Conservation of Energy (Learn to solve any problem)
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- Опубліковано 3 сер 2024
- Learn how to solve conservation of energy problems step by step using animated examples.
Intro and theory (00:00)
The roller coaster car has a mass of 700 kg, including its passenger... (03:15)
The assembly consists of two blocks A and B, which have a mass of... (07:35)
Two equal-length springs are “nested” together in order to form a shock absorber...(09:00)
Find more at www.questionsolutions.com
Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.
Taking my dynamics exam in an hour, been watching these videos all day to cram and be prepared. Your explanations are super helpful, and I like that you pause at parts that could potentially be confusing and go over it in different terms. It is incredibly helpful, thank you for all your hard work! I'm sure you're helping hundreds if not thousands of other students with these videos. Thank you!
I hope your exam went well. Thank you for taking the time to write your comment, it was really nice and I enjoyed reading it. I wish you the best with your future endeavors, keep up the awesome work!
Reading this while writing my dynamics exam in an hour, hope you passed Ndi I also pass
same haha, did u pass?@@NyandanoSipholi
@@mohammadshabih5293 Results come out in 3 weeks
ME RN AT THIS VERY MOMENT
This is the best explanation of this topic, I've binged watch all of your dynamics videos! Loving it. The clearest explanation in the whole youtube!
Thank you so much! Really glad to hear that :)
This is my first ever youtube comment. You have an excellent channel. Explanations are clear and straight to the point. You provide great animations as well as solving each exercise and step clearly. Thank you for your great work, keep going!
Thank you very much for taking the time to write this comment. I really appreciate it :)
This entire playlist is so helpful, thank you!
Glad it was helpful and you are very welcome!
Extraordinary video on conservation of energy and dynamic great effort from the man all hail his hard work knowledge has no limits hatts off to you may your channel reach 1 million subscribers very soon your destination and destiny both are defined in this video keep working hard on it never give up⚡⚡⚡⚡
Thank you so much for your kind comment. I really appreciate it and I wish you the best of luck with your endeavors!
Thank you so much. It is crystal clear now!
Really glad to hear that! :)
Thanks for the great video. To be honest, I actually have the most difficulty in the work done by non-conservative forces portion of this topic. T1 + P1 + integral(F_nc *dr) = T2 + P2. I believe adding this topic could help greatly. Thanks again for the great playlist
I will add that to my list and cover it in the future :)
Sir you are great, and the way you teach is outstanding. Love from Pakistan
Thank you very much, I am really happy to hear that!
I'm so grateful I found this channel. Love you bro💙
Glad to hear it :) 💙
OMG thank you, finally I understand.
You're welcome! I am so happy to hear it helped you :) Best of luck with your studies.
• Some inside understandings from this video :
1) Conservative force is the force in some potential field which stores the kinetic energy of body in potential energy due to to position from reference datum.
For eg. Gravitational force for body w.r.t. earth's surface,
Spring force for a spring due to its elastic deformation w.r.t. mean configuration.
2) Work done by conservative force is path independent i.e. we can take any path or trace any distance between any planar two points in conservative force field with same work.
3) Frictional force is no conservative because, it degrades or consumes the kinetic energy to lower forms i.e. heat which is non-mechanical energy. Hence, work done by frictional force is path dependent.
4) In absence of friction in the path of motion, there is complete conservation of mechanical energy. Hence, both kinetic & potential energies are completely inter-convertible.
5) Larger is the path of motion, larger will be the loss of mechanical energy due to friction which is path dependent & directly absorbs at every local displacement along the path.
6) For bodies in motion in space or plane, there are inherent forces like friction & fluid drag (depends on relative air velocity) which consumes/absorbs the mechanical energy to rise local temperature of both bodies or fluid interfaced surface in contact. Here, total energy is conserved but mechanical energy is not.
👍👍
Thank you for the amazing work! Keep it up!
I have one question I'm a little bit confused on and would appreciate greatly if you can clear it up for me, please.
For the last problem, I get that the final gravitational potential energy for Block A is positive because it is above our chosen datum, and vice versa for Block B.
But is there any reason why in the change in displacement equation you have deltaS_A=(deltaS_B)/3 instead of -(deltaS_B)/3 when solving for deltaS_A?
Would we change the sign to positive anyway when plugging it into conservation of energy equation? I'm confused on that part.
