Physics 8 Work, Energy, and Power (5 of 37) Pushing an Object (Frictionless)

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In this video I will show you how to find the work and kinetic energy pushing an object horizontally on a frictionless surface

[@animusdx]

I watched this vid 5 years ago when I needed it in college and it just randomly got recommended to me right now. The Algo works in mysterious ways...

[@MichelvanBiezen]

Time does indeed pass very quickly. We have made almost 5000 more videos since then.

[@koober_]

You are an amazing teacher and I hope you continue to do this. Back in high school when we were learning physics, I never could have known that the equations they pumped into our heads came from somewhere other than a textbook. Thank you for both explaining the process of problem-solving and derivations of things we take for granted in such a clear way.

[@Zonekiller24]

This is a god tier teacher. Simplicity at its finest.

[@jayakumarelpula189]

Excellent,professor ! Though we know the formulae but explaining how and where to use the same is your forte and we all are fortunate to have you teach us. Thank you!

[@SarunaAlbahr]

I wish I had you as a professor in school. I would have loved physics and, by extension, math. I have always had the unfortunate experience of getting a professor who could not understand the material or could not teach it or both. You have made this AMAZINGLY easy to understand. Thank you!

[@jakubkusmierczak695]

Now I know why Ek=mv^2/2. Every video is more and more exciting. Thank You Mr. Professor.

[@MichelvanBiezen]

You are welcome. Glad it was helpful. 🙂

[@Dkirbs]

here's a good story, i study engineering and recently i was taking a physics class. At first i dont know what the approuch is to physics, i cant seem to figure it out. Unlike calculus that you really need to practice solving problems to get good with it. The first semester of physics really beat the hell out of me. So i tried many strategies and it did not work. many weeks later i found a method that worked well for me. I read a textbook, write down all the formulas, and watch this dud go through problems. And it worked !!! I passed my last quiz ang hoping to pass the midterms exam as well. Like math i try many things to solve my problem. If its hard, maybe theres another way. I thank this guy for his extremely good videos, it helped me a lot.

[@MichelvanBiezen]

Glad we are able to help in your journey through physics.

[@tankieag1117]

You are the first person to make me get that 1/2 mv^2 formula

[@MichelvanBiezen]

That is surprising since that is a basic concept in physics.

[@infernosm3857]

You can also use Work Energy Theorem That is W=∆ KE Where "W"is Work done and "KE" is the change in Kinetic Energy......

[@ahmedal-ebrashy3691]

Wish I followed you when I was 18.

[@sukhmankahlon3496]

I am going to fail my midterm but I love this man.

[@MichelvanBiezen]

I hope you don't fail that midterm. Now that you have found our videos, perhaps you will do better in the future. 🙂

[@andreslopera8724]

What about if the object is on top of a hill, like this........ A spring, with constant of 5500N/m, is compressed by 0.60m and used to launch a 20.5Kg mass from rest from point A across a frictionless surface and down a short hill that is H=2.5m high. The surface becomes rough at point B, and the mass comes to a stop at point C. 1) How much energy is stored in the compressed spring? 2) KE of the object at B? 3) How much work was done by friction between B and C?

[@MichelvanBiezen]

W + KEo + PE0 = KEf + PEf + E lost 0 + (1/2)kx^2 + mgh = 0 + 0 + mg (mu) d 1) (1/2)kx^2 2) KEb = (1/2)kx^2 + mgh 3) W by friction = KEb

[@andreslopera8724]

@@MichelvanBiezen Thanks

[@StudyResearchandAnalysis]

Sir, I have a Doubt in The Concept of Work and Energy, Work = KE (According to Work Energy theorem) So If, Lets say we do a work of 100J and make a 2 kg block move then Basically the Block gains KE of 100J and that we can say as the block will have a velocity 10m/s now. So doesn't this seem to say that Work and Energy are the same ? Further, Like in Derivation of Bernoulli's equation, We Write Work = Pressure x Volume and Then we also Write KE (Kinetic Energy) Work 1 + KE 1 + PE 1 = Work 2 + PE 2 + KE 2 Here why do we Take the Work in the equation ? Hasn't the work been used up in order to Generate a KE ? In Deriving the Bernoullie Equation (I have seen Your Derivation too, But this is a Similar Derivation which I saw = kzbin.info/www/bejne/r6LXdYSirq2EfrM) why are We Taking "Work" in The "Conservation of Energy" . If A Block of Mass 2kg is Pushed and it gains Velocity 10m/s, and someone asks us "What is its energy?" we would say 100J as that is its KE and Not 100J of Work + 100J of KE and Hence 200J, Then why do we do a similar "Wrong" thing in case of Bernoullie Theorem ? (Really Sorry for this long Comment)

[@MichelvanBiezen]

The standard definition of work is the dot product of force and displacement. Work can result in increasing kinetic or potential energy or simply to overcome friction or wind resistance, with no increase in energy. It depends on the situation.

