Physics 12 Final Exam Review of Kinematics and Forces

[music] So, we jumped in way back in September. Cole, one of the things I talked about was how do I add two vectors together? I made it dramatic. I always had the Nerf dart gun and I cocked it. How do I add two vectors together? There was two words that went before that. Yeah. Okay. Draw vam tip to tail. And then the resultant was from the tail of the first vector to the tip of the second vector. And it was trig. In fact, the very first lesson I did with you, Anna, was a math review. We did a lot of trig. Okay. Then follow-up question. How do I subtract two vectors? Yep. Okay. Add the opposite. What we mean by that? Instead of going vector A minus vector B, go vector A plus the opposite of vector B. What do I mean by the opposite? Shahab, what's the opposite of south? What's the opposite of south? You got this. I know. What's everybody? What's the opposite of south? What's the opposite of west? What's the opposite of up? Did I say angle? Everybody. Shahab, did you get that one? Okay, you're back with us a little bit. I don't care that it's first block. We got to work on this. If you give me an angle, I will almost never use the thing on the angle. I will break it into its horizontal and vertical components. If it's a projectile or if it was a ramp for forces, we did per perpendicular and parallel. Parallel and perpendicular. Uh, the best tool for projectiles that I gave you is what I called a tea table. What did a tea table look like? this. I'm going to do the next column in a second, but I'm going to go down here. I had what I called my physics no-brainers. I knew what ax was and I knew what a y was for a projectile. What was ax exactly as a number and I could write that down for any projectile? Zero. In fact, let's put that over here. Ax equals zero. What was a y exactly as a number? And I could write that on the earth it was 9.8. And I'll keep you on the earth if it's a projectile. And then if we were horizontal, we didn't need to do components. But if we're at an angle, we did. I gave you a stupid hack. Does anybody remember which trig function vy was going to be? How could you remember that? I'll let you think of your own way. Yeah. Vy was equal to the velocity they gave you at an angle sin theta. Which trig function was Vx then? Can you remember? Of course you can. Lucas is seeing those for the first time. It was a stupid way to remember. Vy is side and then VX. Of course you can't. You can do the trig, but we said let this is one time we're using it so often we'll come up with a dumb way to memorize it. And then the big thing, do not put a horizontal value into a vertical equation. Don't put a vertical value into a horizontal equation. So I started using subscripts x's and y's in a lot of my equations just to remind myself I better pay attention to component. Oh, and the angle one never went into anything because it was never neither horizontal nor vertical. You're going to have two projectile questions. So your final exam is going to be a booklet style. When you open to page two, page one is going to be a cover page. When you open to page two, you'll see question one A, question one B. It'll be two projectiles. One is going to be horizontally from a cliff. We're going to do one. One is going to be from the ground at an angle. Did I say angle? Did I say angle? Did I say angle? Yeah, Adam, I'm going to keep going until you and Lucas, did I say angle? There we go. I know where to look. Um, I will not be giving you cliff at an angle. I thought about that. That was the one where you had to use the quadratic formula. It's just a lot of writing. I'm not going to worry about that. So, here is a horizontal projectile. Says a projectile is launched horizontally at 67. You know what? I'm going to draw it in. 67 m/s right there. And the cliff is 124 m high. Laura, what does a want me to find? Horizontal or vertical? Range, remember, is dx. It's the horizontal distance. And I'm going to do my tea table over here on the left and I'll do my workings over there on the right. So my two physics no-brainers. I can write ax= 0. I can write a y =9.8. Then I look at that 67 m/s. Horizontal or vertical or angled? No, it's not at an angle. horizontal or vertical. So I can say vx = 67. What's vy initial? This is what made horizontally from a cliff the easiest ones to do. But you had to remember this. And a lot of us struggled to remember this. A lot of us did all sorts of nonsense. I saw people try and do some dumb Pythagoras thing. No, no, no, no, no, no. horizontally from a