Physics 9 Conservation of Energy (4 of 11) Roller Coaster

[@OswaldChisala]

Hi Michel! I noticed that you have a very low dislike value across your videos. It's probably because they're clear, straightforward, and for the time that you present for, really informative. Great job, and keep up the brilliant instruction! :)

[@MichelvanBiezen]

Thank you.

[@Maafyl]

I absolutely love your videos. Thank you so much for doing this. Been binge watching all your videos to review and everything is making so much sense!

[@PolarBear56]

Thank you, so much more clear than my professor

[@denissemorales5602]

Thank you so much for your work, you make me feel capable!

[@MichelvanBiezen]

Glad you find our videos helpful. We are all capable. We just need someone to show us the way.

[@historyisthebest5831]

The work done by or against a conservative force does not depend on path, it only depends on a starting and ending point. And a potential energy can be defined for any conservative force.

[@MichelvanBiezen]

That is an excellent way of looking at it. 🙂

[@Jake-jw3sk]

All I could think about is how you sound like Gru from Despicable Me, lol. Thanks for the helpful video.

[@MichelvanBiezen]

You are not the first viewer who tells us that. 🙂

[@DrachenIvy]

you have saved my physics class thank you

[@MichelvanBiezen]

Happy to help

[@Henryy97]

Thank you, you're a great teacher

[@theniftylynx6897]

thanks for the help, this helped me finnaly understand how to calculate the answer

[@MichelvanBiezen]

Glad we were able to help

[@trishac.199]

This was excellent, thank you!

[@MichelvanBiezen]

Glad you enjoyed it! 🙂

[@honestman276]

Thanks. From Bangladesh.

[@MichelvanBiezen]

Glad to help.

[@shaikaanjum9167]

Amazing lecture ❤️❤️❤️

[@MichelvanBiezen]

Thank you! 🙂

[@rosamar8132]

Are you a professor here in Miami? I would live to take a class with you

[@MichelvanBiezen]

Not in Miami, but on the other side of the country in LA.

[@joeyborja423]

After W is done taking the car to the top, it gained PEinitial then at the end when it came up to its "final"position it did not have PEfinal (due to the change on the reference zero height?). Can you explain why? Seems that the car again has stored energy to get ready to cruise down the loop and thus has to have PE not equal to zero unless it's the end of the loop. I have trouble determining these initial and final values to help me start solving the Energy equation.

[@MichelvanBiezen]

Whenever you solve any energy problem like this, even when you have a pulley problem, you are free to put the "zero height level" anywhere you like for that problem. It doesn't matter where you place it because you will always get the same answer. (Only the difference in height matters between the initial and final point)

[@joeyborja423]

@@MichelvanBiezen that makes sense lol I tried with ground level as reference and it worked! Thank you, sir!

[@LoveAlgeria54]

Dear Michel, Thank you for you video. Just a small comment: The input work is zero because it is the work of external forces and there is only one external force, on the roller coaster car, which is the normal force which is always perpendicular to the motion so its work is zero.

[@MichelvanBiezen]

Not exactly. As the roller coaster goes up and down, the angle changes and the force of gravity is NOT perpendicular to its motion. In the energy equation we do not include the force of gravity as "doing work" because it is already accounted for by considering the change in potential energy.

[@LoveAlgeria54]

I completely agree that we do not include the force of gravity as it is already included in the total energy because its work is: W(mg) = -integral(dU) = -(U2-U1) and within the "work-kinetic energy" theorem it passes to the other side to be added to the kinetic energy and be the total energy and I already said that there is only one external force that works excluding the gravitational force as it is already included in the the total energy. But I am referring here to that most of the books are saying the total energy is conserved: Einitial = Ffinal but they do not even mention that the work of external forces if zero as you did!. The only thing I added is that indeed the work of the normal force = integral [(vect(N).vect(dl))], dl is the elementary displacement, is zero because the the normal force is always perpendicular to the surface where it is opposite to the mg component which is perpendicular to the surface and the parallel component of mg makes to the motion parallel to the surface. I am sure about this. Following the action-reaction newton third law the surface react on the car which just what was applied perpendicularly by mg (of the car) and for that reason we call this force the normal force (means perpendicular to the surface and it is just the reaction to the normal component of mg on the surface). So here I would only to comment and mention this, so we know why the work is zero.

[@yilinlian7995]

Dear sir, what if the constant of friction is 0.4 , how can we solve the question ? Do we need to know the m of the roller coaster or whats the equation ? Thank you sir .

[@MichelvanBiezen]

If there is friction it would be much more difficult, because the slope is constantly changing. This would then require an equation describing the slope of the roller coaster at every point and we would need to integrate the work done to overcome the friction over the entire length of the roller coaster.