[@StudyResearchandAnalysis]

@@MichelvanBiezen Thank You Sir

[@Volumed94]

i love these video lectures. thanks !!!!!!!!

[@iramsiddiqui9678]

thanks sir they are very useful.. short and understandable...

[@abre.ham121]

God bless you

[@albertopoli8896]

Simply fantastic!!!!!

[@DanielFBest]

Unbelievably helpful, professor 🤓

[@DarthDeth]

Thank you so much

[@naviibbyy07]

I wish you were my professor!

[@sbk1398]

Isn't he great? Needs to come to my school

[@zakirhussain-js9ku]

All type of energies manifest themselves through motion of mass, electric and magnetic fields

[@MichelvanBiezen]

I would agree with that.

[@schopenhauer6251]

We were able to use kinematic equations, does that mean the object is moving with constant acceleration?

[@MichelvanBiezen]

In order to use the three equations of kinematics, the acceleration must be constant. (When the 3 equations were derived, that assumptions was made).

[@joshuaronisjr]

Could you do this same example, but where the force isn't constant, but a function of the distance, as would be the case if the 10 Newtons were provided by a spring?

[@MichelvanBiezen]

We have examples like that in the playlists.

[@blazeofwrath5461]

Thank you sir!!!!🤗🙂🙂🙂😁😁🙏🙏🙏🙏🙏🙏🙏!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@MichelvanBiezen]

Most welcome!

[@jayakumarelpula189]

Excellent

[@historyisthebest5831]

Hello sir! When you're moving a lawn mower at constant speed, and there's friction force, does your work done equal the work done by friction? I think since it's not accelerating, all of the energy you put into were removed. Thank you!

[@30kUzi]

I might be wrong but isnt Wnet the sum of all work acting on the system and since the Y components cancel out wouldn't Wnet be W of force applied + W of friction

[@nd685]

Can there be kinetic energy without any work? Suppose an object is moving with constant velocity. There is no net force. Work is depended on force. Hence no work. But there is velocity. And kinetic depends on velocity and mass but not on force according to the formula

[@MichelvanBiezen]

An object that is dropped will gain kinetic energy as potential energy is converted kinetic energy. You can also call that the force of gravity doing work on the object.

[@nd685]

Thanks professor :)

[@jryistheflyguy84]

If you knew the work would convert into KE at the end why not just use 1,000=.5*2*(v)^2 and solve for V ... it gives you the same velocity at the end of 31.62m/s..

[@codosacho5924]

what does kinetic energy actually means in definition not calculations ?? and thankss a lot for your effort and your awesome way of teaching

[@MichelvanBiezen]

Codo, An object has kinetic energy because it is moving. Think of energy in terms of the work that it can do. When you swing a hammer it has kinetic energy, because it is moving. When it hits a nail, the energy is used to drive the nail into the wood.

[@codosacho5924]

Michel van Biezen but the instant that the object stops will it have kinetic energy ??knowing that this object has no height from the surface of the earth

[@MichelvanBiezen]

Codo Sacho Codo, The moment the object stops, it will no longer have any kinetic energy, regardless of its position

[@codosacho5924]

Michel van Biezen so according to the law of conservation of energy where does this energy go ??

[@MichelvanBiezen]

Codo Sacho That depends on what happened, Energy can do work, so sometimes the kinetic energy is used to perform work like pushing an object up a hill or in the case of billiards one moving ball will push another moving ball so the kinetic energy is transferred. But often the kinetic energy is transformed to heat, like in the case where an object falls to the ground from a height. The kinetic energy is transformed to heat which is absorbed by the floor and the object. Heat is stored in an object by the vibrations of the atoms in the object. The more heat the object absorbs, the faster the atoms will vibrate and the hotter the object becomes.

[@ryanbroome23]

How do you know when the acceleration should be calculated or just assumed to be 9.81 m/s^2?

[@hazzimariff9258]

+Ryan Broome we use 9.81 m/s^2 when involving gravity. It is called the gravity acceleration.

[@odwamlambo4564]

hey I've just listen to this video and I have been doing calculations with you but the last calculation when calculating kinetic energy is 998.56 instead of 1000j how did you do it?

