[Music] hey everybody welcome to the fy6 classroom here the F in physics stands for fun today we're talking about Springs now the thing that makes Springs interesting is that you can push on them and pull on them and deform them but they always go back to their original shape now you can stretch these things so much to where they Bend and break and then they don't work anymore but as long as you're kind of reasonable with how much force you apply pushing and pulling they'll typically go back to right where they started now Springs come in all different forms but we usually see them in these metal coils I have here basically just a metal coil of wire and we actually have all sorts of uses for Springs and these two I have right here came off this little motorcycle I had sitting around this one was used in the back brake to release the back brake so it kind of coins Springs it back out this one right here which is just like absolutely brutal this is like such a tight spring I can take apart this one actually came off the kickstand of the motorcycle I can kind of get a little bit in there I came off the kickstand of a motorcycle but also over here on this tool right here I have some Springs right in this area so if you were to compress this tool on these wire cutters by the way not sponsored by Klein tools the spring pushes it back out it has like a restoring force it restores this back to its original state so you can squeeze it together this is actually kind of a weak spring which is good because I don't want to be like fighting as hard as I need to to squeeze this down there's also Springs inside of this pin right here if you open it up there's there's a little spring right there Springs are used in all sorts of devices like very very small things like the spring in this pin but also you could go to very large Springs like what's in a car suspension those are very Springs now when you apply a force to a spring like this into forming it and changing its shape but it's an elastic material which means it will return to its original shape after the force is gone but like I said don't think you can just stretch these as much as you want and it'll be fine they eventually do get to a point where they break depending on how strong the spring is now some Springs like this one right here will break a lot easier than this one over here now a very bitter angry man back in the 17th century spent a lot of time looking at Springs because he probably didn't have Netflix or a girlfriend or anything this guy's name was Robert Hooke his most famous contribution to science was what he found out about Springs his second most famous contribution science was basically talking a lot of smack about Isaac Newton anyways Robert Hooke found out that the relationship between the force you applied to a spring and how much it stretches was a linear relationship so if you were to make a force versus distance graph you would get a nice straight line like this and like I said these things have a limit and when you reach that limit it will start to curve and get this weird-looking line right here but everything before that limit makes a nice straight line showing a linear relationship now there's an equation out there that's called Hookes law Hookes law is F equals KX f is the force that you apply to the spring X is how far the spring is stretched and K is the spring constant now let's talk about that X for a second X is not how long the spring is but how long did you stretch it okay so a spring has like an equilibrium state right now the spring is in its equilibrium position it's not stretching or compressing this is just where the spring rests naturally but if I were to apply a force I have deformed it and I've stretched it just a little bit so I have changed the distance so that's what that X is it's how far did the spring stretch it's not how long is the spring which could be confusing but it's just how far did it get stretched now K which is the spring constant is really just a number that represents how stiff the spring is to hire the spring constant the stiff of the spring the units for your spring constant are going to be Newton's per meter and you're gonna see us work a few problems out with spring constants here in a minute now just uh just as a little side note sometimes you'll see the equation written as F equals negative KX with that negative sign there don't worry about it not a big deal on the AP Physics one formula sheet it's actually written as the absolute value of the force of a spring is equal to K times the absolute value of x now all this comes from like people who are really particular about their signs the positives and negatives and all that kind of thing now I do want to say the sign is important a lot of times the sign is very very important don't get me wrong anyway back to Hookes law I'm just gonna use F equals KX for the equation and we're gonna do a couple practice problems I'll do some easy ones and then they're gonna get harder and harder as we go if you want to learn how to do these pause the video and try them out on your own and then check and see if you solve them correctly if you're not confident in doing these on your own don't worry just give it a try we don't judge here in the FY six classroom and without further ado let's do some practice problems here guys all right here we go one Nova spring is attached to a wall a 25 Newton pool on the other end it causes the spring to stretch by three centimeters what is the spring constant okay so let's just draw a little picture here here's my wall we also know that the spring stretches a distance of three centimeters which we need to convert that to meters which is 0.03 meters and the question is asking what is the spring constant how stiff is this spring all right so let's pull out the Hookes law equation F equals K X and we will plug our numbers into the appropriate spot so F is 25 Newtons K we don't know that so we're going to leave that as the letter K and X is 0.03 and then we're going to need to solve this in our calculator all right so we're going to divide both sides by 0.03 and the five by 0.03 that cancels out over here we're left with 25 divided by 0.03 833 so K is eight hundred and thirty three point three Newton's per meter and typically I'm okay with that answer I'm fine if a student of mine turns that in but if we want to be specific about