Practice: A vertical spring is originally 60 cm long. When you attach a 5 kg object to it, the spring stretches to 70 cm.

(a) Find the force constant on the spring.

(b) You now attach an additional 10 kg to the spring. Find its new length.

Subjects

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Intro to Friction | 40 mins | 0 completed | Learn |

Kinetic Friction | 39 mins | 0 completed | Learn |

Static Friction | 15 mins | 0 completed | Learn |

Inclined Planes | 34 mins | 0 completed | Learn |

Inclines with Friction | 73 mins | 0 completed | Learn |

Forces in Connected Objects (Systems) | 50 mins | 0 completed | Learn |

Intro to Springs (Hooke's Law) | 21 mins | 0 completed | Learn |

Concept #1: Intro to Springs

**Transcript**

Hey guys, so in this video I want to start talking about Springs and spring forces let's check out, so when you push or pull against the spring with a force let's call this force Fa an applied force the spring pushes back this is Newton's Third Law action reaction, you push against something it pushes back against you so the force of the spring will be the negative of the force that was applied on it and again Newton's action reaction that's why the same magnitude but opposite direction so this negative here means just opposite direction, so there's nothing new there what is new is that this also equals to -Kx and I'll talk about what kx is in just a second notice that there's another negative here and again that negative just means opposite direction that negative is you see this more later is almost sort of a conceptual negative to remind you that it's opposite direction but in a lot of calculations we just get rid of that negative to make things a little bit simpler, alright? So, what is K and X? I'm going to start with X, X is the deformation that the spring will experience so the idea is that if you push or pull against the spring the spring will either be compressed which means it's now shorter or it will be stretched which means obviously that it is now longer but X is the deformation It's not the actual length but it's the change in length so in springs what matters is not how long it is but how much longer or how much shorter it gets, OK? So X you can think of it as the absolute value of the final length-the initial length the reason why it's the absolute values because if you get a negative you just think of it as a positive because the direction doesn't matter, OK? So, for example here I have this first one here I was spring attached to a wall, this green thing is sort of like a base that you can sort of drag on the spring and if the spring is just sort of left hang in there it will have no deformation it's the original length of the spring X=zero we call this the relaxed position of the spring, right? Spring is chilling basically without having been compressed, in this situation over here this red dotted line here indicates the original length of the spring and now I'm to the right and that's because we're to the right of the spring here of the original length and that's because we pushed to the right with the force Fa, if I push to the right action reaction spring pushes back this way, force of spring is this way the gap between its original length and its new length we're going to call it's (deformation) we're going to called that X, OK? Now since I have two different situations I'm going to call this X1 we're going to use X2 for the bottom here, here the bottom I was originally the spring was originally here and then I pulled it this way so my applied force is Fa like that and the spring will push back with Fs this way and the gap between its original length and its new length is deformation X which I call X2, OK? So that's how F and X kind of work. K is the spring that's the second variable here K is the spring's force constants, OK? Now there's actually multiple names for this I've seen force constant I have seen spring coefficients I've seen stiffness constant or stiffness coefficient but basically this is the coefficient of how hard it is to compress a spring, K has to do with how stiff or how hard to compress or deform in general a spring is the higher the K the harder it is to the deform a spring so for example if you've ever played I'm sure you have with one of those pens that you push on it and the tip comes out you've seen inside of it there's a small spring the allows that to happens very simple and it's a thin spring and if you've ever played with it it's really easy to squeeze and sort of stretch it out and moves very easily so that spring has a very low K. Now if you had springs for example at the bottom of an elevator shaft that design for the safety if the elevator comes down following the spring maybe tries to hold the elevator in place those springs are going to be very stiff otherwise the elevator will crush them and they wouldn't do anything, right? So that K has to do with how hard it is to deform something, real quick the units for K are Newtons per meter and you can see this from the equation, force is Newton equals KX, X is meters so K has to be Newtons per meter so that this equation is dimensionally consistence, ok? So, if you ever forget the units you can just get it from the equation by isolating by leaving the X... solving the X or the K rather leaving it by itself and seeing that the units and you do unit analysis you get Newtons/meter or you can just remember that, OK? So Force, one more point to talk about the force the force of a spring is a restoring force and it's really important it's a restoring force what does that mean? It's always going to the spring is always trying to get back to its original length if you pull the spring the spring to the right the spring will pull back left and if you post a spring to the right I'm sorry to the left it pulls back to the right and if you pull to the right it pull back to the left, you can see here that we went to the right of equilibrium and the spring is trying to pull back to the left here we are to the left and the spring is pulling to the right, so if you're on the right you can be pulled to the left and vice versa so the spring is always pulling you back the spring force is always pulling the spring back to original length X=0, it starts relaxed it likes to being relaxed and it's going to try to get back to its original length, the force is always opposite to the deformation X, OK?

