Conservation of energy | Work and energy | Physics | Khan Academy


Welcome back. At the end of the last
video, I left you with a bit of a question. We had a situation where we
had a 1 kilogram object. This is the 1 kilogram object,
which I’ve drawn neater in this video. That is 1 kilogram. And we’re on earth, and I need
to mention that because gravity is different from
planet to planet. But as I mentioned,
I’m holding it. Let’s say I’m holding it 10
meters above the ground. So this distance or this
height is 10 meters. And we’re assuming the
acceleration of gravity, which we also write as just g, let’s
assume it’s just 10 meters per second squared just for the
simplicity of the math instead of the 9.8. So what we learned in the last
video is that the potential energy in this situation, the
potential energy, which equals m times g times h is equal to
the mass is 1 kilogram times the acceleration of gravity,
which is 10 meters per second squared. I’m not going to write the units
down just to save space, although you should do this when
you do it on your test. And then the height
is 10 meters. And the units, if you work them
all out, it’s in newton meters or joules and so it’s
equal to 100 joules. That’s the potential energy when
I’m holding it up there. And I asked you, well when
I let go, what happens? Well the block obviously
will start falling. And not only falling, it will
start accelerating to the ground at 10 meters per second
squared roughly. And right before it hits the
ground– let me draw that in brown for ground– right before
the object hits the ground or actually right when it
hits the ground, what will be the potential energy
of the object? Well it has no height, right? Potential energy is mgh. The mass and the acceleration of
gravity stay the same, but the height is 0. So they’re all multiplied
by each other. So down here, the potential
energy is going to be equal to 0. And I told you in the last video
that we have the law of conservation of energy. That energy is conserved. It cannot be created
or destroyed. It can just be converted from
one form to another. But I’m just showing you, this
object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it
has no energy. Or at least it has no
gravitational potential energy, and that’s the key. That gravitational potential
energy was converted into something else. And that something else
it was converted into is kinetic energy. And in this case, since it has
no potential energy, all of that previous potential energy,
all of this 100 joules that it has up here is now going
to be converted into kinetic energy. And we can use that information
to figure out its velocity right before
it hits the ground. So how do we do that? Well what’s the formula
for kinetic energy? And we solved it two videos
ago, and hopefully it shouldn’t be too much
of a mystery to you. It’s something good to memorize,
but it’s also good to know how we got it and go
back two videos if you forgot. So first we know that all the
potential energy was converted into kinetic energy. We had 100 joules of potential
energy, so we’re still going to have 100 joules, but now
all of it’s going to be kinetic energy. And kinetic energy is
1/2 mv squared. So we know that 1/2 mv squared,
or the kinetic energy, is now going to
equal 100 joules. What’s the mass? The mass is 1. And we can solve for v now. 1/2 v squared equals
100 joules, and v squared is equal to 200. And then we get v is equal to
square root of 200, which is something over 14. We can get the exact number. Let’s see, 200 square
root, 14.1 roughly. The velocity is going to
be 14.1 meters per second squared downwards. Right before the object
touches the ground. Right before it touches
the ground. And you might say, well Sal
that’s nice and everything. We learned a little
bit about energy. I could have solved that or
hopefully you could have solved that problem just using
your kinematics formula. So what’s the whole point
of introducing these concepts of energy? And I will now show you. So let’s say they have the same
1 kilogram object up here and it’s 10 meters in the air,
but I’m going to change things a little bit. Let me see if I can competently
erase all of this. Nope, that’s not what
I wanted to do. OK, there you go. I’m trying my best to erase
this, all of this stuff. OK. So I have the same object. It’s still 10 meters
in the air and I’ll write that in a second. And I’m just holding it there
and I’m still going to drop it, but something interesting
is going to happen. Instead of it going straight
down, it’s actually going to drop on this ramp of ice. The ice has lumps on it. And then this is the bottom. This is the ground down here. This is the ground. So what’s going to
happen this time? I’m still 10 meters in the