significant figures we need to look up here and say okay there's only two sig figs here two sig figs here so our answer really should only have two significant figures so we'll say 830 Newton's per meter and that would be our answer right there awesome first one easy all right our second question here has two parts to it this one's more of a conceptual question but you know good one to learn from so a spring has an unstretched length of 10 centimeters and it exerts a restoring force F so it doesn't give us a number it just says it's F when stretched to a length of 11 centimeters for what total stretched length on the spring is its restoring force 3 F so basically if I were to reword this a spring is 10 centimeters long on its own but when you apply a force F to it it goes out to 11 centimeters so the first question is how much is it going to stretch if we triple the force all right so let's look at this real quick so if you think about the equation for Hookes law F equals K X I want to rearrange this equation to where X is only the is the only thing on one side of the equation okay so let's uh let's do that real quick divide each side by K and then it cancels out so I have F over K equals X all right so that's a good first step to the equation or to solving this problem I should say now let's think about it with following what I call the rule of one so if I were to plug a 1 in for everything on the left side what's that going to give me an answer for on the right side 1 right okay so my force is F and now I'm going to triple that 4 so I'm going to make it 3 times greater so instead of a 1/4 the force I want to plug in a 3 for it and what's my answer gonna be well my answer is going to be 3 and what this tells me is that whatever my answer was originally if I triple the force it's gonna triple my answer so let's look at this so originally it only stretched one centimeter with just the force F but now that's gonna stretch three times further so it's going to stretch three centimeters so it's going to stretch three centimeters so the question here is for what total stretch the length of the spring is its restoring force three F so basically it's a really weird way to ask the question how long is the spring if you triple the force well it's going to stretch three centimeters so it was at 10 centimeters and now it's going to add three centimeters to it so the answer is going to be 13 centimeters so if you triple the force and you're stretching it it's going to stretch three times further that originally stretched one centimeter now it's going to stretch three centimeters now for Part B that what compression link is the restoring force to F sword now we're pushing the spring together so we can think about this we can think about this logically right so if you apply a force F to it it's going to move one centimeter if you apply a force three F to it it's going to move three centimeters so what do you think is gonna happen if you apply a force to F to it right yeah try to figure it out two centimeters okay so it's going to move two centimeters but now we're compressing it we're not making it longer we're not stretching it we're compressing it and the thing is this whole law of Hookes law and how it's a linear relationship that works with its compressions or with its stretching so with that said we know that it's to compress two centimeters in so and it started at 10 centimeters and at what compressed link is the restoring force to F so we start off at 10 centimeters and then we are going to subtract 2 centimeters away from it so that gives us a length of 8 centimeters so that's how long the spring is going to be when you compress it with twice as much force as whatever this force was right here okay so this one was more conceptual we did a little math to figure it out in some logic but hey you got to just kind of understand the relationship it's a linear relationship and if you double the force you double the distance the spring stretch also if you triple the force you triple the distance that it is stretched or we can say compressed as well all right cool all right let's look at a third one a spring is stretches five centimeters when a 0.20 kilogram block is hung from it if a 70 kilogram block replaces the 0.2 kilogram block how far does the spring stretch all right so let's just go ahead and make a little list of everything we know we know that the string stretch is spread sheets stretches 0.05 meters I took that centimeters and just converted it to meters that's all I did there we know that there's a mass of 0.20 kilograms and so that's that's like our first mass and then we got a second mass of 70 kilograms and that 70 kilograms is later going to replace the point 2 kilogram block how far does the spring stretch all right so let's draw a little diagram here of what's going on so you got the spring and it's hanging from the ceiling and there's a block hanging from it so there's really like two different ways to go about this problem I'm gonna go about it both ways to kind of show you so first off let's go ahead and use Hookes law to figure this out now we know that there's a force of gravity pulling down on this block and the force of gravity in this situation is going to be equal to the force that the spring has on it okay so however hard gravity's pulling down on it the spring is pulling back with the same amount of force so we know that the force of gravity is mg and the force of the spring is KX and we can plug our numbers in we're going to go ahead and use everything from the lighter block and we're going to figure out what the spring constant is because that's going to be important to figure it out figure out how far it stretches with a heavier block so if we figure out the spring constant we know that the force of gravity is mass times gravity our first mass with 0.20 gravity we're just going to put the number 10 in there round it up from 9.8 to 10 we don't know K that's what we're going to solve for and our spring constant is 0.05 so let's go ahead and solve for this 0.2 times - and then you divide that by point zero five you get 40 so our spring constant in this case is going to be 40 Newton's per meter and we're going to need to use that to figure out how far it's going to stretch so I can use Hookes law again F equals K X and we know that the force of the 70 kilogram block so that's going to be mg