Thank you for your kind words.
In regards to your question, it's a typo, when isolating for SA, it should have been negative. However, at 10:50, I explain that it will be negative, so the negative and negative gives us the positive, which means the previous error isn't too much of a problem since we still get the correct answer. But you are right, it should have been negative when isolating and that was a typo :(
Thank you Sir.
May Lord bless you always.
Thank you very much!
Thanks this video is very useful
Glad it was helpful!
Nice explanation 👌
Thank you very much!
I once more come for your help. I have 3 days left before my Dynamics exam which consists of the following hibbeler topics:
- chpt 12 (done)
- chpt 13 (done)
- chpt 14 (done)
- chpt 15
- chpt 16
- chpt 17
- chpt 18
- chpt 19
- chpt 22
So far I have only covered the first three chapters (12, 13, 14) so I have about 6 more chapters to go. Based on your experience of all these chapters what do you recommend is the best way to prepare for my exam in the limited time of 3.5 days I have left? Like do you recommend for example watching all these videos (which are very good!) and just doing some preliminary and fundamental exercises? Or do you think there's a better way to prepare/study these chapters? Or are there any chapters which you think aren't as important as others? Do you recommend going fast through certain chapters and spending more time on the tough ones?
Since you have covered most of these topics I thought maybe you could tell me what is the best way to prepare for these 6 chapters in a short amount of time.
Any advice is greatly appreciated!
This is heavily dependent on your studying style. But this is something you can do. Read each chapter, (if you already did, glance over it again quickly). Then go to the video that talks about that subject, they are only about 10 mins long. Each video covers 3 to 4 questions. Before I solve each question, pause and try to do it yourself, if you get stuck, un-pause and see what needs to happen next. By the time you're done the 4 questions, you will notice that you struggled with certain topics more than others. Write those topics down, and move on to the next. Once you cover everything, with whatever time you have left, do more questions on all the topics, but focus more on the topics you struggled with. That should be pretty good for you, but again, I don't know where you're at with these topics. Also, please note that I did not cover chapter 22 (vibrations). Another thing to keep in mind is, if you're good at simultaneous equation solving, then don't go through the process of doing it, plug the equations into Symbolab or WolframAlpha and get it done. That's only if you're already good at simultaneous equation solving, if not, give it some practice too.
I wish you the absolute best on your exams and do your best 👍
@@QuestionSolutions Thanks a lot !
@@pure_ligma2387 You're very welcome!
Awesome,best way to clear our concepts with the video problems
Glad to hear that. Best of luck with your studies!
very very heplful thank you so much keep it up
You're very welcome :)
In the first problem, I saw you use TB + VB = TC + VC to find the velocity C in which you got 13.1 or 14(which I got) I Used TA + VA = TC + VC to solve the problem which resulted in the same answer. However, think you have a point that we may run into problems similar to that in which TA + VA = TC + VC may not always be a guarantee.
Yes, that's true. Also, keep in mind that you can arrive at the same answer with a few different methods. Different techniques, equations yield the same answer 👍
Awesome videos on Dynamics, will definitely recommend to other students.
I would really appreciate that! Thank you so much.
Thanks again! I have a question by the last example though. If we assume the system has no potential energy in the beginning, how then can it start to move by itself?
This is more to do with the location of the datum. Since we drew the 2nd datum through the center of mass of the blocks, the displacement from the datum is 0, meaning in this specific situation, at the beginning there is no potential energy. Block B moves down due to gravity, and block A moves up due to block B moving down. Again, the assumption is made with the datum in mind. I hope that helps!
Great video. On the spring example why is the work done by spring positive and not negative when the box is dropped?
I think you might be confusing work with potential energy? 😅 Regardless, the springs were designed to "arrest the motion" so we actually want it to stop it. In most questions (to do with work), this isn't the case, we usually push something and we don't want the spring to stop it, sort of like how friction tries to stop it but it does by applying an opposite force.
Thanks for making these videos! I have a question about the first problem at 7:26
I understand that we need to recalculate the Normal force at B because our velocity at C is higher than the minimum required at loop C, but why is it that we use the radius at loop C for finding the normal force at B?
Actually, what we found was the normal force at C, not at B. At B, we made sure it was 0 for the normal force, which gave us the minimum velocity required for the roller coaster to not fall off. So at the end, we used the radius of loop C to figure out the normal force at C. 👍
Oh I see, I misread the question. I just saw that we are finding the normal at loop c also haha
Thank You so much!