cliff right when you leave the cliff for a split second what's your initial velocity is a number vertically exactly yes and the zero times table right now even Shahab could handle that first thing in the morning right yes that's two things I need three things to find the fourth well why is this wrong if I write dy= 124 why is that incorrect because it is incorrect Got to make it negative. It's a displacement. I ended up below from where I started. Otherwise, I'd be telling the universe that I accelerated down 9.8 and ended up above from where I started. You're going to get an error. And you should What did you say they wanted me to find? What am I really going to spend most of my time finding? I gave you a hint. What am I really going to spend most of my time finding? Time. Which column has three things? I'm going to find time vertically. So, I'll put t equals question mark here. I'm looking for an equation, Laura, that has an a, a vi, a d, and a t in it. All of you have your formula sheets out. There is one. Oh, and I'll say don't waste my time. Yeah, I can go d y = a y t ^2 over a half a t^2. But I'm going to write it that way because I suspect I'm going to be doing some formula manipulation. You could also write it.5 a t^2 Lucas if that's what you prefer. I'm fine with that. But I'm going to get the t by itself. How would I move the a over? How would I move the two over? Or if there was a 0.5 in front, I would divide by the 0.5. How to get rid of a squared square root? I'm going to find the time is equal to 2 d y / a y square. It's going to be the square of 2 * 124 /9.8. And once you've written that down, you want to get out your calculators. I think I have room. If I scooch the screen over a smidge, no, I'll write it here. 2 * -124 /9.8 square root answer button is going to be 5 something 5.03. Anybody else double check me. I'm not done. That's not what they wanted me to find. Laura, what did they ask me to find? They want me to find dx. How can I do that? Well, now I can put the time here, 5.03, because time being a scaler can go in either column, which is why I spend most of my time finding time. Uh, and I want to find dx. Right back at you, Laura. I'm looking for an equation. It's got an a, a v, a t, and a d in it. There is one, and don't waste my time. Yeah, we end up with it's the same equation as you said earlier except this time the half a t^2 vanishes. I get dx= vxt. It's going to be 67 times and I'll just put my answer button and I get 337 m 67. Shut up. Okay, I've got several different versions. On some of them, I'm going to give you the height and say find the range. On some of them, I'll probably give you the range and say find the height, which arguably is a bit easier. Kobe, how would I find the height if you gave me the range? Well, then I'd have a dx here. I'd use this equation. How would you get and you're going to still spend most of your time finding time. How would you get the t by itself? The x waves, no square root, hardly any form of manipulation at all. And then you would walk time over here and you would use this equation plugging in time and that would give you the displacement. You'd get a negative answer because it's giving you displacement, but that would also be there. Okay. The other one you're going to see is some variation of launching from the ground at an angle. Did I say angle? Everybody. Wow. Did I say angle? Everybody. They'll figure it out the first time one of these times. He knows where I'm looking. A projectile is launched from the ground at an angle. You know what? Let's write it in. So, it says it's 36° and it says the velocity is 99 m/s. We're still going to do a tea table. It is a projectile. It says find the range. I'm going to find the dx. So, I would certainly go like this. I would say x y. I would write down my physics no-brainers. I know the horizontal acceleration is zero. I know the vertical acceleration is 9.8. Ooh. Now, to find vx and vy, I'm going to have to do some trig. You can draw out the triangle if you want where that's 99 de that's vx de that's vy initial and this is 36° got someone coming in late sorry YouTube so you could do that but we had and then do the trig it's opposite adjacent hypotenuse but we had a stupid hack what was vy sign. So let's do vx. vx then is going to be 99 cos 36 99 cos 36 and I get I'll carry some extra sigfigs. I get 80.093. I initial is going to be 99 sin 36 as Vy is sin. And I can just backspace and edit. Yay. And I get 58.191 meters/s. Why did I put an initial on the vy but not on the vx? you zero acceleration. So