air, so let me draw that. That’s still 10 meters. I should switch colors just
so not everything is ice. So that’s still 10 meters, but
instead of the object going straight down now, it’s going to
go down here and then start sliding, right? It’s going to go sliding
along this hill. And then at this point it’s
going to be going really fast in the horizontal direction. And right now we don’t
know how fast. And just using our kinematics
formula, this would have been a really tough formula. This would have been
difficult. I mean you could have attempted
it and it actually would have taken calculus
because the angle of the slope changes continuously. We don’t even know the formula
for the angle of the slope. You would have had to break
it out into vectors. You would have to do all sorts
of complicated things. This would have been a nearly
impossible problem. But using energy, we can
actually figure out what the velocity of this object
is at this point. And we use the same idea. Here we have 100 joules
of potential energy. We just figured that out. Down here, what’s the height
above the ground? Well the height is 0. So all the potential energy
has disappeared. And just like in the previous
situation, all of the potential energy is now
converted into kinetic energy. And so what is that kinetic
energy going to equal? It’s going to be equal to the
initial potential energy. So here the kinetic energy
is equal to 100 joules. And that equals 1/2
mv squared, just like we just solved. And if you solve for v, the
mass is 1 kilogram. So the velocity in the
horizontal direction will be, if you solve for it, 14.1
meters per second. Instead of going straight down,
now it’s going to be going in the horizontal
to the right. And the reason why I said it
was ice is because I wanted this to be frictionless and I
didn’t want any energy lost to heat or anything like that. And you might say OK Sal, that’s
kind of interesting. And you kind of got the same
number for the velocity than if I just dropped the object
straight down. And that’s interesting. But what else can
this do for me? And this is where it’s
really cool. Not only can I figure out the
velocity when all of the potential energy has
disappeared, but I can figure out the velocity of any
point– and this is fascinating– along
this slide. So let’s say when the box is
sliding down here, so let’s say the box is at this point. It changes colors
too as it falls. So this is the 1 kilogram
box, right? It falls and it slides
down here. And let’s say at this point it’s
height above the ground is 5 meters. So what’s its potential
energy here? So let’s just write something. All of the energy is
conserved, right? So the initial potential
energy plus the initial kinetic energy is equal to the
final potential energy plus the final kinetic energy. I’m just saying energy
is conserved here. Up here, what’s the initial
total energy in the system? Well the potential energy is 100
and the kinetic energy is 0 because it’s stationary. I haven’t dropped it. I haven’t let go of it yet. It’s just stationary. So the initial energy is going
to be equal to 100 joules. That’s cause this is
0 and this is 100. So the initial energy
is 100 joules. At this point right here, what’s
the potential energy? Well we’re 5 meters
up, so mass times gravity times height. Mass is 1, times gravity, 10
meters per second squared. Times height, times 5. So it’s 50 joules. That’s our potential energy
at this point. And then we must have some
kinetic energy with the velocity going roughly
in that direction. Plus our kinetic energy
at this point. And we know that no energy
was destroyed. It’s just converted. So we know the total energy
still has to be 100 joules. So essentially what happened,
and if we solve for this– it’s very easy, subtract 50 from
both sides– we know that the kinetic energy is
now also going to be equal to 50 joules. So what happened? Halfway down, essentially half
of the potential energy got converted to kinetic energy. And we can use this information
that the kinetic energy is 50 joules
to figure out the velocity at this point. 1/2 mv squared is equal to 50. The mass is 1. Multiply both sides by 2. You get v squared
is equal to 100. The velocity is 10 meters
per second along this crazy, icy slide. And that is something that I
would have challenged you to solve using traditional
kinematics formulas, especially considering that we
don’t know really much about the surface of this slide. And even if we did, that would
have been a million times harder than just using the law
of conservation of energy and realizing that at this point,
half the potential energy is now kinetic energy and
it’s going along the direction of the slide. I will see you in
the next video.

