Yes, I wished I would've taken at least one AP Physics course while I was in high school. You remind me a lot of a physics teacher from my high school, and everybody loved him as well as how well he taught. Wished I would've had that opportunity to take a class from him. Physics in college is a different cat!
I appreciate you and your hard work. Realize that my AP Physics C test is in ~24 hrs and and because of you, I just might pass. Super clear, understood everything.
Woke up at 3:45 AM to study for my physics quiz today, and felt like I needed a good review of things before. Thanks a lo for this really helpful video!
In the first 108 seconds of this video, I learned more than hours and days of physics classes taught in a month. I'm never going back to "49 degrees south of east"!
That annoys me as well. It took me several years as a teacher to realize I need to be very precise in everything I say and write. It is the best way to make sure students understand what I am trying to teach them.
I did some research and U is purely a technical varibale, not a symbol like F for force. It was chosen because all the other letters were taken up in Thermo!
That is my current plan. My hope is to, after completing the AP Physics C: Mechanics Review Videos, to then move on to work simultaneously on the AP Physics C: Electricity and Magnetism Review Videos and continue working on the AP Physics 1 videos which are more demonstration oriented (and therefore more fun to create). Hopefully that's what will happen.
The U in potential energy actually comes from the graph of potential energy, which decreases as it converts to kinetic energy when losing height and regaining as that kinetic energy is converted back to potential energy 👍
Can you tell me about your technique of learning ? U have got such an amazing personality with good humor ! Very less people possess them ! We will always be supporting you ! I told all of my classmates and my relatives about your videos ! I m totally renowned in the class because of your videos ! Everyone's saying "its great " ! Keep uploading videos ! Can you do some more videos relating to nunerical of kinematics
Thank you for all of your support! It is awesome you are letting everybody know about my work; that is one of the best ways to make sure I can continue to do this. I have a whole series of videos about kinematics: www.flippingphysics.com/algebra.html#1d
I have a question: Thinking about a vehicle accelerating from 0 to 60 and then slowing from 60 to 0. Is the energy to accelerate the vehicle from 0 to 60 the same as the energy to slow the vehicle from 60 to 0? ... I contend that more energy is required to accelerate to 60 than is required to slow the vehicle to 0. The reason for my conclusion is that wheel and bearing friction must be overcome to accelerate the vehicle, while the same friction assists with slowing the vehicle. Therefore the energy required to slow down the vehicle is less than the energy necessary to accelerate the vehicle. ... What do you think?
When a car speeds up, the engine is doing positive work on the road that is reflected back onto the car through the Newton's third law pair force of the traction. This positive work is occurring simultaneously with negative work due to parasitic friction in the bearings and parasitic air drag. The net work during the speed-up period is the work done by the engine minus the heat loss through the simultaneous resistive forces. The net work done BY the vehicle as it comes to rest equals the heat absorbed in the brakes plus any heat loss to air drag and friction that also occurs simultaneously. Both of these net work values will be equal and opposite. The NET WORK done accelerating an object from rest equals its kinetic energy once in motion, which also equals the net work (or damage) it can do as it returns to rest. You need to consider all forces that can do work, when keeping track of net work.
13:47 I take this class during my exchange year (i’m a junior) ... i also have AP calculus BC and AP Biology ... but physics is definitely the most fun but hardest class I have here..... the biggest problem is the language in the FRQ part
Question: what's different about the marker and the bottle that makes ones center of mass go down and ones center of mass go up, as you said in the video? Thanks for the videos
The difference is in the physical geometry of the objects. The shape of the base of the bottle makes it so that the center of mass of the bottle goes up. Although one could argue that the marker is in stable equilibrium for very small displacements from equilibrium. You are welcome for the videos.
I really enjoyed this video. As I teach physics it is great to see good explanations of topics. Here is a link to my notes on Energy: th-cam.com/play/PLfm0FJ-UppoEtypWeTXL41a46RoruNbCw.html As well as a link to some practice MCQ AP Questions: th-cam.com/play/PLfm0FJ-UppoFzxoi44hGQ_Z_flv8nqLvV.html It is always great for students to see concepts from multiple viewpoints.
