 An object of mass 10 kilograms, initially at rest, experiences a constant horizontal acceleration of 5 meters per second squared due to the action of a resultant force applied for 30 seconds. Determine the total amount of energy transferred by work in kilojoules. So I have an object that I will draw as a sphere-ish thing. And it would be reasonable for us to treat this mass as being a constant mass, so I will call it a control mass. Meaning I have a closed system, and I will call that initial position, state 1, and our process is going to be describing the movement of that object across the 32nd time period, and I will define the end of that 32nd time period as another state point, which I'm going to call 2. I know the mass is initially 10 kilograms, and if I make the assumption that it's a constant mass, then I would write m2 is equal to m1. And actually just for good measure, I will write that down. 1. Mass doesn't change. Probably could have written that as masses constant, or change in masses zero, and had to write fewer words, but we're not here to write fewer words. We're here to solve a problem, and that problem involves a change in velocity as well as a displacement. I know the velocity at the end of the process is going to be greater than it was at the state, at the initial state, and I could describe the duration here. It's bringing 30 seconds, and we want to know the work required to accomplish this process in kilojoules. There are basically two ways that we could approach this problem, and the difference between the two is really only a matter of perspective. The first way that we would approach it is probably the way that most students would approach this problem having just gotten out of a dynamics course. And we'll call that the dynamics perspective. We approach it by thinking about the work itself. In that work we have the integral of force with respect to displacement, and our force in this problem is going to be written as mass times acceleration, and our mass is constant because we've assumed it, and our acceleration is constant because we were told it's a constant horizontal acceleration. Therefore, my force is constant, and if force is constant it can come out of the integral, and we have force times delta x. And since I have mass and acceleration, I will just substitute those now. Mass, which I know, multiplied by acceleration, which I know, multiplied by our displacement, which we don't know. But we can determine that. We've been studying kinematics for a while now, and we know our velocity is defined as a change in position with respect to time, so I could write dx as v dt and integrate both sides. The left hand side is just delta x, and the right hand side we have to actually evaluate because our velocity is changing. So for that I will step to our acceleration definition. The acceleration is the thing that allows us to figure out how much the velocity changes, and we know it's defined as a change in velocity with respect to time. So that I could write as dv is equal to adt, and when I integrate both sides here, I have delta v is equal to acceleration because it comes out because it's constant for this problem times duration. And this could be written for this problem as a velocity at any point in time, and that velocity at any point in time, by the way I'm doing that because the velocity at the initial state is zero, so delta v is really just the velocity at the time that you're considering is equal to acceleration times time. Then when I plug that in over here, velocity as a function of time is equal to acceleration times time, so this becomes an integral of acceleration times time dt from t1 to t2. Then acceleration comes out of the integral because it's still constant in this problem, and I have the acceleration times the integral of time with respect to time from t1 to t2, and that integral simplifies to 1 over 2 times time squared. Not sure why I said that so bizarrely, but here we are, evaluated from t1 to t2. So acceleration times 1 half times our time to squared minus time 1 squared. The initial time for this is zero because I'm describing t2 as an amount of time that has elapsed since the initial time, so everything's described in terms of time as relative to the beginning of the process, therefore t2 is really just the time at the end which is delta t. So acceleration times 1 half times delta t squared. And now that I've written that out, that representation of displacement, I can plug it into our work equation, and that would be mass times acceleration times acceleration times 1 half times, here I'll zoom in so that's a little bit better quality handwriting, times acceleration times 1 half, see better quality handwriting, times displacement squared. And I will write that as 1 half times mass times acceleration squared times duration squared. We know the mass, we know the acceleration, we know the duration, so now it's just a matter of computing a result. So I will take 1 half times our mass, which was 10 kilograms, question mark? Yes, it was 10 kilograms. And our acceleration, which was 5 meters per second squared, hope this isn't too nauseating for those of you following the scrolling, that was 5 meters per second squared. And I'm taking that quantity squared. So I'm taking 5 meters per second squared, which I will write as 5 squared meters squared, and then seconds squared squared, right? So seconds to the fourth. And then I'm multiplying by duration, which was 30 squared seconds squared. So seconds squared are going to cancel two of the seconds in the denominator, and I'm going to have kilogram meter per meter squared per second squared, which is all well and good, but I wanted kilojoules. So because my destination is kilojoules, and because I encourage you to get into the habit of starting at your destination and working backwards, breaking things down into their primary dimensions until things start canceling, let's do it that way. I will start with a kilojoule and recognize that that could be written as a thousand joules. And a joule is defined as a Newton meter, and a Newton is a kilogram meter per second squared. Now joule cancels joules, Newton cancels newtons, kilogram cancels kilograms, meter squared cancels meters and meters, second squared and second squared cancels seconds to the fourth, leaving me with kilojoules. So if I pop up our friendly neighborhood calculator here and try to get it to wake up, thank you calculator, and I take 0.5 times 