we've introduced these other equations is by using this right here, right? If you know work net and D cosine of theta you can find F Net, OK? So that's just being able to play with the equations so work net which I now know is 15 is F net, D the distance is 5 cosine of theta, so this is 15 cosine of theta remember when you're trying to figure out the angle there you got to slow down a little bit and be careful, theta is the angle between F Net and D what is the direction of the net force here? Well you should know that the 2 forces that are sort of competing here are this F here that pulls you up and MGX and I know that F has to be greater than MGX because this box is moving up, OK? It's moving up so I know that the direction of the net force has to be going up this way, alright? So, because of that I know that this angle here must be 0, If the net force is up and my delta X is up then this angle is 0, right? So we can now calculate our net force right here, by the way another way that you know this thing is going up is the fact that the work going up is greater than the work going down so you know that more energy is being pumped into moving this thing up than it is being pumped into moving down so because the energy is greater going up this thing will end up going up and therefore I can confidently say that these are in the same direction which means the angle I'm supposed to put there is 0, the cosine of 0 is 1 so I end up with F net=15/5 which is simply 3 Newtons, OK? That's the net force acting on this object, alright? So I want to take a quick point here and then I want to give you guys a practice problem to work on, to find the net work there are two ways one of them is if you have the net force that's great but usually you're going to need to first identify all the forces whether it's because you are going to identify the forces so you can find all the little works and add them up or because you can identify all the forces so you can calculate the net force however you want to do it, OK? But the point is that to find net work you first have to identify forces and the the big idea here is that not all the forces do work so I like to draw this little diagram here that a force may cause a work or a force may do work this means 2 things, one you cannot have work without a force so if there are 5 forces acting on an object there might be 5 work done to the object but either 5 or less because some forces don't some forces don't do work, work is F D cosine of theta so right away you see how you need to have a force in order to have work but remember also that if a force is perpendicular it does no work, the work done by perpendicular forces is 0 so identify all the forces and then figure out which ones do work calculate and add it together, OK? I want you to try this practice problem here and then we're going to keep going so let's give this a shot.

Practice: A 3-kg box is on a flat surface. The box-floor coefficient of friction is 0.6. When you pull horizontally on it for 10 m, it moves with 2 m/s2 . Find the net work on the box. (Start by finding the magnitude of all forces acting on the box)

Practice: A 2-kg box is on a rough horizontal surface. When you pull horizontally on it, it moves with 3 m/s2 . The magnitude of your force and the box-floor coefficient of friction are unknown. What is the net work on the box across 5 m?

Concept #2: The Work-Energy Theorem


Hey guys, so in this video I want to talk about the work energy theorem which is basically an equation that's going to connect work with kinetic energy and it's going to allow us to solve problems in a different way let's check it. Alright, so the work energy theorem is this equation right here and it says that the net work or the work net the total work = the change in kinetic energy now change in kinetic energy obviously means K final - K initial, OK? And work net means the sum of all works, alright? So remember we can calculate the work done by a force using this equation and remember that the net work can be calculated using either the sum of each individual work or the work done by the net force you can find the net force by combining all the forces and then find the work done by the net force and remember also that kinetic energy is 1/2MV squared so those are all the questions that we need to talk about the work energy theory so it says here gives us a way to find the initial or final velocity of an object that is acted upon by forces so if there is work done there will be a change in kinetic energy and obviously kinetic energy has a velocity in it so you should you should see a V inside of these guys and these are the kinds of problems we're going to have, you're going to push a box and the moves and we will find some velocities, OK? So, I'm going to do the first example here and I want you guys to try the practice problem.

Example one, a 2-kilogram object has a kinetic energy of 4 at point A and a speed of 3 at point B so I'm giving you two points I don't really give a sense of whether one is to the left or to the right it doesn't really matter but let's start writing this stuff down here, mass=2 I'm going to say that kinetic energy at point A (Ka) is 4 and the speed at point B is 3, OK? Let's just get the information down and then it says find a speed at A, so find a Va notice that I know Ka and then I'm looking for Va well all you have to do is find an equation that relates those two variables and it's pretty easy, the kinetic energy equation ties the two together so not only can I find K if I have V but I can also find V If I have K and that's the idea here, so I'm going to write that Ka=1/2MVa squared, K is given it's a 4, the mass is a 2, so this 2 cancels this 1/2 here and I have the VA is going to be the square root of 4 therefore Va is 2 meters per second, OK? That's it for part B I want to know the kinetic energy at point B so what is Kb? Very similar just backwards I give you Vb and I want to know the kinetic energy at point B, this is even more straightforward Kb is 1/2MV squared at point B as well and we just have to plug in the numbers here, the mass is 2 these two cancel and the velocity is 3 squared so Kb is 9 Joules, OK? That's Va that's Kb 9 Joules right there.