equals K X and now we're going to use the 70 kilograms all right in we're back so the mass is 70 kilograms gravity we're gonna keep that at 10 K we found that out just a minute ago was 40 and X we're gonna figure that out now okay so 70 times 10 well we know that's 700 so 700 divided by 40 oh that's 17.5 so X ends up being seventeen point five meters so that is how far it's going to stretch also let's look at our sig figs here we got 50 oh sorry two sig figs three sig figs two sig figs so our answer can only have the least number of sig figs we have so our our answer we're gonna round that up to 18 meters okay cool now like I said there's gonna be two ways to solve this problem the first way is you know going this route right here also the second way is you can think about it as we got point two kilograms and then you got 70 kilograms so your force is going to multiply and get a lot larger now how many times larger is this force now so we could say 70 kilograms divided by point two kilograms so the force is 300 with 350 times larger than what it was before that means it should stretch 350 times more than what it did so if we have 350 and we multiply that by point zero five to represent the five centimeters that it was originally stretched you would get seven 8.5 you round that up to 18 so you could have done it going the mathematical route like I did here or you could have done it conceptually in thought of the fact that hey how many times larger is the force now so it should stretch that many times greater okay all right now we have a 60 kilogram student is standing atop a spring in an elevator that is accelerating upwards at 3 meters per second squared the spring constant is 2.5 times 2 no the third news per meter by how much is the spring compressed okay so this is well this one's a little bit more tricky and involves a little bit more stuff some things that you've probably learned from the past okay so this whole thing is accelerating upwards at 3.0 meters per second squared and the mass of the student is 60 kilograms and the spring constant is 2.5 times 10 to the 3rd Newton's per meter and we're trying to figure out how much is the spring compress so we're looking for the spring compression now one thing you got to remember about a scale or whenever you're standing on something like this is that it is compressed due to the normal force so however much normal force you have the spring is going to compress that same or have that same force applied to it so you got a normal force going upwards whatever your normal force is is going to be equal to the force of the spring in this situation okay so I'm going to do a little side work over here and I'm going to say that the normal force is equal to mass times gravity and that's true if you are on a surface that is not accelerating upwards or downwards but if you're accelerating upwards then you need to add the force from your acceleration and I could rewrite that as normal force equals M times G Plus a like so and that works just fine and now we're going to come back to our main problem over here and we say that the normal force is equal to the force of the spring okay so I'll rewrite my equation I'm going to go a little lower because I want to start overlapping my normal force is M G Plus ay equals and then the force of the spring is KX alright so now I'm going to start plugging in my numbers and solve for how far the spring stretch this is the easy part right here so 60 g is 9.81 pretend that's a plus sign and then the acceleration is 3 equals K which is 2.5 times 10 to the third x PEX all right so here we go we're going to just go and solve for the left side there's a few more numbers here that's going on remember do the parentheses first 9.8 plus 3 equals twelve point eight multiply that by 60 times 60 you get seven hundred and sixty-eight five times ten to the third multiplied by X now I need to divide both sides by two point five times ten to the third but about two point five times ten to the third now if you're using a calculator that's not the iPhone maybe like a scientific calculator don't forget to put this in parenthesis down here all right so what I'm going to do I know that two point five times ten to the third is the same as two thousand five hundred so I'll do 7th out or sorry 768 divided by two thousand five hundred and you get a zero point three zero seven two so x equals and let's see how many sig figs do we have we have two two and only two here so your answer would be 0.3 one so zero point three one meters [Music] right now moved on to a very light ideal spring of spring constant which also is force constant 250 Newtons per meter is 0.15 meters long when nothing is attached to it it is now used to pull horizontally a 12.5 kilogram box on a perfectly smooth horizontal floor you observe that the box starts from rest and moves 0.96 meters during the first 1.6 seconds of its motion with constant acceleration how long is the spring during this motion well let's start off with drawing a picture here's your box here's your spring and there's a force being pulled that way ok now we know that there is an acceleration of this entire system and it's said it well it didn't say it did it hmm well we know the whole system is accelerating because it started from rest and then it moved so it had to gain some sort of velocity throughout the whole time we're probably going to need to find that acceleration and then let's go down we know that the spring constant K is 250 Newtons per meter we know that the length of the spring is 15.15 meters long when nothing's attached to it I'm just going to underline that as important information I'm not going to put that in my list down here because you know you might put that as a letter X and then that gets confusing with all the different X's that might go on here so I'm just going to underline that no hey I'm going to come back to that later and then the mass of the box is 12.5 kilograms it's perfectly smooth floor so no friction you observe that the box starts from rest initial velocity of 0 meters per second and it moves a distance all right so that's the final position of 0.96 meters and it does all that in a time of 1.6 seconds ok so like I said we're first going to need to find the acceleration because the whole premise of this problem is going to be the force applied is going to be equal to the force of