[@MichelvanBiezen]

Odwa, You used 31.6m/sec for the velocity which is a "rounded" value. It is actually Square root (1000) = 31.6227766 m/sec. (Usually I leave the previous value in my calculator, so I don't have a rounding error.

[@naviibbyy07]

you can round that up

[@pranainarayanvenkatesan6199]

Could we use a as v^2 = u^2 + 2as too ? and we get v^2-u^2/2s as accerelation and then integrate both sides

[@MichelvanBiezen]

There are often multiple ways in which a problem can be done. I always encourage my students to try a different method to see if they get the same answer.

[@leejy2]

for the Final Velocity, can I equateW = 1/2 mv^2 ? 1000 = 0.5*2*v^2 v(final) = sqrt[1000/(0.5*2)] = 31.6m/sIs this a correct solution as well?

[@leejy2]

oh I just saw the ending of the video. Does it mean that the W=1/2mv^2 is always true? or in some cases it will be not true?

[@elzemtu]

2:26 is there a mistake.? I think v=10m/s^2

[@MichelvanBiezen]

Ebru, The answer is correct as stated in the video.

[@elzemtu]

I really dont understant or i am missing something. I think it should be vf^2=vi^2+2ax so vf^2=0+2.5.100 then v=10m/s^2 .

[@MichelvanBiezen]

ebru uyar Vf^2 = 0 + 2 * 5 * 100 Vf^2 = 1000 Vf = Sqroot (1000) Vf = 31.6 m/sec

[@elzemtu]

sorry my bad.I spent your time for trivial algebra and you know sometimes person stubbornly cannot see even if it is so simple.Really thanks for your answers.. .Everything is more clear now.

[@MichelvanBiezen]

ebru uyar Ebru, Don't feel bad. That happens to me as well. Sometimes I look back and wonder how i could have made a mistake on such a simple problem. Once our brain is locked onto a certain concept, even when wrong, it is hard to see the correct answer. That is why learning how to do problems in a certain pattern will help limit those types of errors. That is why I teach math and physics the way I do.

[@CanoManuelGonzaga]

And so if you have 1000J for work, then why solve for KE is KE is uqual to W?

[@MichelvanBiezen]

Work can be defined in a number of ways. In this example work can be defined as causing the change in energy, which in this example is the change in kinetic energy.

[@ryanbroome23]

Bowtie on fleek

[@sreebommineni5123]

Couldn't we just find v by saying 1000J=1/2 mv.v where v.v=1000 and v= 31.622 Am I right?

[@MichelvanBiezen]

That is correct.

[@maxamedxuseen7871]

Your hand writing is small or invisible

[@azizkash286]

i love you

[@ThePianoMan1999]

Is this a level work???

[@MichelvanBiezen]

+ThePianoMan1999 Sorry, I don't understand the question

[@ThePianoMan1999]

What level is this for a 16 year old ? Like gcse level?

[@MichelvanBiezen]

+ThePianoMan1999 The level of these videos range from high school through upper division college.

[@minecraftvidsbyjames]

I've done this at AS level if that helps :)

[@Peter_1986]

+ThePianoMan1999 This is introductory high school physics - however, you do study this stuff during your first year at an engineering program as well, although in an engineering program there is more pressure on having a deep understanding of everything. Michel Van Biezen does a good job in showing where the formulas come from, and I like that. Physics normally starts with Mechanics, which in a nutshell concerns the behaviour of objects when they are affected by forces.

[@fizixx]

👍

[@MichelvanBiezen]

Keep it going.

[@fizixx]

@@MichelvanBiezen 😀

[@DanielFBest]

But you left out the constant of integration

[@MichelvanBiezen]

You would only need a constant of integration if there was some initial work you wanted to add. (there was none)

[@nasradinali]

This is not correct!!! This only applies if the force applied is constant! Otherwise object Will move at constant velocity if we have zero friction.

[@MichelvanBiezen]

Video is correct. Thanks for checking.

[@nasradinali]

Thank you Mr Van Biezen, one of my teacher lacked motivation! So instead you became my teacher!!

[@minecraftvidsbyjames]

"Square root of 1000? That would be about 31.something I supose". Easy maths innit.

[@lolerishype]

Sorry. Physics 8?????

[@MichelvanBiezen]

Not sure what your question is? 🙂

[@Mr1984eL]

Yeaa.... You kinda lost me with this one. Very over-explained.. Still good

[@monazza_]

But it was awesome

[@younique9710]

As the previous lecture, 4 of 20, why we do not include constant C after integration at 4:25?

[@MichelvanBiezen]

You could add a constant of integration, but that would represent any initial work not associated with the force doing work, so we can just assume it is zero.