that's never going to change. Vy is going to be constantly changing because we have an acceleration. Okay. Um Oh, all right. Banner, what do they want me to find? Part A. Okay. So, I'm going to say uh dx equals question mark. What do you think I'm going to spend most of my time finding? Okay, horizontal or vertical? Let's see. How many things do I know in the horizontal column? One, two, cuz I wrote Vx twice. How many things do I know in the vertical column? One, two. Cuz I wrote Vy initial twice. That's not enough. So, here's my question. What else do I know? Yeah. What's Vy final as a number exactly? It's not zero. If you stand on your lab table and do a belly flop onto the cement floor, it's going to hurt. And the reason it's going to hurt is because when you hit the ground, vy is not zero. Vfal is not zero. It can't be zero. By the way, where is Vy 0? There is a place where Vy is zero. The very top. That would have given me half the time and I could have doubled it. A whole bunch of people used that method and forgot to double the time. I'm going to say, hey, what goes up must come. We're starting and ending on the ground. That means we're going to hit the ground at 58.191 or negative answer button if I want to go that way. Hey, that's three things. Now I can find time. Banner, back to you. I'm looking for an equation that has an A, a VI, a VF, and a T in it. There is one. Yeah, one of the first ones we learned, our old friend. Get the T by itself, please. So, I'm going to go Vyf final minus Vy initial and then divided by a Y. You said you were going to subtract first, right? You didn't foolishly say we're going to divide first or anything like that. That'd be silly, right? It's reverse bad mess. So, to type that, I'm going to go negative answer button minus answer button /9.8 8 since I have that 58.191 hopefully sitting on my calculator which is why you want to follow along on your calculator bracket 58.1 answer button minus answer button close bracket divided by9.8 eight. This gives us the total time of flight. Divide it by two. That's half the time of flight. That would be when you're at the top. Or you could have let Vyf final be zero and solve for time. That would have given you the time at the top. Uh I get 11.8756. I'm going to go 11.876. I'm going to throw this on my calculator. What did I say, Mr. Dick? 11.876. Of course, we're not done. This isn't even what they asked us to find. Banner, back to you. What did they ask us to find? Well, now I can put time in this column. I'm looking for an equation that has an a, a v, a t, and a d for the range. Don't write this down. So, here was the suggestion. What's ax as a number? So what would dx be if we use that? I don't think I think it does have a range. Yes. So I ask you again, I'm looking for an equation that's got an a, a v. What was the equation for the range always? The equation for the range always. And if you want to, you can maybe even go up here and you can make a little note. Range dx that was always just vxt the half a t ^2 was what vanished. So we can go dx = vxt the range is going to be what was vx 80.093 or you could retype 99 coast 36 if you wanted to be really really accurate. uh times my answer button 80.09393 Mr. Dick times my answer button and I get 951 m and change. Okay, parents, what does B want me to find? Yeah, I'm going to do B right here. I'll just kind of draw a little line. I guess I should have said that's part A. I'll put a part A right there. So, kind of vaguely be organized. Maximum height. Am I going to solve maximum height vertically or horizontally, do you think? Be stupid obvious. Height. What does the word height suggest? horizontal or vertical, folks? Yeah, we're going to solve it vertically. Okay, that's your clue. So, I'm going to quickly jot down what I know. Uh, I know that a y was -9.8. I know that vy initial was 58.191. Looks like they want me to find a dy, a maximum height. That's only two things. I need three things to find a fourth. Yeah, vyfal is zero. I'm looking for an equation that's got a a VI, a VF, and a D in it. There is one. Squared squared. Get the D by itself, please. Get the D by itself, please. I'll put the y's to remind myself I better use all the vertical stuff. You can see by the way if somehow you forget that vy final is zero. A lot of kids are oh v initial is that 99 cuz that's how fast we were traveling when we started. No, we never use the one at an angle. It's neither horizontal nor vertical. But that was a common mistake. So it's going to be 0^ 2 - 58.191^ squared all divided by 2 * -9.8 8 0^ 