Physics 9 Conservation of Energy (3 of 11) Moving Up An Incline (Friction)

[@benbarris3648]

This guy puts up the best videos! thank you!

[@malikamirvohidova5805]

Thank you so much for the videos. I'm learning a lot through them.

[@oldaccountyeet5765]

I watch your videos every single day. Good stuff!

[@SuperSuadad]

you are the man!!!!

[@arieljackson9309]

Why is the friction term not negative? I thought it opposed motion.

[@MichelvanBiezen]

+Ariel Jackson It all depends how the term is used. In this equation we are equating the total energy of the initial state to the total energy of the final state plus the energy lost. For the left side to be equal to the right side you must ADD the energy lost on the right side.

[@sajjadmahdi1242]

3:35 why is the sign not negative? Isn't the work done by friction force negative if opposed to motion direction?

[@MichelvanBiezen]

If the energy lost due to friction is placed on the right side of the equation, you must make it positive

[@sajjadmahdi1242]

@@MichelvanBiezen so it was originally on the left (negative), and got taken to the right side?

[@MichelvanBiezen]

No, with this equation, energy lost due to friction is always on the right for the equation to be balanced. But you can make it negative and place it on the left.

[@sajjadmahdi1242]

@@MichelvanBiezen Ok, what about Delta PE = -mgh when does it give positive? In Conservation law perhaps?

[@aliabrahimi3711]

At 3:40 don't we have to multiply by the 180 degree angle? because force friction is opposite to the direction of velocity

[@MichelvanBiezen]

ali abrahami If you place the "energy lost" term on the right side of the equation you must add it. If you place the "energy lost" term on the left side of the equation, then you must subtract it.

[@TheLastArchive]

Quick question, wouldn’t you have to add the component of the block’s weight parallel to the slope to the force of friction as it will also act down the slope?

[@MichelvanBiezen]

No, the only force that affects the force of friction is the perpendicular component of the weight (plus any other forces that act perpendicular to the slope).

[@BrickfilmMan]

Thanks! Amazing explanation!

[@jaredronning3020]

I find it easier to account for additional work and lost energy using one term work_other on the left hand side of the equation instead of two terms. Friction would be taken into account as contributing negative work and other forces would be summed in to that one term.

[@Hiep156]

Hey, my lecture notes from my teacher is written like this: E_p0 + E_0 + W_R = E_p1 + E_K1. However, from your video u write W_R on the right side of the equation on the final energy. The difference between your videoes and my teacher's note is the sign. I don't know what it is correct. It is the same example. Who is right?

[@TheSanaHin]

Isn't the angle between friction force and distance is 180..???

[@gaafaralkazaaly1663]

A skier starts from rest at the top of friction-less incline of height 20 m. at the bottom of the incline, she encounters a horizontal surface. Show in figure How far does she travel on the horizontal surface coming to rest

[@jayakumarelpula189]

I like the different colour ink used for the friction formula!

[@mohammedal-maimani9047]

great video and helped allot .. but shouldn't you take the lost energy with minus sign because the friction is on the apposite direction ?

[@MichelvanBiezen]

Mohammed, There are a number of ways in which you can solve these types of problems. I find it easiest to set up an equation where on the left you place all the energy that was either added to the system or the system already had at the start and on the right you place all the energy that was lost by the system or that it ended up with. In order to balance such an equation, all the terms have to be positive (even the energy lost due to heat on the right side).

[@naveednaiemi3979]

First of all thanks for all the efforts you have done to prepare such a helpful videos. Sir, why is initial energy equal to the final energy? Is it always the case or are there some situations that initial energy is not equal to the final energy? thanks.

[@MichelvanBiezen]

It is actually almost NEVER the case. The only reason why we can use that equation is because we also account for the energy lost by adding it to the right side of the equation (as energy used to overcome friction).

[@hassanulger7167]

Sir what shoud we do if it asks for two unknown for example the final velocity and the distance

[@yesplease154]

IF THE SURFACE IS FRICTION LESS CAN IT POSSIBLE TO DO BY VERTICALY PROJECTION MEANS Vy

[@MichelvanBiezen]

If there was no friction, then you would solve the problem exactly the same way except you would eliminate the "energy lost" term.

[@abdoelosta5655]

great explanation

[@Yah11]

What would you do if you had an equation that was not at an inclinde but included friction?

[@Yah11]

nevermind, I figured it out by watching the 5th video :)

[@ahmedal-ebrashy3691]

Wise SIr, I dont understand why mgsintheta was not part of the conservation of energy since it does work on the object besides friction?

[@MichelvanBiezen]

the mgsin(theta) term is part of the mgh (potential energy gained) term.

[@ahmedal-ebrashy3691]

Yes but shouldnt it be added to the friction, acted against the K:E and hence should be subtracted from the KE or what am I missing. Thank you SOOO MUCH

[@MichelvanBiezen]

The friction force is defined as: Ffr = N u where N is the normal force and N = mg cos(theta). The mg sin(theta) component does not add anything to the friction force.

[@galalliatorful]

Sir, i have a problem of choosing which method to use… why we are using total energy conservation instead of newtons second law or impuls momentum? Im really confused

[@MichelvanBiezen]

Typically when the question asks for the acceleration, use Newton's laws. If the the question asks how fast will it travel or how high will it go, conservation of energy is a good method. That said, you can usually use either method.

