Further Mechanics Hookes Law 9231

追加: 2026-05-25

[00:00:00]Okay, so now we're going to be starting Hooke's Law. So, Hooke's Law basically states that tension produced when a string is extended the tension produced will always be proportional to its extension, okay?

[00:00:18]So, you probably learned this in your physics. So, you know that T can be written as K of X. Here, K is the constant, but in further maths, you're going to be learning two new stuff. So, this K can be written as modulus of L divided by L, okay? I'm sorry. Uh modulus of elasticity divided by L.

[00:00:48]Here, the modulus of elasticity depends on the string itself. It should be given in your question, or they could ask you to find it as well, okay?

[00:00:58]So, this L is the natural length. The natural length basically means that suppose it does not have any load on it, okay? This is zero Newton. So, this length should be natural length.

[00:01:14]But, let's say you extended a force on it, and now there is an extension.

[00:01:21]So, this should be L plus X, right? The whole length of the string.

[00:01:28]So, the natural length is this length where there are no forces acting on it, okay?

[00:01:35]And another thing we have to understand is the elastic energy. The elastic energy is basically a potential energy inside the string, which occurs when you pull it, or when you compress it, okay?

[00:01:51]So, this formula can be find find out by using the area of uh the tension to its extension.

[00:02:02]So, this area should give me my energy.

[00:02:05]So, using that formula, I can prove that my energy is lambda square lambda x square divided by 2L.

[00:02:15]Okay?

[00:02:18]So, for energy, we have to always remember that loss in energy is equals to the gain in energy.

[00:02:27]Uh and we also have to remember that initial energy is equals to the final energy.

[00:02:55]So, these maths are really simple. I would like you guys to pause and uh have a look through it.

[00:03:04]These are really really simple maths.

[00:03:41]So, what I am concerned about is the maths after example 16.9.

[00:03:47]So, basically, in your exams, which is on 21st, you will be getting maths like these, okay?

[00:03:55]So, here in this question natural length has been given by two.

[00:04:01]So, this your in your exams, you're going to mark it immediately, okay? So, you're going to mark it as L.

[00:04:08]You're going to mark it as modulus.

[00:04:13]And yeah, that's all about it for now.

[00:04:16]The other end is attached to a particle P of mass 3 kg.

[00:04:21]The particle is pulled to the side so that its distance is AP.

[00:04:26]Okay.

[00:04:27]So, what it says is that natural length was obviously at first it was 2 m long.

[00:04:33]Then you pulled it. After you pulling it, it reaches point P.

[00:04:38]And after point P, the total length is 4 m.

[00:04:42]So, my extension must be the total length minus the natural length. So, my extension must be two, which I have written here.

[00:04:55]Okay?

[00:04:56]So, they have also told me that the coefficient of friction is 0.4.

[00:05:01]So, first of all, I need to sorry, so initially at particle P, the kinetic energy must be zero since velocity is zero.

[00:05:11]And the EPE using the formula lambda x squared over 2L, I calculated to be 60.

[00:05:20]Okay?

[00:05:21]But at A I don't know the velocity.

[00:05:26]I don't know the velocity, so the kinetic energy can be given like this.

[00:05:32]And at A the extension actually becomes zero, doesn't it?

[00:05:46]If the natural length here was two and it compresses back down here the extension becomes zero.

[00:06:02]So the EPE must be zero.

[00:06:07]So work done due to friction when the particle was traveling through here the work done due to friction must be force into the friction.

[00:06:17]Right? Since work done is force into distance. So I found the work done to friction to be this. So we know that initial energy is equal to the final energy. So I wrote basically that and I finally found the velocity.

[00:06:33]After that they asked me to find the initial acceleration.

[00:06:38]So for the initial acceleration I know that force must be tension minus friction.

[00:06:45]Since the tension must be working here and the friction is working here.

[00:06:53]So ten by using tension minus friction equals to 3A I I know the value of tension already using the formula lambda X over L.

[00:07:05]And with that I can find the acceleration.

[00:07:09]So that is 16.9 done.

[00:07:16]Now in this math it is actually mixed with circular motion.

[00:07:32]So in this math what happens here is they have given me a setup of a circular motion.

[00:07:41]So I don't know the tension currently, but I can definitely find the tension using this formula using the information above. Okay? So, we do know that tension must be equal to the centripetal force.

[00:07:57]So, using that, I have found the angular velocity.

[00:08:01]So, for the particle, with using the angular velocity, I can find the velocity of the particle using V equals to R omega.

[00:08:13]And with that velocity, I can find the kinetic energy as well.

[00:08:20]So, at that point, if I know the kinetic energy is 30 J, I can also find the point uh the elastic potential energy at that point using this formula.

[00:08:35]Using this formula, I can finally end up saying that the total energy of the system of the particle must be 40 J.

[00:08:43]Okay?