2ar minus 58.191 squared close bracket / bracket 2 * -9.8 8 173. And then if I did ask part C, part C would be a onemark question that says, what's the velocity at the top? What is the velocity at the top exactly? It's not zero because that would mean it was standing still at rest. The vertical velocity is zero. But Malcolm, what's always there? What never changes? What's constant? You answered this about 5 minutes ago. Yeah, Vtop. Hey, all that's left is Vx. I can tell you exactly what is for that split second at the very tippy top. It's traveling at 80 093 which I would probably write as 80.1 meters/s cuz remember we're always moving sideways. I had several versions of my kinematics test but I know on some of them I asked that question as like the part C and a lot of students for one mark really struggled with that vx because vy is zero. So all you have left is vx. It's horizontal. That was a very quick review of kinematics. Forces key concepts need to know Newton's laws. Newton's first. If you're accelerating, there's an unbalanced force in the direction of the acceleration. If you're not accelerating, forces must be. I'm looking for it. It starts letter B. Yes. Newton's second. That was actually F= MA. But we turned it into winner minus loser equals MA. It's Fnet equals MA. We said winner minus loser equals fnet. Uh Newton's third forces come in pairs. Technically forces come in opposite pairs. Remember this is a job for a free body diagram. Remember winner minus loser equals ma. Mg parallel if we were on a ramp mg parallel which I symbolized with two lines. Which trig function? So mg parallel was mg sin theta and then mg perpendicular was of course it is mg perpendicular was equal to mg cos theta. There's you're going to see two of three questions depending on which version of the exam. You're going to get an Atwood machine but with a ramp. You're going to get something moving on a ramp, but I don't ask you to find a. I ask you to find time or vi or vf or d. Spoiler alert, you're going to need to find the acceleration or you're going to get a lawn mower question. So, let's look at this first one. It says if m1 equals 12 kg, you know what? I'll even label that 12 kg and m2= 16.5. I'll even label that theta equ= 61°. Let's make that a little larger so I can read it. 61° and it says there are coe there's a coefficient of friction of.24 mu =.24. It wants me to find the tension in the rope. Sh I'm not. What do I have to find first before I can find the tension before? Yeah. But what am I going to bend my efforts to finding? We said that when there was more than one mass, you found the something of everything by writing an equation for everything. Okay, first we want to find the acceleration because remember my terrible tension deficit disorder joke. If you write an equation for everything, the T's are going to cancel. I did the stupid you can stretch out losing tension joke. Then I'll go look at an individual mass to find an individual force like tension. Probably this one because I suspect it's going to be way less cluttered. Okay. To find the acceleration, what might this be a good job for? Okay. What are the forces acting on m1? Get the obvious one. Hello. Get the obvious one. Okay. And since there's more than one force, I'll call that M1G. What else? Tension. Is there more than one rope in this question? No. So, I don't need to go tension one and tension. If there was more than one rope, that would be a three mass question. I'm not going to give you one on the final, but just to jog your memory. That's the two forces acting on that mass. That's the easy one. Now, let's go to the tougher one, m2. What are the forces acting on it? Get the obvious one. And what I told you when there's a ramp is I always draw MG a little bit through the base because it gives me one extra 90° angle that I can use as a reference. And I always put the MG on the pointy side of the ramp. Pause. Now I broke this up into components, but not vertical and horizontal. Parallel and excuse me, perpendicular. So I'll change colors. I would draw a dotted line. Here's perpendicular. And by perpendicular, I mean perpendicular to the ramp. And I would stop right there. This is going to be mg. Oh, M2G. I should have put a two here because it's the second mass. M2G perpendicular. And then parallel. Remember, parallel is parallel to the ramp, not parallel to the base. Parallel to the ramp. That's going to be dee d M 2G parallel where that's a 90° angle. And then I went like this. Color, color, color, color, color, color, color, color. Those two angles