Work/energy problem with friction | Work and energy | Physics | Khan Academy


Welcome back. Welcome back.
Welcome back. I’ll now do another conservation
of energy problem, and this time I’ll
add another twist. So far, everything we’ve been doing,
energy was conserved by the law of conservation. But that’s because all of the
forces that were acting in these systems were conservative
forces. And now I’ll introduce you to
a problem that has a little bit of friction, and we’ll see
that some of that energy gets lost to friction. And we can think about
it a little bit. Well where does that
energy go? And I’m getting this problem
from the University of Oregon’s zebu.uoregon.edu. And they seem to have some nice
physics problems, so I’ll use theirs. And I just want to make
sure they get credit. So let’s see. They say a 90 kilogram
bike and rider. So the bike and rider combined
are 90 kilograms. So let’s just say the mass is 90
kilograms. Start at rest from the top of a 500 meter
long hill. OK, so I think they mean
that the hill is something like this. So if this is the hill, that
the hypotenuse here is 500 hundred meters long. So the length of that,
this is 500 meters. A 500 meter long hill with
a 5 degree incline. So this is 5 degrees. And we can kind of just view
it like a wedge, like we’ve done in other problems.
There you go. That’s pretty straight. OK. Assuming an average friction
force of 60 newtons. OK, so they’re not telling us
the coefficient of friction and then we have to figure
out the normal force and all of that. They’re just telling us, what
is the drag of friction? Or how much is actually friction
acting against this rider’s motion? We could think a little bit
about where that friction is coming from. So the force of friction is
equal to 60 newtons And of course, this is going to be
going against his motion or her motion. And the question asks us, find
the speed of the biker at the bottom of the hill. So the biker starts up
here, stationary. That’s the biker. My very artful rendition
of the biker. And we need to figure out the
velocity at the bottom. This to some degree is a
potential energy problem. It’s definitely a conservation
of mechanical energy problem. So let’s figure out what the
energy of the system is when the rider starts off. So the rider starts off at
the top of this hill. So definitely some
potential energy. And is stationary, so there’s
no kinetic energy. So all of the energy is
potential, and what is the potential energy? Well potential energy is equal
to mass times the acceleration of gravity times
height, right? Well that’s equal to, if the
mass is 90, the acceleration of gravity is 9.8 meters
per second squared. And then what’s the height? Well here we’re going to have
to break out a little trigonometry. We need to figure out this side
of this triangle, if you consider this whole
thing a triangle. Let’s see. We want to figure out
the opposite. We know the hypotenuse and
we know this angle here. So the sine of this angle is
equal to opposite over hypotenuse. So, SOH. Sine is opposite over
hypotenuse. So we know that the height–
so let me do a little work here– we know that sine of
5 degrees is equal to the height over 500. Or that the height is equal
to 500 sine of 5 degrees. And I calculated the sine of
5 degrees ahead of time. Let me make sure I
still have it. That’s cause I didn’t have my
calculator with me today. But you could do this
on your own. So this is equal to 500,
and the sine of 5 degrees is 0.087. So when you multiply these
out, what do I get? I’m using the calculator
on Google actually. 500 times sine. You get 43.6. So this is equal to 43.6. So the height of the hill
is 43.6 meters. So going back to the potential
energy, we have the mass times the acceleration of gravity
times the height. Times 43.6. And this is equal to, and then
I can use just my regular calculator since I don’t
have to figure out trig functions anymore. So 90– so you can see the whole
thing– times 9.8 times 43.6 is equal to, let’s
see, roughly 38,455. So this is equal to 38,455
joules or newton meters. And that’s a lot of