@@jiseongkwon66 You're very welcome!
Hi there :) for the second example, why are the potential energies of the springs positive? I think maybe I am confusing this with the previous video's concepts, but would the potential energy not be negative as it is opposing the motion of the block? Many thanks x
I think you might be confusing work with potential energy? 😅 Regardless, the springs were designed to "arrest the motion" so we actually want it to stop it. In most questions (to do with work), this isn't the case, we usually push something and we don't want the spring to stop it, sort of like how friction tries to stop it but it does by applying an opposite force.
keep it up keep it up keep it up, you are the best!
Thank you so much! Happy holidays :)
Thank you so much for your videos - they are sooooo Helpful.
Time marker at 9:30 - why is it not 3Sa + Sb = l +2a? you have the right-side just = to "l + a". Also how did you get the original equation to start with?
Yes, good catch! That should be 2a, that's a typo, but it doesn't matter to the solution since fixed distances turn to 0. To understand how to get these equations, please see: ua-cam.com/video/IudPPGIV5QM/v-deo.html
Hi Mr. Steven. well appreciated for your effort. I need to point out something. In 7:55 I don't get why you did consider the s as 0.2 when you write the gravitational potential energy because I see that two springs compress for 0.2 m which makes 0.4 m when we add them up. So don't we need to write 0.4 m for the gravitational potential energy?
There is 2 springs but neither of them compress more than 0.2 m. 👍
Nice video
Thanks so much!
Thank you very much
You're very welcome! :)
Hello, I have to say that I truly appreciate the effort that you have put into your videos. However, upon further inspection, is it perhaps the case that you did not cover section 14.4 "Power and Efficiency"? Regardless, these videos have been invaluable for my learning, thank you.
You are welcome, and you are correct, I did not cover power and efficiency, I don't think it requires a separate video on it, its not too hard to understand :)
@@QuestionSolutions Fair enough. Keep up the good work 👍
@@DASmallWorlds Many thanks!
thank you this was helpful
Glad it helped! Keep up the good work.
Hi at 8:57, how did you get 286.7? I followed the same calculation and I got 268.7. Great video by the way. They are super helpful!!
You probably made some numerical error, I re-did it and got the same value as shown on the video. So maybe redo it and pay really close attention to exponents and negative signs :)
Sorry I have so many questions video but for the spring position equation why did you include 2(Sa-a)? I’m still rusty on these pulley problems do you have any recommendations to really get the concept of when you count some sections vs when not? I’m practicing to no avail. 9:40
No worries, see if this video helps you out: ua-cam.com/video/IudPPGIV5QM/v-deo.html
It's all about pulleys.
you make this stuff look like a cakewalk
😅
TYSM :))
You're welcome! :)
thx u bhai
You're welcome!
The second problem, I notice that you have T2 the Final Kinetic Energy as 0. I thought it starts at rest at the dadum and as it falls it should have some speed before it lands on the spring. Why would final kinetic energy be 0.
It falls and stops on the spring. There is no movement. 👍
9:02 Are we allowed to separate the energy equation into two equations: one for mass A and one for mass B?
Or we could only use one since it is a complete system? Thanks for the reply.
You can solve these questions in multiple ways, but in this problem, it's a single pully system, so you only need one single equation to solve it.
@@QuestionSolutions OK thank you so much! I watched the video about momentum and impulse and we could separate the single system of pulley into two momentum equations. That’s why I went back to this question to ask whether we could use two equations for energy. Thanks for the explanation!
@@user-km8dq2rv6h Yeah, you can do these questions in so many ways. Do whichever method you're comfortable with. All will lead to the same answer. Best wishes with your studies and keep up the good work!
When I think of conservative forces, I would think of objects, bodies, affects apply the same amount of force on each other. For instance, let say I was applying a force on the wall. In order for the wall to stay still, it would have to apply the same amount of force on me. If I was applying a force on a spring, than in order to come to a complete stop at that instant position, it must apply the same amount of force on me. This is, I think, where Newton Third Law comes in. It says for every action, there is an equal and opposite reactions. Let alone I also notice that the force that applied on me, whether is on a spring or due to gravity, would have to go same distance as previously in order to return back to it’s zero point.