are the same. We proved it once and then we said we're not going to bother proving it. We'll just jump straight to there. What did I do next? Well, I guess I got to pause. Give me a second. Sorry, YouTube. What do I usually do next? Are we sinking into the ramp like quicksand? Are we flying through? After I did gravity, in this case with components, but even when it wasn't with components, I always tried to do the normal force next. What's the normal force going to look like? It's going to be the same size as mg perpendicular. There's my normal force number two. Which force am I missing? Friction. Which way problem? I don't know which way this 16.5 is going to slide. I mean, if this was 1 kilogram, I'm pretty sure this is winning. But they're pretty close. So before I can figure out which way to point friction, I need to do a comparison. Which two forces are really having the tugofwar in this diagram? M1G down the ramp. So get your calculators out. We don't even write this down. We'll just do a little bit remembering. Let's quickly crunch M1G. So M1G is 12 * 9.8. M1 is pulling with 117.6 Newtons, which means M2 is getting tugged with 117.6 Newtons up the ramp. Now, let's find the force down the ramp, which is going to be M2G parallel to line. Okay, so it's going to be 16.5, don't forget, time 9.8 times the S of 61. I'm lucky I've got a multi-line display. I don't know if all of you do. Otherwise, you'll have to just remember the 117 points. Oh, you know what? Down the ramp is winning. So, which way if if M2 is sliding down the ramp, which way am I going to point? Friction up the ramp. I'm going to call it friction force number two to go with mass number two. Terren, is that okay? You look a little befuddled. Um, by the way, I have multiple versions of this. I'll bet you in at least one of them it's sliding down the ramp. And I'll bet you in at least one of them it's sliding up the ramp. So, you can't just, oh, I'll see what other people saw and do it the same way. Um, okay. Now, we've got everything set up. Who's winning? That's not a force. M2 is a mass. Which force specifically from my free body diagram is winning? M2G parallel. Anything I might even say, hey, that's the acceleration. Anything down the ramp is going to be winner plus. Anything that ends up pointing up the ramp is going to lose minus. So, down the ramp, here we go. And again, Shay, remember, we just walk along the rope. These are the forces we're interested in. I don't care about that one. I don't care about that. They're not along the rope. They might show up later in disguise, but for now, it's going to be winner. M2g parallel minus friction force 2 minus tension. Did I put tension on here? You know what? I forgot to put the second tension on here. We should do that, shouldn't we? That would be a sloppy mistake. And then this tension here may if I follow it, follow it. When it gets over here, it's going to be pointing down the ramp. So, it's going to be a winner plus. And this M1G when I follow it, follow it. When it gets here, it's going to be pointing up the ramp. It's going to be a loser minus. Equals. If I had more than one mass, what was the modification I had to do on my right hand side? M all A. And then usually here, hey, if you want to, you can join me. I did the dumb stretch because alley why. And then because I know I'm going to find a I often on this line right away just said, you know what? m1 + m_sub_2 from our formula sheet. Friction is what times what? I might make a little note here that this is mu * normal force number two. And now I'm going to start to do some trig. So I'm going to say a equ= m2g two lines sine I'll just put a theta there to make it generic and I'll bring the angle in later minus mu I don't know normal force number two oh but look look look I don't know the force the same size as normal force number two which one Tristan you're telling me that this force and this force force are the same. They're not even in line with each other. It's not mg. We're on a ramp. This is why you've never ever heard me say friction is mu mg. It's mu times the normal force. I don't know the normal force. Oh, but look look look look. What is the normal force going to be here? Banner m2g perpendicular. Perpendicular. Which trig function? Yeah, of course it is. So it's going to be mu mg cos m2 g cos theta and then we have minus m1 g all divided by m1 + m2. I'm going to just try and keep this on one line. So I'm going to just move this over. Yay for digital ink. I've scrolled down. What was m2? 