Physics 9 Conservation of Energy (7 of 11) Atwood Machine

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In this video I will show how to calculate the final velocity of an atwood machine.
Next video in this series can be seen at:
• Physics 9 Conservati...

[@oceanusprocellarum6853]

This is actually really helpful, with the confusing potential energies! Thanks a lot!

[@MichelvanBiezen]

Glad it was helpful!

[@ta2000-g7g]

Thank you professor ❤, it was exactly to the point.. concise , clear , and perfectly helpful!!

[@MichelvanBiezen]

Glad you found it helpful. 🙂

[@elmanahmed6117]

Very informative, professor! Thanks . Keep up the good work.

[@ozandabak]

If I can pass my exam I swear you will go to heaven :D thank you :P

[@robertktw]

im used to solve for acceleration or tension in the string with Atwood machines, didn't think about Conservation of Energy before. Thank you for showing this! Happy Holidays!

[@arboretum8378]

Fantastic video, thank you!

[@MichelvanBiezen]

Our pleasure!

[@subdelnormalbravoarge2117]

great explanation! just a question about a physics problem i found. if the cube m1 (the one with less weight) has an elastic force pushing it down. that gets added to the formula as well? it would be something like: m1gh = m2gh + 1/2 (m1+m2) V^2 + 1/2 K x^2. is that ok?

[@historyisthebest5831]

Sir I have a question.. If m2gh is 0 then that means m2 is at zero height. So should the two masses stop and have a final velocity of 0? And thus final kinetic energy for both mass equals 0? But why is there kinetic energy?

[@gagaaicee4367]

You're a lifesaver! Thank you!

[@mochigojo9753]

I have a question!! Does the sign of h matters? Cos just like what you did I answered the problem in our exam just like this. But some of my classmates treated h as a displacement so one mass' height is negative and such. And since it was moving downward, the velocity was negative. I didnt do those stuff.

[@MichelvanBiezen]

You can do the problem either way. It is more a matter of preference or how one wants to approach the problem.

[@amrobaia5731]

i really need to know why we didn't use kinematics, like vf^2=vi^2 +2a(xf-xi) ? and thanks

[@MichelvanBiezen]

One can use Newton's second law to find the acceleration and then the equations of kinematics to find the velocity and time (which is shown in earlier playlists), but this is a different technique, using the conservation of energy equation.

[@kawaiionion1022]

Hi sir, could you please explain to me how we know that the height for the 2kg will be 2m. I am a bit confused regarding this. Is it an assumption we can make? If so could you please explain it. Does this mean that the height lost by the one block is equal to the height gained by the other? I saw another comment ask something similar. The comment said that if both blocks were suspended 2m above the ground - the lighter one would go to 4m. I don't really understand why. Practically it does seem right that it will go up 2m but what is the theory behind it?

[@MichelvanBiezen]

Since the 2 blocks are connected with a string, however one block moves, the other block must move the exact same distance, since the string has a fixed length.

[@iftikharakhan1256]

Great Examples! Thank You! :)

[@kwakuappiah9996]

excellent vid...thanks

[@jhonatanibarra7356]

thanks for your help

[@ariannekho]

Great video, very informative. I was wondering why would m1 go to the same height of m2 was at before (2 meters)when we let go of the system? like since m1 is lighter why doesn't it go even higher than 2 meters? thank you :)

[@MichelvanBiezen]

+Arianne Kho They are connected, so they will travel the same distance.

[@user-ec8yk1ed8c]

May I ask why would they travel the same distance just because they are connected?thanks

[@jvemon5343]

It's not very clear to me why is the initial PE based on the second object, whereas the final PE is based on the first one - that is, what dictates which of the two objects should be used to calculate the PE? At first glance I would have used second object for both the initial an final PE. I am guessing that the PE of an entire system is the sum of the PE of each of its objects. Initially the second object has PE because it is not on the ground, whereas the first object doesn't have PE, so naturally we simply use the second mass to calculate the PE of the system. At the end it is the other way around, where the first object is the only one that has PE so that's why we don't use the second object at all. Would that be correct? Thanks!

[@MichelvanBiezen]

The PE is calculated to a reference height. You have to pick a point and call it zero height. (That point can be picked anywhere, but traditionally we pick the lowest point like the surface in this example). In the beginning only the large object has PE because ii is above the surface. The small object is sitting on the surface and thus has zero height. (PE = mgh) At the end it is reversed.

[@jvemon5343]

Suppose the exercise begins with both masses halfway through (so the second mass has a height of 1m and the first mass as well) - is the initial PE still just (m2 * g * h) ? Or do we have to add (m1 * g * h) as well? Assuming that the reference height is still the surface below.