Sir when energy is nonconservative than force constant or not if it is non constant force then y in spring we use non constant force formula as we know spring force comes under conservative force. Or energy
The conservative vs non-conservative classification of a force has nothing to do with whether it is constant or not. There are examples of constant forces that are non-conservative, such as kinetic friction. There are also examples of non-constant forces that are conservative, such as universal gravitation. What it means for a force to be conservative, is that the work done by it, is independent of the path, and that any energy exchange because of the force is reversible. Work done by a conservative force only depends on the initial and final states within that force field. Therefore, a conservative force has value in keeping track of its potential energy. By contrast, a non-conservative force means that energy is either introduced from another source to make it work (like human pushes and pulls), or energy is irreversibly removed from the mechanical energy domain (like friction). A spring force of an ideal is reversible, if you neglect friction, and if you don't overload it to the point of no return (called yielding). The work you do compressing a spring will equal the work that the spring will do on an object as you release it and let it return to the unstressed state. It doesn't matter if you repetitively compress and restore the spring, or if you just compress it once. The work done by the spring force is independent of path. With kinetic friction, an object does work on the friction agent as it comes to rest (i.e. heat generated in the two materials that rub together). But when it gets to rest, it will not bounce back and speed up the other way. If you make an object move against friction, the friction acts against its motion regardless of which direction it moves. Work done on the agent of friction is irreversible, and cannot be restored to mechanical energy with 100% efficiency.
For Ue=1/2kx^2, can this be only used when the situation is elastic? Also, if so, what does it mean by Elastic Potential Energy? Thank you! Comment: Your explanation....if someone does not understand, that is weird #having_question_does_not_mean_no_understanding Thank you!
@@danielkwon3062 The equation only applies if your spring is linear-elastic in the region of loading. This means it is a spring that obey's Hooke's law, and reversibly restores all energy you add to it, when deforming it from the unloaded position. Most materials exhibit a region that is either linear-elastic, or can be approximated as linear-elastic, where the stress vs strain curve is linear. For a brittle material, it ruptures immediately after reaching its strength limit, and the material is usually linear-elastic up to that point. For ductile and malleable materials, overloading the material will take it to a point of no return, called yielding, where it will permanently store some of its deformation. The stress vs strain curve will also be concave-down in this region. Upon unloading it, it will return upon a line that is parallel to the original elastic line, but when unloaded, there will be permanent deformation stored in the material. The 1/2*k*x^2 equation, and correspondingly F=-k*x equation, will both be invalid if you load a spring beyond the point of no return (the yield point).
@@FlippingPhysics One theory is that it's because electric potential energy is V and U looks very similar to V (and historically, they were the same letter).
@@bbrk24 That seems to be a reasonable guess, since V stands for Count Alessandro Volta's name that evolved into Voltage (i.e. electric potential energy per unit charge). The U and V distinction would be what gets used to distinguish "potential energy" and "potential", with the latter being potential energy per unit property of matter that causes the kind of force involved. U would be the choice for potential energy just by being V's alphabet neighbor, if this notation evolved after the two were distinct letters.
If i were riding in car 80mph and threw a pencil up in the air inside the car, shouldn't the pencil fly back immediately just as if i were to do it outside the car?
@Brandon Tashi But it's not Force times displacement, it's force dotted with a displacement vector. It matters what direction the force is in relation to the displacement
@@legoblox01 In the special case of the force and displacement being aligned, the dot product is simplified to a simple integral of force relative to displacement. In the special case of force being constant, the integral simplifies to a simple dot product of the force and displacement vectors. When both these special cases happen in the same situation, work is a simple product of force and distance. In the general case, work is a line integral of force dot product with the infinitesimal displacement vector, along the path that the object moves. You need the integral to account for force not being constant, and a dot product to keep track of it not necessarily being aligned with displacement. You make an infinite number of force dot displacement dot products along each infinitesimal displacement of the object's motion, and then add them up. When you add up an infinite number of infinitesimal values, that is integration.
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You taught me more than my physics teacher.
sorry or you're welcome. (you choose)
IFKR! this guy is really good
I am in college and these videos help greatly with calculus-based physics. Thanks a lot!
You are welcome. AP Physics C is basically a generic calculus-based introductory college physics course.
Yes, I wished I would've taken at least one AP Physics course while I was in high school. You remind me a lot of a physics teacher from my high school, and everybody loved him as well as how well he taught. Wished I would've had that opportunity to take a class from him. Physics in college is a different cat!
@@Folinic damn that was 5 years ago I’m interested where you’re at right now
I appreciate you and your hard work. Realize that my AP Physics C test is in ~24 hrs and and because of you, I just might pass. Super clear, understood everything.
Woke up at 3:45 AM to study for my physics quiz today, and felt like I needed a good review of things before. Thanks a lo for this really helpful video!
Best of luck with your quiz today!
(I hope you get a nap.)
I've got my apc mechanics final tomorrow and this is really helping me out, thanks!
You are welcome. I hope the final went well!
In the first 108 seconds of this video, I learned more than hours and days of physics classes taught in a month. I'm never going back to "49 degrees south of east"!
Glad to help you my friend.
I love the fact that you are so precise. I've got books where the displacement is simply denoted by "d", and that annoys the hell out of me.