Indeed! In a "perfect" world, this is how it would be, but don't forget, we lose energy as sound, heat, etc in the real world (though for most questions, we ignore this part). 👍
@@QuestionSolutions you got a point there
hi, can you clarify when to use principle of work and Energy, conservation of energy, linear impulse, conservation of momentum and angular impulse? like what kinds of situations required certain equations?
This is dependent on what's given, how the question is formulated, etc. Sometimes, you can get the answer using multiple approaches. You will get a feel for things by doing a lot of questions and instinctively know which to use. On the other hand, exams are rarely formulated to use just one method. Most of the time, you will need to use 2 or 3 methods to get to an answer.
@@QuestionSolutions thank you I appreciate it.
1:06 would the force of a bendy spring be conservative? how about a spring that doesn't have the same k throughout?
For springs, it's conservative because it has stored energy when it's compressed or elongated. Even if it's a spring that doesn't follow Hook's law, if we stretch it out and hold it, it will snap back to it's original position.
i would like to ask question number 2 . Do we consider the kinetic energy to be zero when it hit the spring, because when the block hits the spring the velocity automatically would be zero?
or maybe the kinetic energy got transfered into another form of energy like sound and thermal hence the KE would be zero?
I think you're referring to T2 in the problem? If yes, the kinetic energy is transferred to the spring which compresses it. So the energy is absorbed by the spring itself during the compression. And yes, there is minor amounts of other forms of energy like heat and sound as well. @@bagasnoor7593
wait sir so according to what i learnt last year is that the Mechanical energy is only conserved if the are only conservative forces and in the last example you used the principle to solve s_b while the was the tension in the pulley which is non conservative
These are considered "simple" pully problems, so you can still solve them as you normally would. A lot of real world factors are ignored when it comes to pully problems :)
At 4:49 why is N and W both positive? Shouldn't they both be negative for pointing downwards?
So we wrote the equation with respect to the n-axis, which means we assumed down to be positive. You can do the opposite as well, you'll still end up with the same answers.
hello, can I ask regarding to the question number 2. I don't understand why potential energy for A and B is positive ? hope you can answer my question. Thank you :)
I mean potential energy for spring
So the spring gets compressed and it tries to exert a force upwards to get back to equilibrium, so it's positive. Almost always, springs are positive. 👍
@@QuestionSolutions Thank You !
@@syairaezreen259 You're very welcome! ❤
wish me luck tomorrow is my engineering dynamics final
I wish you the best of luck with your exam! You got this. 👍
why does SA not equal to negative 1/3 of SB (10:00 part of the video)
So notice how we drew our position coordinates all facing down. That means we assumed down is positive. Once that's done, we need to follow our positives and negatives until the end of the solution. So SA is positive downwards and SB is positive downwards during the solving part. This is more to do with pulley problems than conservation of energy, so please see: ua-cam.com/video/IudPPGIV5QM/v-deo.html
When do you use this method instead of the Principle of Work and Energy?
It depends on what is given. You'll probably have to use more than 1 method to solve questions. Sometimes, you can even use either/or to solve as well. It all depends on what values are given in the question.
@@QuestionSolutions alright thanks! Definitely subscribing :)
@@gassyroyalnova2582 Many thanks :)
In the third problem, should the (delta-Sb)/3 be negative because when you move Plus Delta-Sb to the other side it should be negative. Why is it positive.
In the conventional sense, you are right, however, all we are saying is, when SA moves a certain distance, let's say the distance is "x", then the distance of SB is 1/3x. In other words, it moves 1/3rd the distance of SA. When we write our conservation of energy equation, that's when we have to think about positives and negatives based on the datum we draw. I explain this at 10:45. 👍
@@QuestionSolutions I also heard you say in certain problems like this, although there is already a dadum at the wheels of the pulley, sometimes it help to draw another dadum at the boxes two (in Conservation of energy) so that it won’t mess up the whole concept of the problem.
@@darrylcarter3691 That is correct. It's super helpful with positive and negative signs, otherwise, it can be a pitfall for students. One positive or negative sign wrong means the whole answer is wrong at the end 😢
@@QuestionSolutions another thing I did is this, after I saw you work out the problem, I draw some paths of the block delta-Sb and delta-Sa. When I derived the length of the rope into the change of displacement, I got 3(delta-Sa) + delta-Sb = 0. But since B is below the dadum and A is Above, I replaced plus delta-Sb with Minus delta-Sb and keep delta-Sa positive.