[@MichelvanBiezen]

Then you will have to add the PE of m1 as well since then, both objects will have potential energy. (or you can set the reference h = 0 at the location where both masses start and then neither mass will have PE initially.

[@sivakumarsiva1743]

This can be solved using equation of motion If we solve for acceleration we get 5.88ms-2 Initial velocity is 0 and height is 2 So v2-u2=2 as V=root of 2 as V= sqrt 2×2×5.88 V= 4.8 ms-1 It can be easily solved like this rater than using conservation of mechanical energy

[@vyomgupta8877]

the formula you applied for calculating the acceleration has been derived from this method only, but surely no offence.

[@albertopoli8896]

Sometimes I mistake the Vf , for example,in this case I have thought that Vf=0, because the mass arrives at ground and it stops here, so velocity is zero because it’s at rest......

[@MichelvanBiezen]

Yes that is a common misconception. The final velocity is the velocity at which it strikes the ground.

[@kristjandeda5166]

Hi, can I use this method even if the pulley has a given mass and radius?

[@MichelvanBiezen]

No, these videos assume the pulley has no mass. If the pulleys have mass then you need to use this different techniques: Physics - Mechanics: Moment of Inertia and Rotational Kenetic Energy (2 of 3) Pulley System kzbin.info/www/bejne/a5uYnmaYrKhjhLc or take a look at this playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 kzbin.info/www/bejne/jnnOdX2thpmLpMk

[@iamdanish99havocx79]

hi, since the work done by 6 kg mass is negative, why wont the final equation be: m2gh= -m1gh+1/2mv^2

[@MichelvanBiezen]

In order to write the energy equation, you must have positive values on both sides of the equal sign. KEo + PEo = KEf + PEf

[@cocothetimeless8382]

does the W stand for work done by non conservative forces?

[@MichelvanBiezen]

W represents the work put into the system by an external force (not including gravity which is accounted for with the potential energy).

[@_tasneem7378]

Sir, can you do one since that Atwood machine be inside an accelerated elevator!!

[@MichelvanBiezen]

We have one like that in the playlist (not in an elevator) but the atwood machine accelerating, which is the same thing Physics - Mechanics: Applications of Newton's Second Law (10 of 20) pulley with constant velocity kzbin.info/www/bejne/fpLJe5t-Yttko5o

[@MohamedAshraf-ch8my]

great explanation, thank you

[@MichelvanBiezen]

You are welcome!

[@wardaavery4401]

Hi professor Michel Van Biezen, how do you know for sure that m1 will have the same height as m2? Because I was trying the same kind of problem but only with a different mass and height (block B had an initial height of 10 m and a mass of 4,90 kg en block A had an initial height of 0 m and a mass of 0,55 kg. So I was also thinking that block A will rise till 10 m above the ground. But my solution was wrong. And I don't know why? I hope you will explain it.

[@MichelvanBiezen]

It depends on the question. After the large block slams onto the floor, the small block will continue moving upwards due to its kinetic energy. Therefore when the small block reaches a height of 10 m, you'll have to use the conservation of energy to determine how much higher it will go: v = square root (2 g h)

[@wardaavery4401]

Okey, I think I understand it much better. So what I did was: For the first situation I calculated the velocity (same methode as the video) and by situation 2 when block B slams the floor I used this equation: mA g 10 = mA g hmax + 1/2 (m total) Ve^2 (Ve is the velocity that I calculated in situation 1). (mA= mass of block A en h max= the maximum height that block A will get. Is this plausible? Thank you

[@wardaavery4401]

hmmm, I think that I did something wrong, because my solution is not right.

[@farahjamal3257]

Thank you very much!! i had the same problem though.

[@user-pf9jg8ek9s]

They have the same acceleration

[@tomsakmens5571]

You can add aditional question to this problem - after m2 lands, how high will the m1 "jump" due to inertia? :D

[@MichelvanBiezen]

I believe we have an example video like that.

[@fikrihaziqahmad7995]

Professor how do you know that there is no work while the pulley is moving?

[@MichelvanBiezen]

If the pulley has no mass (or the mass is so small that we can ignore it) and if the pulley has no friction, then it doesn't require any work to turn the pulley.

[@sakshamjethi4520]

Amazing explaination

[@leonitasn9527]

Sir , what if I am asked to find the maximum height of m1 block , how can i calculate it or its just the same as 2 meters ?

[@MichelvanBiezen]

When the large block reached the ground the small block will be at a height of 2 meters moving up with some velocity (found in the problem). Then think of the small block as a projectile moving up vertically, how high will it go (above the 2 m) with that initial velocity.