That annoys me as well. It took me several years as a teacher to realize I need to be very precise in everything I say and write. It is the best way to make sure students understand what I am trying to teach them.
Your videos and editing are amazing!
Thank you for the compliment!
I did some research and U is purely a technical varibale, not a symbol like F for force. It was chosen because all the other letters were taken up in Thermo!
Thanks. Could you please provide links for your sources?
Will there be a series of videos like this for APPC: E&M in the future? Thank you for all you do!
That is my current plan. My hope is to, after completing the AP Physics C: Mechanics Review Videos, to then move on to work simultaneously on the AP Physics C: Electricity and Magnetism Review Videos and continue working on the AP Physics 1 videos which are more demonstration oriented (and therefore more fun to create). Hopefully that's what will happen.
Flipping Physics have you made any of them yet? The test is in a couple days and anything would be great
No. The E&M videos are not done. Sorry.
Flipping Physics np thanks for your work
The U in potential energy actually comes from the graph of potential energy, which decreases as it converts to kinetic energy when losing height and regaining as that kinetic energy is converted back to potential energy 👍
You are not the first person to tell me this, however, I have searched for evidence this is true and been unable to find any.
Can you tell me about your technique of learning ? U have got such an amazing personality with good humor ! Very less people possess them ! We will always be supporting you ! I told all of my classmates and my relatives about your videos ! I m totally renowned in the class because of your videos ! Everyone's saying "its great " ! Keep uploading videos ! Can you do some more videos relating to nunerical of kinematics
Thank you for all of your support! It is awesome you are letting everybody know about my work; that is one of the best ways to make sure I can continue to do this. I have a whole series of videos about kinematics: www.flippingphysics.com/algebra.html#1d
this was so helpful thank you!
I have a question: Thinking about a vehicle accelerating from 0 to 60 and then slowing from 60 to 0. Is the energy to accelerate the vehicle from 0 to 60 the same as the energy to slow the vehicle from 60 to 0? ... I contend that more energy is required to accelerate to 60 than is required to slow the vehicle to 0. The reason for my conclusion is that wheel and bearing friction must be overcome to accelerate the vehicle, while the same friction assists with slowing the vehicle. Therefore the energy required to slow down the vehicle is less than the energy necessary to accelerate the vehicle. ... What do you think?
When a car speeds up, the engine is doing positive work on the road that is reflected back onto the car through the Newton's third law pair force of the traction. This positive work is occurring simultaneously with negative work due to parasitic friction in the bearings and parasitic air drag. The net work during the speed-up period is the work done by the engine minus the heat loss through the simultaneous resistive forces. The net work done BY the vehicle as it comes to rest equals the heat absorbed in the brakes plus any heat loss to air drag and friction that also occurs simultaneously. Both of these net work values will be equal and opposite.
The NET WORK done accelerating an object from rest equals its kinetic energy once in motion, which also equals the net work (or damage) it can do as it returns to rest. You need to consider all forces that can do work, when keeping track of net work.
good luck tomorrow yall
Absolutely!
EXCELLENT!!!...... Period!
+Philip Y Thank you!
13:47 I take this class during my exchange year (i’m a junior) ... i also have AP calculus BC and AP Biology ... but physics is definitely the most fun but hardest class I have here..... the biggest problem is the language in the FRQ part
(I’m from Germany)
Question: what's different about the marker and the bottle that makes ones center of mass go down and ones center of mass go up, as you said in the video? Thanks for the videos
The difference is in the physical geometry of the objects. The shape of the base of the bottle makes it so that the center of mass of the bottle goes up. Although one could argue that the marker is in stable equilibrium for very small displacements from equilibrium. You are welcome for the videos.
you made me love physics!
awesome!
I really enjoyed this video. As I teach physics it is great to see good explanations of topics. Here is a link to my notes on Energy:
th-cam.com/play/PLfm0FJ-UppoEtypWeTXL41a46RoruNbCw.html
As well as a link to some practice MCQ AP Questions:
th-cam.com/play/PLfm0FJ-UppoFzxoi44hGQ_Z_flv8nqLvV.html
It is always great for students to see concepts from multiple viewpoints.
Sir when energy is nonconservative than force constant or not if it is non constant force then y in spring we use non constant force formula as we know spring force comes under conservative force. Or energy
The conservative vs non-conservative classification of a force has nothing to do with whether it is constant or not. There are examples of constant forces that are non-conservative, such as kinetic friction. There are also examples of non-constant forces that are conservative, such as universal gravitation.
What it means for a force to be conservative, is that the work done by it, is independent of the path, and that any energy exchange because of the force is reversible. Work done by a conservative force only depends on the initial and final states within that force field. Therefore, a conservative force has value in keeping track of its potential energy. By contrast, a non-conservative force means that energy is either introduced from another source to make it work (like human pushes and pulls), or energy is irreversibly removed from the mechanical energy domain (like friction).