10 times 5 squared, I'll put that into parentheses, 5 carat 2 times 30 carat 2 divided by 1000, I get 112.5 kilojoules. So looking at this by the work perspective, the work associated would be 112.5 kilojoules. That's option one. And it's perfectly fine. In this problem, we have, we actually have the option of approaching it from either perspective. The other perspective would be the thermodynamics perspective. And in a thermodynamics perspective, we are going to apply an energy balance. I know it is very exciting. The energy balance is going to relate how much energy is required to accomplish the thing that you want to how that energy is described when it's coming in. So I am going to write this as delta u plus delta k e plus delta p e on the left. And that static energy of the system is going to be related to energy in and energy out. Because I have a closed system, because we have a control mass, it's constant mass, I'm going to say e in could be q in and or work in. And e out could be q out and or work out. I am going to recognize that I have no change in potential energy. That's not even an assumption that we have to make because we were told it's horizontal acceleration. And I'm going to assume there are no changes in internal energy. Because we were given no indication as to anything meaningful about the possibility of the internal energy changing. It doesn't say the ball is initially at 20 degrees Celsius and ends at 22 degrees Celsius or something like that. But the ball starts to melt as it goes through a process. So I'll write that down as an assumption. Delta u is zero. And while we're at it, let's throw in no heat transfer. If there's no heat transfer, I can write that as adiabatic. Remember that adiabatic is a word that means no heat transfer. And then I'm going to only consider one work. Why only one work? Because there's only one opportunity for work here. It's the opportunity to change the kinetic energy. You could also think of this as like a network. You can describe this as a network in doesn't really matter. You could either call it a network in or just get rid of a workout and call it a work in because we have no opportunity to describe a workout. So the work in must be equal to the change in kinetic energy of our system. So work in is going to be kinetic energy 2 minus kinetic energy 1. And remember that the total kinetic energy of a system is going to be represented as one half times mass times velocity squared, that would be at state 2 minus one half, one half times mass one times velocity one squared. And velocity one was zero. So this really is just one half times the mass because it's a constant mass. We've already assumed that times the velocity at the end squared. So we know mass, but we don't know velocity. What to do about that? Well, we don't know the velocity, but we do know the acceleration. So we could take the same kinematic approach to describing velocity. I could say the acceleration here is going to be dv dt. Therefore, dv integrated going to be the integral of a dt. Therefore, delta v is equal to acceleration comes out of the integral because it's constant times the integral of dt, which is acceleration times duration. And this is v2 minus v1, which I'm calling zero. Therefore, v2 is equal to acceleration times delta time. So we could calculate a number or we could plug it in symbolically, at which point I have one half times acceleration times, excuse me, got a little bit ahead of myself, one half times mass times acceleration times duration quantity squared, which you know, just for convenience, I could write as one half. Maybe I could write it one half times mass times acceleration squared times duration squared. And hey, we have the same equation. Isn't that neat? Huh. Well, I could plug in all my numbers here and get the exact same answer. I wish we could point out that we have the same answer. Or if I wanted to be just a little bit more, I don't know, interesting with our second approach here, maybe just for fun's easier, I'll actually calculate a velocity state too. So calculator five meters per second squared multiplied by 30 seconds. Seconds are going to cancel leaving me with a quantity in meters per second. That's 150 meters per second. So I could say 0.5. Again, this is purposely inefficient just to calculate it another way. One half times 10 kilograms times 150 meters per second. And that's squared. It's 150 squared meter squared per second squared. And I have the exact same unit conversion as earlier because I'm describing kilograms meters squared per second squared. So I know that in order to get to the quantity that I want, I'm going to have to divide by 1000. And if I take 0.5 times 10 times 150 squared divided by 1000. That's a big number. Oh, great, because I took 150 raised to the fifth. Okay, let's try that again. Hey, look, it's gone. It never happened. Hey, we got 112.5 kilojoules. Interesting, right? Right. So the point here is that we can look at how much work is required to accomplish a change in kinetic energy and say that that work is the amount of work that's required or we can look at the amount of work actually associated with this process. And in this problem, because there was only a change in kinetic energy and a work, those two approaches are just as convenient and don't make a difference in terms of our calculation. You could think of this like if I were to go buy five standard t-shirts and I know that they cost $100 a piece and I wanted to know how much money was required to accomplish that transaction. Let's assume that there's no tax. I could either determine the price of a single standard t-shirt and multiply by five because I calculated all five or I could look at the amount in my wallet at the beginning of the process and look at the amount in my wallet at the end of the process. That's the distinction between these two analyses. And the reason that the distinction is important is because what if there isn't just kinetic energy and work happening, right? What if we had, I don't know, a heat transfer term or we had a change in potential energy as well, we could calculate the work that work associated with this process and plug that into my energy balance to calculate something else. If I had a change in kinetic energy and a heat transfer term and a work term, the energy balance alone would have one equation and two unknowns. So the definition of work would give me one of my unknowns. Does that logic make sense?