Part C says find the net work from A to B? So, what is the net work from A to B so network I can calculate by doing the sum of all works or by finding the work done by the net force and now there's a third way you can do this and that's the big idea of the work energy theorem is that it gives you a third way to find work net which is by finding the change in kinetic energy, OK? I'm not able to for this problem here I'm not able to find the net work because I don't know all the forces that are acting on this and I don't know the net force so I'm going to have....But I do know velocities and kinetic energies and stuff so I can do it this way ,so if I'm going from A to B. then the network between A to B is just the change in kinetic energy so it's kinetic final or kinetic B- kinetic initial kinetic A and I know these numbers kinetic B we just found it's 9 and kinetic A was given it's a 4 so the net work is 5 Joules, OK? That's it so all we've done is play around with the kinetic energy equation and with the work net equation and the answer is in there somewhere, right? So, cook I want you guys to try practice number two let's give that a shot.

Practice: A 4-kg object has speed 6 m/s at point A, and speed 10 m/s at point B.

(a) How much work was done to it between A and B?
(b) If –32 J of total energy is done to the object between B and C, what speed does it have at C?

Concept #3: The Work-Energy Theorem,ually


Hey guys, so now that I've introduced the work energy theorem I want to talk about how that's actually a very important conceptual point, OK? So, it says here that the work energy theorem is also very important conceptually, it's very useful conceptually So it's very useful if you remember that work net is the change in kinetic energy as you're solving physics problems because there's a really important consequence here, well remember kinetic energy is a 1/2MV squared so if your V doesn't change if your speed and not velocity but here it's speed the number, right? So if your V doesn't change if you're V is constant then kinetic energy is constant as well because the mass usually doesn't change so if this number stays the same this number stays the same, so if the V is constant the change in kinetic energy must be 0, there's no change it stays the same and look at the work kinetic energy theorem here, it says that work net is the change in kinetic energy, well if the change kinetic energy is 0 the net work is 0 as well so the idea is that if your speed is constant your net work is 0, OK? Now I said speed not velocity and why is that? Well imagine if you're going round the circle and your speed is always 5 but your velocity is technically changing because your direction changes but on the kinetic energy equation you're plugging in a V here as a 5 so it's a 5 this way and it's still a 5 this way and the kinetic energy is exactly the same if you remember kinetic energy is also a scalar, it doesn't care about direction so it only really depends on speed so if your speed is constant your work net is 0 in fact if you go around a circle with a constant speed uniform circular motion the net work done on you will be 0 and if you remember that you can save a lot of headaches and solve problems much easier, I want to do example 2 and I want you to try practice 2, now practice 2 is a little tricky but I think you have a good chance of getting at least most of it and it's a good sort of mental exercise, alright? So, let's do example 2 here.

I'm going to show you first you know, the answer I'm just going to tell you right now the answer is 0 but I'm going to show you what you might do what you might be doing here wrong if you don't realize that work net is delta K, so it says here a 5 kg box so typical problem solving steps you're going to write your variables on a rough horizontal surface let's draw this here, it's rough so there's a coefficient of friction and your push on it I'm going to draw that as a pull with a constant force, my force is constant cool so that the box moves with a constant speed, so the box is going to move with a constant velocity of 8 meters per second that's constant as well and you might be thinking well how can I have a force and not have an acceleration? Sum of all forces= MA, if I have a force I'm supposed to accelerate, constant speed tells me that the acceleration is 0 how is that possible? Oh it's because there has to be a friction which in case of box kinetic friction that's exactly canceling with my force, so in a regular problem solving....In a regular steps of solving this problem you might have thought about all of this stuff and now if you want to find the net work maybe you're thinking the net work here is the work done by F + the work done by kinetic friction obviously MG does no work and normal does no work but look at how long that analysis is and if you were to calculate all the stuff you would've found in fact that the answer is 0 and that's because whatever you have here will be a positive and that number will be the same number here as a negative or you could have just seen that it says constant velocity or constant speed and then right away you would have known that the answer is 0 and you wouldn't have to do any of this crap, OK? So, I actually want you to do this I want you to scratch this out so when you're reviewing this you remember that you don't have to go to all this, OK? So, the answer would have just been 0 so if you get this on a test it most likely will be multiple choice for those of you that have that and then you'll be able to kill right away, alright? So, I want you to practice number two let's give that a shot.

Practice: You lift a 3 kg object from the floor to a height of 2 m. Find the:

(a) work done by you;
(b) work done by gravity;
(c) net work done on the object. You then walk horizontally with the object for 10 m.
(d) How much work do you do?