potential energy. So what happens? At the bottom of the hill–
sorry, I have to readjust my chair– at the bottom of the
hill, all of this gets converted to, or maybe I should pose that as a question. Does all of it get converted
to kinetic energy? Almost. We have a force
of friction here. And friction, you can kind of
view friction as something that eats up mechanical
energy. These are also called
nonconservative forces because when you have these forces
at play, all of the force is not conserved. So a way to think about it is,
is that the energy, let’s just call it total energy. So let’s say total energy
initial, well let me just write initial energy is equal
to the energy wasted in friction– I should have written
just letters– from friction plus final energy. So we know what the initial
energy is in this system. That’s the potential energy
of this bicyclist and this roughly 38 and 1/2 kilojoules
or 38,500 joules, roughly. And now let’s figure out the
energy wasted from friction, and the energy wasted from
friction is the negative work that friction does. And what does negative
work mean? Well the bicyclist is moving 500
meters in this direction. So distance is 500 meters. But friction isn’t acting
along the same direction as distance. The whole time, friction is
acting against the distance. So when the force is going in
the opposite direction as the distance, your work
is negative. So another way of thinking of
this problem is energy initial is equal to, or you could say
the energy initial plus the negative work of friction,
right? If we say that this is a
negative quantity, then this is equal to the final energy. And here, I took the friction
and put it on the other side because I said this is going to
be a negative quantity in the system. And so you should always just
make sure that if you have friction in the system, just as
a reality check, that your final energy is less than
your initial energy. Our initial energy is, let’s
just say 38.5 kilojoules. What is the negative work
that friction is doing? Well it’s 500 meters. And the entire 500 meters, it’s
always pushing back on the rider with a force
of 60 newtons. So force times distance. So it’s minus 60 newtons,
cause it’s going in the opposite direction of the
motion, times 500. And this is going to equal
the ending, oh, no. This is going to equal the
final energy, right? And what is this? 60 times 500, that’s 3,000. No, 30,000, right. So let’s subtract 30,000
from 38,500. So let’s see. Minus 30. I didn’t have to do that. I could have done
that in my head. So that gives us 8,455 joules is
equal to the final energy. And what is all the
final energy? Well by this time, the rider’s
gotten back to, I guess we could call the sea level. So all of the energy
is now going to be kinetic energy, right? What’s the formula for
kinetic energy? It’s 1/2 mv squared. And we know what m is. The mass of the rider is 90. So we have this is 90. So if we divide both sides. So the 1/2 times 90. That’s 45. So 8,455 divided by 45. So we get v squared
is equal to 187.9. And let’s take the square root
of that and we get the velocity, 13.7. So if we take the square root of
both sides of this, so the final velocity is 13.7. I know you can’t read that. 13.7 meters per second. And this was a slightly more
interesting problem because here we had the energy wasn’t
completely conserved. Some of the energy, you could
say, was eaten by friction. And actually that energy
just didn’t disappear into a vacuum. It was actually generated
into heat. And it makes sense. If you slid down a slide of
sandpaper, your pants would feel very warm by the time you
got to the bottom of that. But the friction of this, they
weren’t specific on where the friction came from, but it
could have come from the gearing within the bike. It could have come
from the wind. Maybe the bike actually
skidded a little bit on the way down. I don’t know. But hopefully you found that
a little bit interesting. And now you can not only work
with conservation of mechanical energy, but you can
work problems where there’s a little bit of friction
involved as well. Anyway, I’ll see you
in the next video.