[00:08:45]Also, there shouldn't be a dot here.

[00:08:50]Okay, now this type of maths might come in your exam.

[00:08:55]So, basically in this question, they have given me an inclined plane.

[00:09:00]Okay?

[00:09:02]So, in the inclined plane, we have to understand how to resolve this, which we already do know from mechanics.

[00:09:11]So, at this point, this particle has tension T.

[00:09:16]And we do know that there is a force acting downwards, right? So, as it moves upward, it must have a friction as well.

[00:09:24]And the uh resistive force here as well.

[00:09:27]Right?

[00:09:30]So, when the particle is at rest, there is no friction acting on it.

[00:09:35]So, the tension must be equals to the resistive force.

[00:09:41]I mean the opposite force. So, using that I have found the extension of the string.

[00:09:47]So, when it is pulled down further, now there must be friction acting on it. So, I have found the work done due to friction.

[00:09:57]So, when it is pulled down and it is released from rest, at the low point the kinetic energy must be zero.

[00:10:06]And I have taken that low point to be PE to be zero, okay?

[00:10:11]And at that point my EPE should be maximum.

[00:10:15]So, at that maximum EPE I have calculated the EPE to be this.

[00:10:21]So, when it goes up and it goes to rest again, my kinetic energy became zero again.

[00:10:31]And at when it when the particle goes down and goes up, it's EPE actually becomes zero.

[00:10:42]Why? Because there is no extension at the highest point.

[00:10:47]And the PE I can just calculate using uh the mechanics formulas I you used to do in M1.

[00:10:58]And if you realize they have asked me to find the distance of the particle from A.

[00:11:04]And that distance must be this distance here.

[00:11:10]And finally using initial energy equals to final energy, I found my distance to be 1.4.

[00:11:18]But you have to understand that this distance is actually the distance from here to here.

[00:11:27]So, the distance from A must be the total distance minus D.

[00:11:35]So, the total distance I have found is using the initial questions I have seen. So, 2.4 from here, the extension from here, and this 0.5 I have found is from here.

[00:11:54]Okay?

[00:11:55]I hope that makes sense.

[00:11:59]Now, this is also kind of like circular motion.

[00:12:04]So, they have given me the natural length of 1.5, elasticity of this.

[00:12:14]Okay, so this is the natural length.

[00:12:17]And you have to understand that it is a light elastic string.

[00:12:22]Since Okay, so when this is an elastic string, you have to understand that this string when circulating can extend to.

[00:12:39]So, I have to find the extension first.

[00:12:41]I found that vertical component So, I found the extension using the vertical component where I found the tension first.

[00:12:53]So, after doing that, I know the value of tension, I know lambda, I know natural length, I found the extension.

[00:13:01]So, now I can find the radius.

[00:13:04]The radius, I can just use simple trigonometry and find out.

[00:13:11]And I can equate the centripetal force and the horizontal component to give me the omega of this. I hope that makes sense.

[00:13:22]In this question, it is very, very similar.

[00:13:26]So, the only thing different here is they have given me two particles.

[00:13:34]And the angle is actually the opposite.

[00:13:37]You can have a look through it.

[00:13:45]So, in this question what happens is I found the tension first using resolving these two.

[00:13:52]And then I resolved these two and equated them to find the tension to be this.

[00:14:00]And after that, I found the extension.

[00:14:04]And I have already found the value of sin theta using this.

[00:14:09]So, resolving horizontally now, I can see that T cos theta equals to FC.

[00:14:16]So, I can find the value of cos theta using this value of sin theta.

[00:14:20]And then finally, I can find the omega in terms of A.

[00:14:25]Okay?

[00:14:27]So, this is the final question.

[00:14:32]So, this is actually really easy.

[00:14:34]First of all, they have asked me to find the tension, which I have found using the first formula.

[00:14:42]This was pretty easy.

[00:14:43]And after that, we have applied energy method.

[00:14:48]So, in this question they have asked for acceleration.

[00:14:51]So, we know that F is equals to MA.

[00:14:54]So, we just subtracted 26 minus 7.

[00:14:58]Uh we found 26 and 7 from these two tension and we have found acceleration.

[00:15:07]So, at maximum speed, we know that at maximum speed, our acceleration must be zero, right?

[00:15:15]That's why we when we differentiate DV by DT, we get maximum speed. And at at So, basically, acceleration must be zero, right? So taking these again I put this equals to zero and then I found the extension.

[00:15:32]They have asked me to find the distance AC.

[00:15:35]So AC must be 0.6 + 1 over 2. Why 0.6?

[00:15:42]Because our natural length is 0.6.

[00:15:48]You can have a look through it and I hope that should be enough.

[00:15:52]I'm going to take one last marathon before the exam solving all the past papers. So make sure to go through them.

[00:16:03]I'm really sorry that I have to rush so much, but I hope it will make sense.