16.5 * 9.8 sin 61. Just put it right there, please. Thank you. Uh minus what's mu? I don't know. What's mu with you? 24 * 16.5 * 9.8 cos 61. You know what? I'm not going to have enough room to do this on one line. Dumb. And all righty. Let's rewrite it, Mr. Dick. Maybe you guys do, but I got to write large so you can see. A equals what was 16.5 * 9.8 sin 61us.24 * 16.5. Don't forget the 9.8 cos 61 minus and what was 12? * 9.8 all divided by 12 + 16.5. All of you want to try typing that in because this is the level of typing we expected you to be able to handle in your sleep. Oh, brackets are on the top bracket. 16.5 * 9.8 * sin 61 to close off the top minus.24 24 * 16.5 * 9.8 cos 61 close off the not the top close off the trig function - 12 * 9.8 Now I close off the top any typos Mr. Dick that looks good to me divided by bracket 12 + 16.5 I know it's got to be less than 9.8 because the object isn't in freef fall. So I kind of got a little built-in error check. I'm a little worried here. Oh, it is acceler. I I was worried because there was a possibility that we still got a negative answer. What that would have meant is friction was so strong nothing was moving at all. I was a little worried because I made these numbers up. It's accelerating but barely. Is that right? And y'all get.1758426. Okay. So, I was I was panicking for a second doing some of the math in my head going, did I cut it too close? So, this is sliding down the ramp, accelerating down the ramp slowly. I'm going to write 0.17 six, but I'll store this on my calculator. >> We're done. This isn't even what the question asked us to find. Oh, Shay, what did the question ask us to find? Go read at the top. To find an individual force, I'll write an equation for an individual mass. Shay, which mass would I use if I was clever, m1 or m2? Definitely. Which way is m1 accelerating? Up or down? Which way is the winning direction? Same answer. So, which force is winning? Which force is losing? equals m1a. Right? Right? An individual force for individual mass. So to find tension, I'm going to say tension minus m1g equals m1a. And then sometimes to get the t by itself, you had to do a swappy dance. This is not one of those cases cuz the t isn't negative. I just got to plus the m1g over. I'll get tension equals m1 a + m1 g. It's going to be 12 * my answer button + 12 * 9.8 12 times my answer button + 12 * 9.8 820 [snorts] if I round off properly newtons. I probably wouldn't add a part B because you can see we have to do a fair bit of writing and free body diagramming and equation solving. So that to me is a good lengthy question. What else might you see? Hey, I might give you a ramp, but instead of saying find A, I'll ask you to find Well, what's this one asking us to find? How fast is it traveling? You know what this is asking us to find? VF. Okay, Shahab, what's that eight? What is it more specific in terms of the physics concept? If they're asking me to find VF, what is it? What is it? VI. What's that? 6.4. What else do I need if I know VI and T? If I want to find VF, well, displacement. But hey, let's clue in. What am I probably going to want to find instead? Because this is what we've been finding on these. This is a blank I'm going to need to fill in. And then it's going to be, oh, you know what? Which equation eventually will I end up using? VF= VI plus A. Okay? Or I could have given you VF and said find VI. Or I could have given you VI and said find D. Then you would use D equ= VI plus have A T². Or I could ask you to find the time. But all these options. All right. We going to spend most of our time finding A. What's this a good job for? What are the forces acting on this block? Get the obvious one. Oh, you know what? Let's draw in. It says the angle is 28°. Says the mass is 5.2. Oh, and it says there's a coefficient efficient. This one's a little easier. I know which way we're sliding based on those little lines. Looks like we're sliding and the arrow. We're sliding down. So, I know friction is is going to be up the ramp. But it says that that's uh what did it say, Mr. Dick? 12. Okay. What are the forces acting on this? Get the obvious one. So, I'm going to have M1G. I'm going to go That's terrible. Kind of slanty there. I'm going to try and go through the base. And I always put the mg on the same side as the ramp. And even though I said M1, only one mass, so I can get away with just MG. Okay, let me pause for a second. People at the door. Sorry, YouTube. Lots of interruptions. Is there a ramp? Say yes. So I pause. I'm going to do components for gravity. Perpendicular and parallel. It's going