100 Comments

  1. @Oneill9293 haha oh well, I was the one who signed up for it – thinking anything's possible as long as you put ur mind to it. xP And still believing! hehe 🙂 Best of luck to u too m8.

  2. Dear Mr. Khan
    Please stop doing Physics explanations. Indeed, energy is ALWAYS conserved. For this problem should talk only about Mechanical energy or kinetic energy. This idea of ''energy wasted by friction'' etc is not correct. The answer is correct but the explanation is wrong, You are a great explainer ,but if you provide misleading explanations, we professors in engineering have to spend a vast amount of time correcting these misconceptions with only limited success. Prof Arun Srinivasa

  3. energy IS conserved its just converted into heat, that iswhy you don't get the total KE equal to PE at the beginning. you should've stressed on that

  4. u just need that pen that writes on the screen to accomplish the perfectness of teaching physics online, (+a calculator in all vids :P) , other than that ur (y)

  5. @ArunSrinivasa ….if the majority of college professors were actually effective at explaining concepts like this, then Mr. Khan would have no need to do these videos. I'm definitely not judging you (because I have no idea what type of teacher you are), but the truth is, many professors are TERRIBLE at explaining things and care more about showing off their knowledge than they do about the students actually learning.

  6. i think that way it's easier to stream for those with weaker internet connections. otherwise that kinda opposes the 'knowledge for all' purpose. yes it is harder to see but you still get the concept if you listen along as he writes.

  7. im just trying to help… if you're familiar with the E=(mv^2)/2 you'd know that v=the square root of 2gh… much easier and much less time-consuming than the way that the great Khan demonstrated.

    I'm NOT bashing his videos or way of thinking, he's a genius, just saying.

  8. How can the friction (kinetic friction) be in the opposite direction of the motion or distance ? Think about it we are not pushing a box where the friction is in the opposite side of the motion. Friction comes from the sliding of the two surface in contact with each other. Here while the wheel is rolling, friction will be in the same direction as the distance or motion. This was one question I missed in MCAT! Like if you agree 🙂 so the comment can stay on the top!

  9. I think it's because these are older videos (i.e. he did the earlier concepts again later on). However, I'm not sure if I heard him say something like that once, or if I'm just guessing.

  10. Static friction is in the direction of motion if you have a wheel (that's how we are propelled forward, e.g. walking, bicycling). Therefore, this problem is wrong. The non-conservative force should be added instead of subtracted. Very good problem nonetheless. Thanks Sal.

  11. He does that because force is given with respect to the incline plane. If force was in the horizontal direction then you'd be correct.

  12. @toorbabray No if you push an object then you wood be pushing it in a given direction. Friction is a vector quantity so the friction would have to be opposite the direction you push because the friction would be a retarding force (sllowing the box down) . Of iit went in the same direction then relative to the box it would be adding force not taking it away which isn't possible because friction generates heat from converted energy.

  13. lol ok ok i got it you're the 5th person to comment on that 😛
    i wasn't paying attention as I was studying for an exam at that time lol

  14. These are earlier videos from the 2011 ones. ANd I know it's hard but they're just as helpful

  15. why dont you work out what youre going to say and write before making the video? It makes it hard to follow the process when youre continually scrubbing things out and making it messy

  16. hmmm… very unrealistic… as the speed increases, the air drag increases as well.. it would be interesting to calculate it as well

  17. Now I am studying to my Physics Final Exam from these videos which are very useful and clear to understand. I want to say a BIG THANK YOU! I can understand physics with these videos easily rather than 1600 pages of PHYSICS BOOKS. I am very happy to see people who are making good works to help people. From other side of the world with respects!

  18. @7:45 The Work done by friction = force * displacement * cosine(angle) ? But why did he skip the cosine part? I was thinking it shouuld be Wfric = (60N)(-500m)(cos5degrees) ? Can anybody explain this to me? Big thanks 🙂

  19. I don't get it, shouldn't the gravity acting on the bicyclist be 9.81m/s^2*Cos(5) because its at an angle?!?! Please help.

  20. An electric motor takes a car weighing 800kg up to point A 140m above the ground. If the total resistive forces = 0.25MJ, how much energy does the electric motor use to get the car to point A?how will you work this out?
     
    will it be (800)(0.25)(140)

  21. Sal I don't know if you're still reading these but man I have been learning so much from you- Trig, Pre-Calc, Calc, and now Physics. You're a huge inspiration, man. I'm considering becoming a teacher and I just wanted to acknowledge and thank you. Thanks for all of the hard work you do, day in and day out.

  22. is the friction of rolling maximum static friction ?

    and why it is static friction why the velocity of contact surface is equal zero?

    I really want to know why.

  23. i had no interest in maths and physics i used to waste a lot of my time on the internet but now cause of khan academy i have started liking and understanding these subjects a lot better and it becomes very interesting. thnx a lot man

  24. I have learn so much, words are not enough to express my gratitudes .I pray may GOD blessed you all the Tutors whom have put their effort here.

  25. My new PHYSICS SOLVING APP.More then 150+ formulas,Solves for any variable you want,Covers up all physics.download now.https://play.google.com/store/apps/details?id=com.physics.lenovo.myapplication

  26. All these smart people in the comments. I need help 🙁
    I never took this in HS, I took maths I needed to graduate :'(

  27. I didn't take last year of physics in high school and the introductory physics course at my university is way harder than a high school class would've been lol. My final is tomorrow…. and your videos are perfect for reviewing!! How do you know so much!?

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