[@methadonemd3547]

If I was looking for the velocity at different points along the curve, would my change in h be the difference between the starting height and the height at different points along the curve or would my change in h be the difference between each incremental point? Like difference between A and B and B and C or always difference between A and each point?

[@MichelvanBiezen]

Yes, you can pick any point along the way and apply the same technique.

[@AngieKris]

I keep questioning something related to work. So if you pick up a suitcase and walk, is there work done? Because my teacher says there’s no work, but isn’t there work regarding walking? The question didn’t really specify if there’s work done on you or on the suitcase, which makes it confusing to answer.

[@MichelvanBiezen]

The best way to check is to ask: did you add energy to the object (did the speed increase or did the height increase?) If the answer is no, then you did not do work. (picking up a suitcase is doing work, but walking with it at constant speed and keeping it at a constant height is not doing work)

[@AngieKris]

Michel van Biezen thank you!

[@dypproductions8860]

thanks, you saves my xbox series

[@kuldeepsingh7250]

Very good concept.i am from india but i can understand

[@MichelvanBiezen]

Glad you found us. Welcome to the channel!

[@studygroupprepusmle]

Thank you for your videos. They are helping make sense of my physics course. Can you explain why you used height difference for h? why 10m vs 20m when the initial height is 20m for (mgh)initial ?

[@MichelvanBiezen]

+Sarah Groleau When calculating the potential energy, you can put the reference height anywhere you like. So I placed the reference height at 10m above the ground to make it easier to calculate the potential energy. (h = 0 at 10 me above the ground)

[@enriquesoler3150]

i have a question, could i solve this problem by keeping the initial height 20m and the final height 10m?

[@MichelvanBiezen]

Certainly. Only the difference in height matters and you can make the reference height any number you want.

[@valerkand]

Is it possible to solve this problem with consideration of friction?

[@MichelvanBiezen]

Yes, but it would be MUCH more difficult and requires much more complicated math.

[@jade67-c9e]

Hi Michel, I am lost, where did you get 9.8m/s^2? I am new to physics.

[@MichelvanBiezen]

That is the acceleration due to gravity for all object on the surface of the Earth.

[@bhaveshpanchani8173]

Thank you very much, finallyy i understand it

[@68WienerCleaner]

I LOVE YOU 😘❤️💕

[@MichelvanBiezen]

Thank you. Glad you found our videos!

[@sweet294]

what if i decided to use PEf? would it give the same results?

[@MichelvanBiezen]

Lindsy Y. Yes, you can have the reference height at any point and with a PEf not equal to zero you will get the same result.

[@statichallways7413]

love u sir

[@MichelvanBiezen]

Thank you. Glad you found our videos! 🙂

[@lew1003]

why did you put initial height (10) when its actually 20 metre under the square root ????

[@MichelvanBiezen]

We only need the difference in height between the initial position and the final position.

[@lew1003]

@@MichelvanBiezen thank you

[@liuche4303]

is there PE in the end?

[@MichelvanBiezen]

That depends where we place our h = 0 point. To make it easier, we placed the h = 0 point at the height of the end point and therefore there is no potential energy at that point. If we had place h = 0 at the ground level, there would have been PE at the end.

[@liuche4303]

@@MichelvanBiezen if we don't set the endpoint to 0 we have to add the Pef at the end which is Egi=mghf+Ekf and when we set end height to 0 we only have to calculate mghi=1/2mvf^2

[@liuche4303]

is that correct

[@MichelvanBiezen]

Yes indeed.

[@liuche4303]

@@MichelvanBiezen Thank you so much

[@SuperSuadad]

Thank you!

[@Itsmeodie]

thank you so much

[@Lucerozeus]

thank you sir!

[@chiragsaxena7251]

Sir how are you determining that final height will be 0?

[@MichelvanBiezen]

You can place the reference height as any value. It makes it easier to call the final height zero. (less calculations). Only the difference in height is important.

[@Beastboss29]

What if the initial velocity was not equal to 0?

[@MichelvanBiezen]

Then add the initial kinetic energy term on the left side of the equation. (see other examples).

[@PurpleHippoLoverXD]

very helpful!!

[@sajdahamad9980]

thank you

[@gpl3046]

🐐

[@MichelvanBiezen]

Glad you liked it.

[@redface6816]

Thank u

[@MichelvanBiezen]

you're welcome.

[@jashangill8277]

Kamla

[@generalsn1per406]

Among us

[@MichelvanBiezen]

Thank you. 🙂

[@owyvia]

Thank you so much

[@MichelvanBiezen]

You're most welcome

[@jashangill8277]

Kamla