the spring so we need to find how much force is applied to figure out four supplied you would say mass times acceleration it's going to equal to K X we don't know this acceleration we do know the mass but we don't know the acceleration so we got this other information here that we can use to find the acceleration now go back to your kinematics equations and you'll remember the equation x equals x naught plus V naught T plus one-half a t-square so this kind of like side work to find the acceleration so our final position was 0.96 our initial position was zero our initial velocity was zero our time one point six plus one-half a our time is one point six squared alright cool well this zero goes away zero times anything is zero so that part goes away as well then we're left with zero point nine six equals one-half a times one point six squared got one point six times one point six equals two point five six multiply that by point five which is one-half one point two eight so you got zero point nine six equals one point two eight a divide both sides by 1 point 2 8 divided by one point two eight so 0.96 divided by one point two eight the acceleration is zero point seven five meters per second squared so we got to use that and plug it in over here so we got twelve point five times zero point seven five equals 250 and then we need to figure out how much it stretches we got twelve point five times 0.75 times there we go nine point three seven five equals 250 times X divide both sides by 250 we're gonna get about 250 we get zero point zero three seven five zero point zero three seven five that's how far it stretches and meters and the question is asking actually how long is the spring the entire spring so you're gonna have to add it to that point one five so we'll do is zero point one five plus zero point three seven five point one five you get an answer of 0.1 875 meters and like I said if you were to turn that into me I would count that as fine but if you want significant figures you got to look up here we have three sig figs three sig figs 3-6 6-3 6-2 sigfigs so you can only have two significant figures in your answer so we could have we would have to say zero point one nine meters as your answer [Music] all right so a three kilogram brick rest on a perfectly smooth ramp inclined at 34 degrees above the horizontal the brick is kept from sliding down the plane by an ideal spring that is aligned with the surface and attached to a wall above the brick the spring has a spring constant of 120 Newton's per meter by how much does the spring stretch with the brick attached now it says it's a perfectly smooth surface so we're gonna ignore friction and let's think about this we got mass equals 3.0 kilograms we got an angle of 34 degrees we've got a spring constant of a hundred and twenty Newton's per meter and by how much does the spring stretch so we're looking for X how much does it stretch and we need to get a smaller marker here now let's actually zoom in on this picture and get an idea of what's going on here [Music] all right so you know that there's a force of gravity pulling down on the brick but the spring is working horizontally or I shouldn't say horizontally but in the same direction that the ramp is all right so we usually say that the ramp going up the ramp is the X direction and perpendicular to the ramp is the Y direction so let's think about this let's break this into a right triangle right here so you got the force of gravity in the Y direction and we got the force of gravity in the X direction that's how I like to denote which ones which one of these directions there are okay so you got FG X which is parallel to the plane and whatever F G X is however much force there is pulling it down the ramp the spring is applying the same amount of force upwards so with that said we're going to be able to say that whatever F G X is it's going to be equal to force on the spring all right so how do we figure out what F G X is well if we got FG x equals force of the spring so let's come over here we're gonna use our sohcahtoa and we're going to figure out okay well this is the angle of the ramp so that's going to be equal to this angle right here and this angle is if we're worried about the FG x that's opposite of our angle so that gets rid of cosine that's not going to be useful to us and we do know what F G is we can figure that out actually so we know the hypotenuse so that gets rid of Toa here so we're gonna use sine so we say sine theta equals F G whatever the opposite is which was F G X divided by F G and we're going to rearrange this equation to where its F G X so we could say F G at sorry F G sine theta is equal to F G X there you go now you can plug F G sine theta in up here so we're going to say sine theta is going to be equal to force of the spring and we know that the force of gravity is mass times gravity so mg sine theta is equal to the force of the spring and will say mg sine theta is equal to K X now we're at a point where we can plug these numbers in alright so let's go ahead and say 3 times G which is 9.8 sine of 34 is equal to K which is 120 and we don't know what X is that's actually what we're going to be solving for so let's go ahead and solve for this in our calculator if you're using an iphone like me you want to turn it over to where it has the bigger menu there when you want to do the sine 34 first so you do 34 and push sign right here get point five five Allah make sure it's in degrees if you push radians it'll say Radian right there but we don't want that multiply that by three you multiply that by nine point eight you get sixteen point four four but we're going to end up dividing both sides by 120 so all of this divided by 120 you get zero point one three seven 137 meters and then finally with our sigfigs let's look up here we've got two sig figs and three so we can only have two sig figs in our answer so we'll say x equals zero point one four meters all right well guys that was uh that was a lot of fun I hope you learned a lot about Springs and also how to calculate and solve for spring problems and we'll see you in the next video bye all right guys if you liked this video make sure to hit that like button especially if I saw one of your homework problems for you also if you want more content and need more help in the future go ahead and hit that subscribe button see you later [Music] you