Guys, let's check this one out here. So we've got this 10 kilogram suitcase, that's the mass of the box. And we're also pulling with some applied force of 50 newtons. Now we're pulling it across the floor, and it's going to be pulled some distance of 20 meters. But which way is it going to be pulled? Well, even though we're pulling it upwards like this, what happens is it's just going to move along the floor. Right? It actually wouldn't fly up like this. So what's going to happen is it's going to travel some displacement here. Let me draw this a little bit bigger and we know that this delta x, or d, it doesn't matter which letter we use, is going to be 20 meters. So we want to do this first part here is we want to figure out the work that's done by the pull. Right? So your pull here so the work that's done by f. Well, remember, the work that's done by any force is always going to be fdcosθ. So the work done by f is going to be fdcosθ. Remember that this angle is the angle between f and d. So let's check this out. I have the magnitude of the force. It's 50. I have my x or my d. Right? It's just equal to 20 meters. So I can just plug this into my work equation. This is going to be 50 times 1 sorry. 20 times the cosine of the angle that's between these two vectors. We can take a look here. They're actually not parallel. Right? The force actually points this way, but your displacement points along the floor like this. So this is the angle that we use between these vectors, and we know that this angle here from the problem is equal to 37 degrees. So we're just going to plug this in, 50 times 20 times the cosine of 37. You go go ahead and work this out. What you're going to get is you're going to get 800 joules. So that's the answer to part a. Alright. So now in part b, what we have to do is we're going to have to decompose our force into its components, fx and fy, and then calculate the works done by each one of those components. So for part b, we have the work that's done by fx. Remember what we said, the work done by any force is that force, so fx times d cosine of theta. Now we actually don't have fx in our diagram just yet. So what we're going to do is we're going to take this force and we're going to break it down into its components like this. So this is going to be my fx and I know to figure that out, I'm just going to use f times cosine of theta here. So this is just going to be 50 times the cosine of Actually, let me go ahead and write it somewhere else. So I've got that fx is just equal f cosine theta. So it's just equal to 50 times the cosine of 37. So if you go ahead and plug this in, you're going to get 40. And the same thing is going to happen with fy except now we're just going to use sine, right? So fx is fy is f sine theta. That's basically this vector like this. This is my fy. And this is going to be 50 times the sine of 37. So you're going to get 30 here. So those are the components. It's just basically a 30-40-50 triangle, which makes sense. So now what we have to do is we have to calculate our work. So we're going to take the magnitude of fx, which we know. This fx is really just equal to 40, remember? So this is going to be my 40. Now the distance that it travels is still going to be the 20. That doesn't change. But now what's the cosine between the between or what's the angle between these two vectors? Well, now what happens is your fx, remember, actually points along the horizontal in the same direction as your displacement. So what's going to happen here is you're going to have 40 times 20 and the cosine between these two angles is actually going to be or sorry, between these two vectors is actually 0. Right? Because basically your fx points to the right and your distance also points to the right here. So what happens is you get a cosine of 0 and then we know that equals 1. If you plug this in, you're going to get 800 joules. So we get the same exact number but and that's basically because the only force that's actually doing work on this object is this fx here. That's what gives you the 800 joules. One way you could also think about this is that we use fdcosθ when we calculated part a. And in part b, what happens is you already kind of have a cosine of theta that is absorbed in this term here. It's kind of wrapped up inside this fx because really, this is already f times the cosine of theta. So you're doing fdcosθ here, but you're doing f times cosine theta times d when you're doing it this way. So you're just going to get the same exact number here. Alright. So to finish off the problem, we're going to take a look at part c. This is going to be the work that's done by fy. So this is going to be my fy times d cosine of theta. So we take a look. We got fy. Remember this? That's just the 30 that we calculated over here. Times the displacement which is 20. But now what's the cosine or what's the angle between these two vectors? Well, basically, with this, fy is vertical, but we know our displacement actually points to the right like this. So the angle between these two is actually 90, and so, therefore, the cosine of 90 is going to be 0, and the work that's done is just basically going to cancel out to 0 joules. That makes sense because there's a component of your force that's pulling vertically, but if you're traveling horizontally, then there's no work that that force does. Alright, so that's it for this one, guys. Let me know if you have any questions.
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Intro to Calculating Work
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Intro to Calculating Work practice set
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