 Problem one. A 30 kilogram object initially traveling at 5 meters per second experiences a constant horizontal acceleration of 10 meters per second squared due to the action of a resultant force applied for 10 seconds. Determine the amount of work done to accelerate the mass in kilojoules. So let's pick out these pieces of information. A 30 kilogram object, therefore, the mass of our object is going to be 30 kilograms. Initially traveling at 5 meters per second, so let's call it an initial velocity or v1, 5 meters per second. Experiences a constant horizontal acceleration of 10 meters per second, so constant acceleration of 10 meters per second. Due to the action of a resultant force applied for 10 seconds. So I'm going to say that the time period is 10 seconds long. Determine the amount of work done to accelerate the mass in kilojoules means that we're looking for a total work. So this would be a capital W as opposed to say specific work, which would be a lowercase w kilojoules per kilogram or a power, which would be w dot kilojoules per second. So we have a couple of different options for approaching this problem. I prefer to approach it using an energy balance. It's thermal, if in doubt, do an energy balance. So we have a mass experiencing a horizontal acceleration. So if I were to establish that object, the mass of that object as my system, and do an energy balance on it, I'd be able to talk about the different forms that energy is appearing. So we have the change in energy of the system is equal to the energy into the system minus the energy out. Now the problem didn't indicate any ways for the energy to be leaving, so we're going to be ignoring that for the moment. That leaves us with delta E and energy in. So if I ignore everything else except for the information that was given in the problem, the energy change of our system would be expressed as a change in velocity. So that's going to be our kinetic energy. So my change in energy in my system is going to be the change in kinetic energy of my object. And then the energy into the system would be the work done to accelerate the mass. So my energy in is just going to be a work in. And I got this change in kinetic energy by remembering that when we talk about the change in energy of a closed system that could be kinetic energy, it could be potential energy, and it could be internal energy. In this case, I'm ignoring potential energy, it's a horizontal acceleration, and I'm ignoring internal energy changes. That leaves us with just kinetic energy. It's a closed system because it's assumed that the mass of our object doesn't change. So we have the change in kinetic energy of our system is equal to the work in. So we could calculate the work in by figuring out what the kinetic energy of our object is at the end of the process. Call that kinetic energy 2 minus the kinetic energy at the beginning of the process, kinetic energy 1. And then I know that I can, let me zoom in a little bit, a little bit easier to write, I know that I can calculate kinetic energy by taking 1 half times the mass of our object times the velocity squared. So this would be 1 half m2 v2 squared minus 1 half m1 v1 squared. And now I know that 1 half can come out, and the mass doesn't change. Therefore, m2 is equal to m1, they're both 30 kilograms, so that can also come out of my equation here. Or rather, I can bring it out front. So I have 1 half times the mass times v2 squared minus v1 squared. So in this equation, I know mass that was given, I know v1 that was given. So if I had the velocity at state 2, I could calculate the work in. But I don't have a velocity at state 2. However, I know enough information to be able to calculate it. Because I know that it's initially traveling at 5 meters per second and it accelerated constantly for 10 seconds at a rate of 10 meters per second squared. So every second, the velocity increased by 10 meters per second. So you could use that to deduce that, well, 10 meters per second per second multiplied by 10 seconds means an increase in velocity of 100 meters per second. Therefore, v2 is 105. But if you wanted to use the equations of motion, we could do that too. We could step back to the fact that acceleration is a change in velocity with respect to time. So I could rewrite this as being dv is equal to acceleration times change in time. I can get rid of the differentials here by integrating. So the integral of dv would be delta v. It's an excellent triangle. And then in this case, the acceleration is constant. So the acceleration comes out of the integral. I'm left with a times the integral of dt. And that would just be a times delta t. Now, I know that delta v is going to be v2 minus v1. And delta t was given 10 seconds. So I'll leave that as delta t. So I could write my v2 as being v1 plus acceleration times change in time. So this would be 5 meters per second plus 10 meters per second squared times 10 seconds. My second squared drops to just seconds. So I have 5 meters per second plus 10 times 10 meters per second. So this would be 105 meters per second. And just to be safe here, I'm going to try to get into the habit of double checking these calculations with a calculator. Shouldn't rely on my mental math too often. So let's see here. 5 plus 10 times 10 is 105. Awesome. So I know that the velocity at state 2 is 105 meters per second. I can use that with the equation I just came up with and calculate a work in. So my work in is going to be 1 half times the mass, which in this case was 30 kilograms, times the quantity v2 squared minus v1 squared. Remember that because the squareds are on the individual velocities here, that's not mathematically the same as v2 minus v1 