@@darrylcarter3691 I am very glad to hear you're trying different methods, which is a really good thing! Keep it up 👍
7:32 how would you find normal force at B?
We have to set it to 0 to get the required force for each segment.
in the second question I thought gravitational potential energy going down is always positive work done ? why is it minus in this case
Can you provide a timestamp? I will take a look and let you know. Thanks!
@@QuestionSolutions at 8:01 it says -2(9.81)(0.5+0.2) but i thought if gravitational energy is going downs it suppose to be + because is positive work done
@@willgggg900 I go through this at 1:55, but in simple terms, if something goes below the datum, so purple line in this case, it's negative. Why? Because work must be done to bring the object back up to the datum. I hope that helps! 👍
in the previous videos, u mentioned that spring compression will make negative work because the slow down the motion(on the conservative part) why do not we apply this concept to this problem
To which problem? Please use timestamps so I know where you're referring to. Thanks!
@@QuestionSolutions 8:05
Love your videos, but I have a suggestion for you to change the name of your channel to something like:
Pure Concept
Learn the How
or something to the same as it. just an opinion
Thank you for your suggestion 👍
Why is the normal force 0 at the top of the loop?
That's the lowest value at which the roller coaster can "just stay" without falling.
@@QuestionSolutions Thank you so much for replying. I also wanted to ask when is the elastic potential energy of a spring negative?
@@ibrahimhashmi6716 You're very welcome! To answer your question, that really depends on where your initial conditions exist. For example, if you have a spring already compressed, and you say the expansion of the spring is positive potential energy, then anything that pushes the spring to compress past the datum would be negative. I encourage you to google your question, there are a lot of articles on this topic :)
Why is the N-force at 4:15 zero?
That's when the cart is about to fall off. Otherwise, a force is applied from the bottom which keeps it on the track, and an opposing normal force is applied in the opposite direction. So when the cart is about on the cusp of falling, N has to be 0.
Why the normal force at B = 0?
That's the lowest value at which the roller coaster can "just stay" without falling.
4:37 why is the Normal force =0?
That's when the cart is about to fall off.
For the first one, i dont understand why Nc wouldnt also be zero since it is following same rules as B “right before it leaves”
Please kindly provide a timestamp so I know where to look, thanks!
@@QuestionSolutions 4:21 Nb is 0, later u solve for Nc and it gas a value. I don’t understand why Nb and Nc aren't both 0 since they behave similarly.
@@DietCokeIsGoodForYou So both cannot be zero, one of the 2 will be zero. Because if we make B zero, then C won't be zero, and if we make C zero, then B won't be zero. We just test each case. In this problem, we don't have to make C zero, since we show that if it makes it through loop B when N is zero, it will also make it through loop C. So we only need to show that. Remember, we just need the minimum height, so as long as it makes it through loop B and C, we are good.
Why is Normal force equals 0 in the first question?
That's when the cart is about to fall off.
10:38 shouldn't it be negative delta S/3?
If you listened like 10 seconds more, you would have heard me explain why SA is positive and SB is negative for that section 😅 Regardless, SA would be positive because block A moved above the datum so to bring it back down to the datum, the work done is positive while SB moves below the datum, which means work has to be done on the system to bring it back up to the datum, so it's negative. 👍
Yeah i got that sir i heared it thats the principle regarding datum
But transposing in algebra, could you help me to understand sir why
3d delta Sa +deltaSb=0 doesnt change i polarity
@@menglimarrero4296 So we drew all the position coordinates downwards, so the green and pink arrows. We assumed that down is positive for that part. Please watch: ua-cam.com/video/IudPPGIV5QM/v-deo.html
Thank you
I got it
3SA+SB=0
3SA+(-SFinal-Sinitial)=0
SA=-(-SB-0)/3
SA=SB/3
Shi hai
What does that mean? :)
@@QuestionSolutions NOTHING
🐐🐐
😅
Nice
Thank you!
why N=0 should not be = mg ??
What time are you referring to? Please provide a timestamp, many thanks!
@@QuestionSolutions at loop B 5:01
@@ziyadgames5340 If N is not zero, the car won't fall off. We are looking for the minimum velocity, which means we need to find the point at which the car is about to fall off, which would be when N=0. 👍