A spring force of an ideal is reversible, if you neglect friction, and if you don't overload it to the point of no return (called yielding). The work you do compressing a spring will equal the work that the spring will do on an object as you release it and let it return to the unstressed state. It doesn't matter if you repetitively compress and restore the spring, or if you just compress it once. The work done by the spring force is independent of path.
With kinetic friction, an object does work on the friction agent as it comes to rest (i.e. heat generated in the two materials that rub together). But when it gets to rest, it will not bounce back and speed up the other way. If you make an object move against friction, the friction acts against its motion regardless of which direction it moves. Work done on the agent of friction is irreversible, and cannot be restored to mechanical energy with 100% efficiency.
For Ue=1/2kx^2, can this be only used when the situation is elastic?
Also, if so, what does it mean by Elastic Potential Energy?
Thank you!
Comment: Your explanation....if someone does not understand, that is weird
#having_question_does_not_mean_no_understanding
Thank you!
That is the equation for elastic potential energy which I explain in this video:www.flippingphysics.com/intro-pee.html
That video helped a lot! Thank you!
@@danielkwon3062 The equation only applies if your spring is linear-elastic in the region of loading. This means it is a spring that obey's Hooke's law, and reversibly restores all energy you add to it, when deforming it from the unloaded position.
Most materials exhibit a region that is either linear-elastic, or can be approximated as linear-elastic, where the stress vs strain curve is linear. For a brittle material, it ruptures immediately after reaching its strength limit, and the material is usually linear-elastic up to that point. For ductile and malleable materials, overloading the material will take it to a point of no return, called yielding, where it will permanently store some of its deformation. The stress vs strain curve will also be concave-down in this region. Upon unloading it, it will return upon a line that is parallel to the original elastic line, but when unloaded, there will be permanent deformation stored in the material. The 1/2*k*x^2 equation, and correspondingly F=-k*x equation, will both be invalid if you load a spring beyond the point of no return (the yield point).
U r such an amazing person 😀
thanks. just trying to share my gifts with the world.
Flipping Physics thats great !!
Why is the symbol for gravitational potential energy a capital U?
Honestly, I do not know. My best guess is because a capital U looks like a container for energy. Other than that, no clue.
@@FlippingPhysics One theory is that it's because electric potential energy is V and U looks very similar to V (and historically, they were the same letter).
Bbrk24 Any evidence of that anywhere?
@@bbrk24 That seems to be a reasonable guess, since V stands for Count Alessandro Volta's name that evolved into Voltage (i.e. electric potential energy per unit charge). The U and V distinction would be what gets used to distinguish "potential energy" and "potential", with the latter being potential energy per unit property of matter that causes the kind of force involved. U would be the choice for potential energy just by being V's alphabet neighbor, if this notation evolved after the two were distinct letters.
If i were riding in car 80mph and threw a pencil up in the air inside the car, shouldn't the pencil fly back immediately just as if i were to do it outside the car?
Nope. The pencil will follow the same path as the ball in this video.
www.flippingphysics.com/skateboarding.html
it all depends on whether or not you have a front windshield... :) and a good Eye Doctor....
Is it possible to purchase a button-down Flipping Physics shirt, as seen in your videos?
Sorry, I do not sell those shirts. This is what I sell:
redbubble.com/people/flippingphysics/works/24843428-flipping-physics-billy-bobby-and-bo
When talking about the integral for work why is there no dot-product?
@Brandon Tashi But it's not Force times displacement, it's force dotted with a displacement vector. It matters what direction the force is in relation to the displacement
In first case work done by constant force , so dot product
In second case, the force is variable since it depends on x , so integrate
@@legoblox01 In the special case of the force and displacement being aligned, the dot product is simplified to a simple integral of force relative to displacement.
In the special case of force being constant, the integral simplifies to a simple dot product of the force and displacement vectors.
When both these special cases happen in the same situation, work is a simple product of force and distance.
In the general case, work is a line integral of force dot product with the infinitesimal displacement vector, along the path that the object moves. You need the integral to account for force not being constant, and a dot product to keep track of it not necessarily being aligned with displacement. You make an infinite number of force dot displacement dot products along each infinitesimal displacement of the object's motion, and then add them up. When you add up an infinite number of infinitesimal values, that is integration.
we take these concepts and more in grade 11 physics course in united arab emirates
It was awesome
As all learning should be.
Flipping Physics can u make Some videos on circular motion
That is what I am currently working on: www.flippingphysics.com/algebra.html#cm
Flipping Physics I checked the page it's really helpful 😄