to go d de. I stop right about there. And there is parallel to the ramp. Here's mg perpendicular. Here's mg parallel. That's the right angle. Color. Color. Color. Color. Color. Color. Yeah. Yeah. Yeah. Are we sinking into the ramp like quicksand? Are we flying at off the ramp like Superman? So, there's going to be a normal force. Normal means at right angles, too. So, the normal force is going to be, in fact, it's going to be the same size as mg perpendicular. I'll eyeball it. There's going to be my normal force right there. I'm close. Which one am I missing? Acceleration is not a force. Which force am I missing? Which force am I missing, folks? Someone say it out loud. Go ahead. Friction. The force of friction. Which way are we sliding? Shahab down the ramp. So, which way is friction going to point? And I'm going to assume it's shorter than mg parallel. Although, yeah, you know what? I think it Well, no. I could be slowing down. I might get a VF of like 2.6. could be slowing down. I don't know. We'll find out. I'm going to Oh, which way do we think the winner is? Down the ramp. I'm going to guess that's the winner. I'm going to guess that's the loser. If I've guessed wrong, I'll get a negative acceleration, but it was a pretty easy fix. I could crunch them ahead of time, but I'm going to go. Here we go. Winner minus loser equals ma. Down the ramp, winner. Up the ramp, loser. I don't care about the ones that are perpendicular, not part of this right now. Oh, by the way, friction is what times what? I'm going to remember that this is mu times the normal force. And I'm trying to find a. Normally, I would divide by m. I have a feeling there's an m that's going to show up here. I think I'm going to be able to cancel out the m. I'm going to hold off on that. If you did divide by m, that's fine. you would find well you would just anytime you multiplied by m you would divide by m later but I think the ms are going to cancel um which trig function to okay so it's going to be mg s I'll put a theta there not the 28 I'll generalize it minus now friction is mu * the normal force I don't know the normal force Tristan. Oh, but look look look look. And the force is the same size as a normal force and it's not mg. Of course it is. That equals ma. And yeah, my suspicion Ally was correct. Shahab, is there an M in the first term? Is there an M in the middle term? Is there an M in the last term? Can little kids and grown-ups ski on hills together? M's cancel, which is nice because I've already got the A by itself. Now, if you hadn't done that, your equation would look like this. If you have a fraction, which I have now, and there's an m in every single term as well as on the bottom, then you can cancel them out. But if you're not sure, don't. Cuz one of the most common mistakes kids make is they cancel stuff in fractions that they're not allowed to cancel. Those of you that are in pre-calc right now doing trig identities probably can attest to that. Go forward. Go forward. Right there. And I said, "Ha ha ha. All right. A is going to be 9.8 sin 28 minus shahab. What's mu? I don't know what's mu with you.2* 9.8 cos 28. Remember common silly mistakes forgetting the second 9.8 because your brain has already written 1g. It's very easy to think I included gravity. It shows up in both should get an answer less than 9.8 cuz we're not in freef fall. 9.8 sin 28 close bracket minus.12 * 9.8 8 * cos 28 close bracket you get 3.56247 and change I'll write 3.56 m/s squared should I have this is not what they wanted me to find what did they want me to find as a physics concept VF so I now know this is 3.56 but I'm going to use my answer button looking equation. It's got a VF, a VI, a T, and an A in it. There is one. Yeah. And the nice thing is the VF is already by itself. Technically a vector. Technically a vector. You know, I'm not that fussy on the vector notation. Uh, it's going to be 8 plus answer button times 6.4. And we are actually speeding up. 8 * answer button times 6.4. No, not 8 times. That would have been dumb. 8 plus answer button times 6.4 30.8 Right. So again, Adam, as far as I'm concerned, if I give you any two of d, vi, vf, and t, you can find the acceleration and then find a third one. Find a fourth. Okay. Oh, or I could give you VF, VI, T, let you find A, and then working backwards, I could have you get the coefficient of friction by itself, get the mu by itself. There was I did one question like that. It was a ski question. Um, that to me is fair game. You'll see that in your ultimate review, I think, if you go look at forces. Last