[@leonitasn9527]

this was smart , i tried to do it using conservation of energy and ended up with a mess . Thank u so so muchhhh . God bless u

[@CupsAreUseful]

great! so helpful.

[@MichelvanBiezen]

Thank for the feedback. Glad to be of help.

[@007myzorro]

Amazing Michel. What does you do in LA if I might know? Friendly Bernard van lierde. PS Visualising all your amazing videos is an beautiful challenge..

[@MichelvanBiezen]

I work as an engineer and also teach at a local college and a local university. Have fun with the videos.

[@007myzorro]

I m Very glad for you Michel. A little audacious perhaps (you are not obliged to answer), but how much earns an engineer in the states on the average, I have no notion of it. Thanks Bernard vl.

[@sahildasaur2072]

Can you please do an example where both masses start from the same height ??

[@MichelvanBiezen]

That should not make any difference in how you approach the problem. Both masses will have initial PE and only one mass will have final PE.

[@sahildasaur2072]

Thanks so much! I tried the problem again and got it :)

[@katyushah.beilschmidt6705]

won't PE inital have a different height than PE final?

[@MichelvanBiezen]

When the large block is at its lowest point, the small block will be at its highest. (Notice that the mass has changed in relation to the PE).

[@kgkg8563]

Sir I really appreciate your work. Please keep making.

[@MichelvanBiezen]

Thank you. My wife and I plan to keep making videos for as long as we can.

[@mechboy5954]

The Atwood machine is named after George Atwood who studied Newton’s laws of motion using this particular machine

[@adityagawhale]

Isn't the final potential energy negative as the block is moving up while it's weight is acting downwards

[@MichelvanBiezen]

Potential energy only depends on position (and not velocity) of the object in a force field and on where you place the reference zero potential point, which is arbitrary.

[@adityagawhale]

Thanks When is mgh negative though ?because I read that if a body is being displaced upwards mgh is negative

[@MichelvanBiezen]

On the surface of the Earth PE is positive above a reference point (typically the ground), and negative below that point (below ground). But is space PE is zero at infinity and less than zero everywhere else (like on the surface of the Earth). But in both cases, PE increases when moving upward.

[@adityagawhale]

Michel van Biezen thanks a lot !👍🏻🙂

[@adityagawhale]

Michel van Biezen thanks a lot !👍🏻🙂

[@gregorybeard5294]

good job

[@seoulpeach5548]

can i use the same method to solve a problem where one of the block lies on a inclined plane ?

[@MichelvanBiezen]

We have many examples of blocks on inclined planes. (there are additional forces you must contend with)

[@anshudeo9224]

work done by gravity in atwood machine will be what ??

[@brentwang6061]

Could I have solved by using kinematics?

[@MichelvanBiezen]

Yes, but it is a lot easier with energy.

[@baphnie]

Michel van Biezen I was wondering why tension didn't come into play. This makes sense, but I'm definitely still getting used to looking at a problem from different perspectives/units.

[@michelebianchi7861]

how about finding acceleration by treating the whole as a system? and then finding velocity ?

[@realmotivation.7184]

If height is not given how to solve

[@MichelvanBiezen]

The height relative to the ground is not needed, howevver the change in height must be known.

[@aizharajoyfeliciano9167]

Is energy conserved?. If so, then why?

[@MichelvanBiezen]

Energy is conserved if there are no retarding forces such as friction (and if there are no collisions)

[@dannggg]

Professor how come there is no initial potential energy for M1? Thank you, Dang Le (BTW thank you for your awesome videos. I have a better understanding of physics and calculus)

[@dannggg]

+Dang Le nevermind professor i thought deep about it and I remembered its initial h = 0 which means no potential initial energy..I learned that from you! lol

[@elmanahmed6117]

Yep, u got it. The h of m1 is 0 . P.E =mgh anything multiplied by zero is zero.

[@oscarkholopha2959]

at ground level the height is 0 m

[@Yajna-14]

great.. but there is also another method of doing this

[@MichelvanBiezen]

Yajna, In physics there are usually several methods for doing the same problem. I try to pick the method that makes it easier to understand how to solve the problems.

[@vcortez15]

what would be different equation when you have m2 at the bottom and m1 in the air like m2?

[@MichelvanBiezen]

Javi Cortez, There would be no motion.

[@lieselwong459]

slay

[@MichelvanBiezen]

Glad you liked it. 🙂

[@vcortez15]

Michel van Biazen

[@onii-chan486]

you are a good man, my good sir.

[@MichelvanBiezen]

Thank you kindly