Let's do a quick example here and see how this works out, I have a 1 meter long spring that is laid horizontally so very similar to these diagrams here with one of its and fixed so there's sort of a wall and you have a horizontal spring, there is usually a little bar here or sort of a platform that you can drive the spring or sort of push a box against the spring against this, Ok? So it has an initial length of 1 meter as you do this I want you to remember that what matters in springs is not the actual length but the changing length so it's not the length initially and final but the X the change in length, OK? Now you're going to come here and pull with a force of 50 which will cause the spring to stretch to a final length of 1.2 meters and remember what matters is not the length initial or final but the change in length, in this case if I go from 1 to 1.2 the change is are just 0.2 meters, OK? This is my applied force over here and I want to know what is the spring's force constant? force constant is K, your book or professor might use a different word a different terminology instead of force constant but it's the same thing, right? So what is K? Well right now we only have one equation that's deals with springs we're going to have more later but for now it's just force of the spring is the negative of the force applied or the -KX. the force applied and the force of the spring are the same, I'm just going to write a 50 here and we're going to ignore the negatives, K is always a positive so I can just ignore the negative you can think of it in terms of this negative cancels with this negative but I think it's easier to just think in terms of the negative is just a conceptual negative if I'm calculating a number I'll drop the negatives and I'll get just positives, OK? And that works better, X is 0.2 right here and now we can solve this, K is 500/0.2 and you get to 250 Newtons...Newton/meter rather, OK? Now values, typical values of K that she would see in physics problems are usually going to be in the hundreds maybe in the thousands but it's usually in the hundreds ok? It's usually not really a 2 or 5 or 10000, it's going to be a few hundred so that's generally if you remember that it might give you a little more confidence when you're solving these problems but again they can really give you whatever number want, right? Professors. So how much force is needed to compress it to 0.7? So I want to know how much force I need so what is the Fa needed to compress it to 0.7? Now really important is this word here, compressed 2.7 because I said compressed 2.7....Compress 2.7 means that my length final is 0.7 which means my X is 0.3 I started at 1 I'm going to 0.7 and the X is 0.3 had I said compress by 0.7 it would have meant that my X is 0.7 or if I said a compression of 0.7 then it would also have meant that the X is 0.7 So be very careful here, am I telling you how much it's changing by? or am I telling you what the final length is, OK? In this case we're dealing with this one here but you have to be very careful, X is 0.3 and I want to know F the equation I get is FS-FA=-KX, I want to know the force is so I'm just going to say Fa equals Kx drop the negatives. K we're talking about the same spring so K is 250 and the X is 0.3 So this is 75 Newtons, OK? 75 Newtons is the force needed to get a compression of 0.3, this should make sense, a force of 50 got me and X of 0.2, a force of 75 gets me a compression of 0.3 a little bit more force a little bit more compression or deformation technically, right? In fact this is 75 which is 50 percent more and this is 50 percent more and that should make sense because this is a linear relationship if one doubles the other one will double as well, OK? so that's it for this one I want to quickly talk about Vertical Springs So let's look at how this works if you will attach a mass to a vertical spring and let the mass come down slowly, I'm going to talk about this little detail here towards the end but basically if you put a mass what's going to happen is the weight of the mass will stretch the spring, a spring if left alone won't self-stretch, right? So you need a force to make it stretch and in a situation like this, this object has a mass it's been pulled down by the earth as a weight of MG and that MG is what's going to be responsible for pulling the spring down and it will stretch the spring until they reach equilibrium. Now remember equilibrium means acceleration=0 and that also means that the forces will cancel, OK? So here's the idea before you apply or before you put a mass here this thing is an X=0, it's relaxed because it's X=0 the force of the spring is also 0, 0 and 0, OK? Now let's go through this real quick if you have a spring here you put an object slowly let it fall, once it falls a little bit it's going to have a little bit of a compression or a deformation it's going to stretch a little bit if it stretches down the spring is going to pull back up trying to restore itself to its original length, OK? And the idea is now you get a little bit of an F... little bit of an X and you get a little bit of a force of a spring, let's play with some numbers here, let's say MG is 100, Ok? You start with the force of 0 for the spring but then you let it fall a little bit and now the force of the spring let's say is 10 and you have 100 going down and a force of spring out of 10going up, right? Now you let it fall a bit more this 10 becomes a 20 it's still weaker than MG so if you release it falls a little more and so on so forth and so this number is eventually 100, right? Why? Because this X kept growing so this F keeps growing, right? And the X will keep growing, this thing is going to keep getting pulled down until these two numbers are the same, once they both are 100 once the force of the spring reaches 100 because it's been deformed so much that it's pulling back with a full 100, now these forces are the same they cancel and if you let go it doesn't actually stretch anymore because the forces cancel, OK? So you can imagine that the if you were sort of graph of this force it would look like this and eventually you reach 100 as your deformation gets bigger and bigger, this is sort of a force deformation, here right? So once it gets 100 they will exactly cancel each other and because they were exactly cancel each other, I can say that KX which is a spring force equals Kx and MG which is the gravity will equal each other and that's the equation that you need to remember so there will reach equilibrium and I will have that KX=MG, OK? This is what I like to think of as the vertical spring equilibrium equation, OK? Now that's not the official name of the equation but I give it a name because I really want you to remember that every time you have a spring at vertical equilibrium with a little mass it's just hanging there stopped this equation will be true, now this is only true if you let the mass come down slowly so the idea is that you put a mass in here and then you hold on to the bottom of the mass and you don't just release it you sort of let it come down slowly until it's reached equilibrium once its reached equilibrium if you keep moving your hand down this thing will stay there, right? Now if you just release what's going to happen is instead of getting to equilibrium it's going to pass equilibrium and sort of go up and down it's going to oscillate around the equilibrium point that's not what we want that's a future chapter, OK? So for this equation to work you have to let the mass come down slowly. My last point is that this setup also applies if you have a mass on top of a spring, so imagine I have a spring like this and then you have a mass and top of it if I release the mass is going to do, right? But if I release slowly and just kind of level slowly compress and eventually I'm going to release and that's not going to move anymore, it works exactly the same KX=MG still applies in that situation as well because the forces are canceling just the same. Cool? So that's it for that I want you guys to do this practice problem and remember the stuff we talked about, the X is the difference in size or length not the lengths themselves, remember that this equation applies if you have vertical equilibrium and all the other stuff we talked about let's try to do this and see what you get.