quantity squared. You can't just take 105 minus 5 squared. Introduce them individually. So this would be 105 meters per second squared. Yeah, there we go. Minus 5 meters per second squared. So 1 half times 30 times 105 squared minus 5 squared. So 0.5 times 30 times quantity 105 raised to the 2 minus 5 raised to the 2 gives me an answer of, whoops, I did not type in my parentheses correctly. So let's try that again. That's an H. There we go. Squared minus 5 squared parentheses. There we go. So I have 165,000. Now, what are the units on this? Well, I have kilograms times meters per second squared minus meters per second squared. So this is going to end up being kilogram meters squared per second squared. Now, what is kilogram meters squared per second squared in terms of useful units to me? Well, I could figure that out. I could take kilogram meters squared per second squared and convert that to, let's say, I wanted an answer in kilojoules because the problem told me I wanted kilojoules. So I know that a kilojoule is going to be a kilonewton meter. And then I know that a kilonewton is 1,000 newtons. And I know that a newton is a kilogram meter per second squared. So I have newtons, newtons cancel. Kilonewtons, kilonewtons cancel. Meter squared cancels meters meters. And second squared cancels second squared. So I could convert from kilogram meters squared per second squared to kilojoules by dividing by 1,000. So let's say my work in here is just 165 kilojoules. That gives me my answer. So that was one of the ways we could have solved this problem. We solved it with an energy balance. And it is thermo, should be doing energy balances all the time just out of habit. But there are a couple other ways we could approach this problem. So let's say we didn't do the energy balance. Let's step back for a second. Let's say that we did, let me, that's a race over a year. Let's say that I hadn't done that energy balance. How else could I have solved this problem in you erasing very carefully? Beautiful. One more race. Let's say I had used the definition of work instead. So I know that I can define work as being the integral of force with respect to distance. So when we talk about, well, work is force times displacement. That's when the force is constant. In this case, my force is constant. So my force is going to come out of this integral. I'm going to be left with force times displacement. Let's call that delta s. So I know my force in this case, because the acceleration and the mass are both constant. It's going to be mass times acceleration, both of which are known. So if I could figure out the displacement of our object over the course of this 10 second process, I could figure out the work done. So how would I go about calculating the displacement? Well, that again is going to come from our equations of motion. I know that I could write, let's see, how would I do that? Let's say I wanted to use the definition of velocity is the change in displacement with respect to time. And then let's rearrange that to solve for, that would be ds is equal to v dt. And then integrate both sides. I'm left with delta s is the integral of velocity with respect to time. In this case, the velocity isn't constant, so it can't come out of the integral. But I do know that I could calculate the velocity at any point in time by using this equation. I already came up with that equation. I can plug that in here. So this would be v1, our initial velocity, plus acceleration times dt, excuse me, times delta t dt. And that's what I'm integrating. So this is going to split into the integral of v1 dt plus a times the integral of delta t dt. So this would be v1 times delta t because the v1 comes out of the integral. So I'm left with v1 times the integral of dt, which would be delta t, plus the acceleration times... So this is the integral of delta t dt, which is going to be 1 half delta t squared. So I could use this to calculate a displacement and plug that in here. So if I had chosen to do that, let's see, I did not leave myself enough space. Also, that arrow is terrible. I can't just leave that. Okay, much better arrow. That'll solve my space issues. Okay, so I could write this as being 30 kilograms times 10 meters per second squared times 5 meters per second, that's our initial velocity, times 10 seconds plus 10 meters per second squared divided by 2 times my delta time squared, which is 10 seconds squared. Beautiful. Let's jump back to the calculator, turn it on and attempt to type that in. So we have 30 times 10 times parentheses 5 times 10 plus 0.5 times 10 times 10 squared. And look at that, we got 165,000 again. And this would be in kilograms times meters per second squared times meters per second squared. So this is going to end up being my same units again. So regardless of the way that I did this, I'm left with 165 kilojoules. That's the same unit because kilograms times meters per second squared, that's this mass times acceleration. And then down here I have 5 meters per second times 10 seconds, so the seconds and seconds are going to cancel. So this is going to be meters. So I guess over here too, this would be 10 meters per second squared times second squared. This is also going to be in units of meters. So my final units for that 165,000 number that I had earlier are going to be kilograms times meters per second squared times meters. That's kilogram meters squared per second squared, which is the same unit that I came up with earlier. So I use the same conversion to get it back to kilojoules. That's why it ended up being so convenient. Anyway, either solution would have worked, whichever one is easier for you to think about is the solution that you should use.