[@galalliatorful]

thank you so much, it makes sense now you are such an amazing professor i have my fe exam in two days and im wondering what videos would be useful for me regarding statics,dynamics and mechanics of materials since dynamics is my weakest subject in the exam

[@MichelvanBiezen]

Since I don't know what was part of your course, my recommendation is to look at the playlists and see which ones match. They are all in order and can be found from the main home page on this channel.

[@gideonbennett4187]

Thank you ☀

[@MichelvanBiezen]

You’re welcome 😊

[@nestorcotton]

can you explain how at 3:58 you assigned mgcos(theta) and mgsin(theta)

[@MichelvanBiezen]

+Nestor Cotton There are some other videos that explain that. Basically you separate the weight due to gravity (a vector) into a component parallel to the incline and a component perpendicular to the incline.

[@Sophia-wq7df]

Thank you SOOOOOOO much!!

[@fahamidaani9597]

initially if there was any energy lost then should I write that in E0

[@MichelvanBiezen]

No, the energy lost is during the process, not at the initial condition.

[@eshapierra4593]

what is the miu thingy with the value of 0.2? where does he gets it from?

[@MichelvanBiezen]

That is the coefficient of friction. You can discover more about that here: PHYSICS 4.6 FRICTION kzbin.infoplaylists?sort=dd&view=50&shelf_id=4

[@bimalr]

what about the mg sin(theta) component acting along the incline down? Should it not be added to Friction? Please explain this...

[@MichelvanBiezen]

The friction force only depends on the perpendicular component of mg (namely mg cos(theta))

[@warrencoombs4646]

Since it is going up the incline would the force of gravity component parallel to the plane also contribute to the work? The video shows only Ff doing work as umgcos30, but isn’t mgcos30 (force of gravity component along the incline) also doing work against the block moving up the incline assuming no applied force. So should the equation be something like this: Ei=Ef + W Ei=Ef + Fd 1/2mv^2 = mgh + (mgsin30 + umgcos30)d 1/2mv^2 = mgdsin30 + (mgsin30 + umgcos30)d Am I missing something here?

[@warrencoombs4646]

This video seems to suggest that you have to consider the work lost on the block due to gravity kzbin.info/www/bejne/inrFXquparKBY5o

[@MichelvanBiezen]

The work done to overcome gravity is accounted for by calculating the difference in potential energy.

[@danielderese3170]

that means we dont have to consider the force mgsin(theta) ..( the force of gravity in the x direction)

[@MichelvanBiezen]

+Daniel Derese mg sin (theta) is in the equation. (It shows up as the potential energy gained) Thus you cannot ignore it.

[@illusionarybox3191]

+Michel van Biezen mg sin theta isn't a energy lost?

[@SoheeChoi-d1n]

Hi, so mgsin(angle) is same as friction? or is it different? Thank you!

[@MichelvanBiezen]

mg sin(theta) = component of the force of gravity parallel to the incline. mg cos(theta) * mu is the force of friction.

[@VictorLikome]

you are great

[@nofarmarom1306]

can you solve that problem just by plugging in 1/2v^2/g - mgcos(theda)(friction)=h ? or do you have to go through all these steps and factor out d and g?

[@MichelvanBiezen]

+Nuphar Marom You don't need to follow all he steps that I show you, but there is a reason why I do them. Try it your way an see if you get the same answer.

[@albertmunoz4045]

Very useful thank you for videos

[@farukakkas3991]

thank u teacher great job :)

[@Waveb.8336]

would an equivalent form of writing the equation be W+ ΔK +ΔPE - E lost =0 or W+ ΔK +ΔPE + E lost =0

[@MichelvanBiezen]

That depends on the definition of delta KE and delta PE

[@Waveb.8336]

Michel van Biezen delta KE would mean (K final- K initial) and same for Delta PE

[@MichelvanBiezen]

Then it would not work since you need the initial energy on the left side and the final energy on the right side., also it must be - energy lost on the left side.

[@MichelvanBiezen]

But it is better to think about it logically. It really does make sense that the initial energy that you started with + any work put into the system must equal the final energy you end up with + any energy lost due to friction, etc.

[@md.mominulislam5068]

sir, for frictional force can we use mg.sin30° instead of u.mg.cos30°

[@MichelvanBiezen]

+Md. Mominul Islam u * mg cos(30) is the friction force.

[@md.mominulislam5068]

Sir, why we shouldn't use the final potential energy as mgh.cos30°, because the plane is inclined. Could you plz explain this....

[@MichelvanBiezen]

+Md. Mominul Islam Since h is the height gained (not the distance along the incline), PE = mgh

[@faridabasawad8800]

What if they ask you to find the final velocity of the object when it is at the top

[@MichelvanBiezen]

The final velocity is zero once the final height is reached.

[@danimal3618]

thank you kind sir

[@MichelvanBiezen]

You are welcome. 🙂

[@tamarplotzker5231]

why is there no work if there is mass and acceleration

[@MichelvanBiezen]

Since the force of gravity is taken into account with the potential energy, there is no additional force present to do work.

[@harveenkaur4684]

Very helpful thank you!1

[@MagisterMasekoGameplay]

Why is g not negative?

[@muazuimranwanka3006]

what if the object is initially at rest and has a final velocity of 0?

[@MichelvanBiezen]

Then the object would not move

[@abeeralraisi3548]

excellent explanation thanks a lot

[@sydneybrake7640]

He's like the bob ross of science

[@sivakumarsiva1743]

I am getting answer as 58

[@jensmalzer6344]

does ur solution still work if im gay?

[@MichelvanBiezen]

That is not a relevant question.

[@jonstokes1832]

Nope

[@RiaziMohandesi]

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