100 Comments

  1. Thanks so much man, I have an ETA this week and I never really understood this process. This helped me so much!

  2. can you explain the formula or P.E+K.E+U?. which is change on potential E and change in kinetic E added to change in internal energy? what does the internal energy change, and how to solve for it?

  3. sal you forgot to that mgh is 1*10*10 when you removed the m for the 0.5mv2 you made it no 100 but 10 joule the v is equile to squire 20 

  4. Question: Why would you use this formula if you could use V^2=Vi^2+2ax?  V^2=0+2(9.8)10, V=14m/s Seems easier. Could You show me another example where it might make more sense to choose an energy formula instead of velocity? 

  5. Question, once the object hits the ground, its not moving, so K=0. But, energy needs to be conserved, and it isnt transferred into potentional because PE=0 at h=0. Where does the 100J go?

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  7. If you're travelling through a series of hills in an automobile will you conserve fuel by accelerating through the downhill then coasting through the uphill or by maintaining a constant speed limit?

  8. I still don't get how you can't find the V (before the ice comes in, when it's free falling) by simply calculating through the acceleration. Like this
    V_initial = 0, a=g=10m/s, d=10m, what is V_final?
    V_final should be 10m/s and it should take the object 1 second to hit the ground.
    I would also like to ask a follow-up question please. Imagine the same premise before the ice (when 1kg free falls 10m). Instead of just having ground at the bottom, you have a scale (in Newtons let's say). If the object would be just standing on it (F=ma) the scale would show 10N. Ok… But the object is falling from 10 meters directly on the scale. QUESTION: what would be the MAX value that the scale would show on impact? (in Newtons). Is there a way for that KE to tell us or do we use some other formula?

  9. how did you make The velocity units sudenly from  Jules Into M/s on 3:30 ??? that doesnt seems right
    And then on 3:48 you say m/s^2 like acceleration even if you write m/s which still i didnt get how you got them from the Jules…

  10. Thanks for the videos! I was really lost on my homework so I decided to look for help and this is exactly what I needed.

  11. I think there is a miscalculation while he was calculating the kinetic energy the formula is (KE=1/2*G*H)
    so what he did is that he multiplied 2 both sides to get rid of the denominator and didn't multiply it by the mass which is (1KG) and the answer he got is (14.2). while I was working it out i multiplied 1/2 times 1kg then divided( .5) by both sides then i did the sqr. root to get v by it self, so I ended up with 7.5 (half of what he got). Somebody correct me if I am wrong.

  12. hmmm better than a teacher,, well i guess teachers get bored kind of doing the same thing over and over again for 10 years,, im so unlucky we only have one physics teacher,, if your facing a teacher and cant understand him and he dorsnt even help properly do it yourself,, thats what smart people do if you want to ace the test,, study on your own

  13. Any body believe that energy can not be created or destroyed. I don't think so. If so, why does it exist? It means that energy can be created right?. If energy can not be created, then it will not be there. Think about it…..

    I will change the definition of Conservation of Energy to a better understanding by saying that " Energy can not be added or subtracted ". It makes more sense maybe.

  14. Conservative force/energy = Equal force/energy (before=after)..Its bcoz
    (i)before the motion the object at rest i.e zero
    (ii)after the motion the is also at rest i.e zero

  15. These videos are an amazing help, but that pen!!! If he wasn't talking, I'd be lost! I do have a much better understanding though!

  16. I'm getting really fed up with this poor resolution. Why didn't you fix your system before recording so many bad quality videos? It's nearly impossible to learn anything when the handwriting is inscrutable.

  17. You made it so simple, and I'm able to apply this underlying knowledge to more complex problems. Thank you for helping me.

  18. Is this not false after all the universe was created in the big bang so matter and energy can be created from nothing

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  20. I ALWAYS LOVE your videos, you speak well and have exceptional penmanship. BUT in the last few videos I have watched you are using a thicker font which makes everything hard to read, follow and understand.. PLEASE go back the the thinner font when writing so that it is easier to follow along and read!! Thanks

  21. so going down the ice it accelerates at freefall due to gravity?.. but has to overcome gravity twice on the way down? and still at freefall speed? also if its traveling a longer distance wouldnt the energy be converted from velocity to distance? im so confused. youre using gravity to accelerate it, but not to decelerate it on the ice as it has to travel upwards?

  22. Uhhh… I found about 3 videos on YouTube with over 100k Views that talked about the conservation of energy. This is bizarre. The law of energy conservation is by far the most important physical law. How are there not more videos on this?

  23. Something here I didn't get, how is the final velocity of a free falling object when it hits the ground is equal to the final velocity of the same object when it slides the ice mountain, even when H is the same, this path it took is longer, and there is an X component and a Y component.
    Please someone tills me what I missed(other than the friction-less surface) cause the difference I feel isn't only from friction.

  24. When gravitational potential energy is given to an object during work to some height and the object is dropped, gpe gets converted to ke. If there is no air resistance, where does the ke go after the object hits the ground? The earth? How if a great enough force wasn’t applied over a distance to the earth to transfer such amount of energy?

  25. If energy can not be created ,then since the population is rising ,will there be enough energy for all of us?

  26. This was helpful thanks.
    I'll even send this to my friend william.
    He thinks that the derivation is easy but personally, I am still a bit confused as to how to derive to the final from initial.
    Its like you think they are rivets but they are actually nuts. I'd throw in The Washers atleast. But the added number of components, sure complicates assembly. Double check, triple check everything. Making sure they dont have a nut lose. usually they are all lose. From the vibrations. Re-tightening them just takes more time..

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