Practice: A vertical spring is originally 60 cm long. When you attach a 5 kg object to it, the spring stretches to 70 cm.

(a) Find the force constant on the spring.

(b) You now attach an additional 10 kg to the spring. Find its new length.

0 of 2 completed

A spring with a force constant k stretched by some length d exerts a force of F. If the spring were stretched to 2d, the force would be
a) F
b) 2F
c) 4F
d) 0.5F

a) Two springs with spring constants k1 and k2 are connected in parallel. What is the effective spring constant keff? In other words, if the mass is displaced by x, find the keff for which the force equals F= -(keff)x.

Two springs with spring constants k1 and k2 are connected in series. What is the effective spring constant keff? In other words, if the mass is displaced by x, find the keff for which the force equals F = -(keff)x.

A vertical spring has a length of 0.225 m when a 0.25 kg mass hangs from it, and a length of 0.75 m when a 1.975 kg mass hangs from it.Find the spring constant k in N/m.

A vertical spring has a length of 0.225 m when a 0.25 kg mass hangs from it, and a length of 0.75 m when a 1.975 kg mass hangs from it.Find the unloaded length of the spring in cm

A block with a mass of 0.28 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.0 N on the block. When the block is released, it oscillates with a frequency of 1.1 Hz.How far was the block pulled back before being released? Express your answer with the appropriate units. |Δx| =

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