one. Lawnmower questions. Here, instead of the ground being at an angle, did I say angle components? The force is at an angle. Did I say angle? Oh, did I say angle, everyone? There you go, man. I thought I teed you up, Lucas, but no. Okay. Hey, what's this going to be a good job for, Lucas? Back row. What are the forces acting on it? Get the obvious one. So, let's go. Mg. What else? What do I usually do after mg? Is the normal force going to be the same size as gravity this time? In fact, I would probably pause and I would say little arrow got chopped off a little bit. Sorry. Uh, here's my F applied 575 newtons. It's at an angle. Did I say angle components? I would say there's going to be an FX and an FY. Is that going to make the normal force bigger or smaller than mg? Why? Part of the force is lifting up. Now, the other one, I've got variations of this. In some of them, I have you pushing down at an angle. Then you would have FY pointing down and FX going that way. Hey, that would make the normal force bigger because you're pushing down. The ground has to push up harder to cancel out gravity and the y component. But here, Lucas, I would argue that the normal force is going to be smaller. And I'm going to make it stupid obvious and exaggerated. I'm going to say, hey, there's FN. So I notice >> not the same size as MG. >> I'm missing one more force, Lucas. What? Now that I've done the components, what? Yeah, better get that. Okay, it wants the acceleration. Lucas, what's the tugofwar between in this free body diagram? Which two forces? Yes. Who's winning? Probably FX cuz I would have contrived this to give you an acceleration. Or I might have said it's slowing down just to give you a hint. Then I would know friction was winning. You know what? Or or I mean I could always crunch it too. So my starter equation is this. Winner minus loser equals MA. Oh, and friction is what? Times what? Okay, I'm going to have to figure out the normal force in a second. FX, which trig function? What's the other force component? FY. And you can see, hey, yeah, it is opposite over hypotenuse. It is s fy is sign. So I ask you again f(x) which trig function? What? Of course it is? It's going to be f applied cos theta minus mu. And I'm going to write times the normal force equals ma because this one is tough enough that I'm going to go find the normal force over here. And then I'll walk it into that spot. How do I find the normal force? Are we sinking into the ground like quicksand? Are we flying like Superman? Then really what that means is everything down equals everything up. What's everything down mg? What's everything up? There are two things pointing straight up. Normal force and FY. I want to get the normal force by itself because that's what I need to find friction. Lucas, how would I get the normal force by itself? So the normal force is going to be mg minus f_sub_y. f_y. Okay, the normal force is going to be mg minus f applied sin theta. And then you have a choice if because I tend to do things algebraically. I could walk this whole expression and put it here in brackets. And here, Cole, I would have to put it in brackets. Um, or I can crunch it. And I know some of you like crunching it. So, I'm going to do both. I'm going to first of all get an answer here. It's going to be m 87 * 9.8us. What was f applied, Lucas? 575 time sin 56 and I get a normal force of double check me folks 375.9 and change Okay. And then you could walk that there or you could walk this expression there. Now, Lucas, what was this asking me to find? I probably on this line might have done that. I don't think the mass is going to cancel here because the normal force is too cluttered. So a is going to be 575 if applied cos 56 minus what's mu I don't know what's mu with you.25 25 times the normal force. You could just put your answer button there. Divided by what was the mass? Mass is 87. Okay. Or you could go a = 575 cos 56 minus.25 25 bracket 87 * 9.8 - 575 sin 56 all divided by 87. It's really up to you. I don't care. So if I went bracket 575 co 56 closed bracket minus 0.25 25 * the normal force. Answer button. Mu * the normal force closed off the top divided by 87. Do you all get 2.62 if I round off properly? And if I were to go bracket 575 cos 56 close off that minus.25 here I would need to open up a bracket because there's a minus sign there. 87 * 9.8 8 take away 575 sin 56. Close off the sign. Close off the mini bracket. Close off the top. Divided by 87. I hope I get the same. I do 2. What was it? Six two m/s squared. That's a review of kinematics.