Well done. I use your videos in my classes and I support you on Patreon. I hope enough others do that you are able to continue making videos for years to come. Yours are among the best available. You address the subtle details that many leave out or simply don't realize are there. Thanks
Thank you so much for this video! It was intimidating at first but the students' explanations were very clear and and diagrams helped a lot. I will be watching this video many times, haha.
I have a question about the displacement we used for W_b. I might be wrong but isn't the displacement in the case be d and not circumference/2. Can someone please help me figure out what I'm missing out on? Thank you!
Some version of this question has been posed by several people on this video. Here is my best attempt at answering it, however, please realize a TH-cam comment is a very difficult way to answer these questions. There is a reason I make TH-cam videos with visuals, equations, demonstrations, etc. The equation for a work done by a changing force (and if the force is changing direction, then the force is changing because force is a vector) is work equals the integral of force with respect to position (and only the force in the direction of the change in position). This means that, even though the work equation has displacement in it, because work equals the integral of force with respect to position, that displacement is for each infinitesimally small change in position, which means, as we move around the semicircle, the displacement is in the direction of the force for each infinitesimally small step around the semicircle, and the displacement used in the work equation is the distance around the semicircle. I hope that helps! I have a video about work as an integral here: www.flippingphysics.com/integral-work.html
@@FlippingPhysics I was about to ask that question! Thank you Mr. P for a succinct answer. Just want to ask how answering such questions are not so approachable on TH-cam platform?
Because the work done by a non-conservative force is path dependent. It matters how you got there. If you drag a box to the top of a mountain, friction will do more negative work when you take a zig zag path, than when you take a path straight up the mountain. The work done by gravity will be equal to mg(change in height) regardless of the path taken.
Some version of this question has been posed by several people on this video. Here is my best attempt at answering it, however, please realize a TH-cam comment is a very difficult way to answer these questions. There is a reason I make TH-cam videos with visuals, equations, demonstrations, etc. The equation for a work done by a changing force (and if the force is changing direction, then the force is changing because force is a vector) is work equals the integral of force with respect to position (and only the force in the direction of the change in position). This means that, even though the work equation has displacement in it, because work equals the integral of force with respect to position, that displacement is for each infinitesimally small change in position, which means, as we move around the semicircle, the displacement is in the direction of the force for each infinitesimally small step around the semicircle, and the displacement used in the work equation is the distance around the semicircle. I hope that helps! I have a video about work as an integral here: www.flippingphysics.com/integral-work.html
I have a doubt, Work done by a force is the scalar product of The force vector and displacement vector. While calculating friction you took it as Distance and not displacement. Am i wrong, I am new to this concept so could ya help out
Some version of this question has been posed by several people on this video. Here is my best attempt at answering it, however, please realize a TH-cam comment is a very difficult way to answer these questions. There is a reason I make TH-cam videos with visuals, equations, demonstrations, etc. The equation for a work done by a changing force (and if the force is changing direction, then the force is changing because force is a vector) is work equals the integral of force with respect to position (and only the force in the direction of the change in position). This means that, even though the work equation has displacement in it, because work equals the integral of force with respect to position, that displacement is for each infinitesimally small change in position, which means, as we move around the semicircle, the displacement is in the direction of the force for each infinitesimally small step around the semicircle, and the displacement used in the work equation is the distance around the semicircle. I hope that helps! I have a video about work as an integral here: www.flippingphysics.com/integral-work.html
Some version of this question has been posed by several people on this video. Here is my best attempt at answering it, however, please realize a TH-cam comment is a very difficult way to answer these questions. There is a reason I make TH-cam videos with visuals, equations, demonstrations, etc. The equation for a work done by a changing force (and if the force is changing direction, then the force is changing because force is a vector) is work equals the integral of force with respect to position (and only the force in the direction of the change in position). This means that, even though the work equation has displacement in it, because work equals the integral of force with respect to position, that displacement is for each infinitesimally small change in position, which means, as we move around the semicircle, the displacement is in the direction of the force for each infinitesimally small step around the semicircle, and the displacement used in the work equation is the distance around the semicircle. I hope that helps! I have a video about work as an integral here: www.flippingphysics.com/integral-work.html
sir i watched your video which was uploaded 5 year from now ,Those video was more realistic because u do expermient in class with ur homies and that was intresting 😀😀.this video i did not find that much intresting
Two mistakes I found in this vedio I could be wrong btw: 3:54 the W= -(-mgh) wouldn't that just make is mgh and not -mgh the negative shouldn't be there in front of the prenthesis 9:49 I understand the point you are trying to make here but my physics teacher would torture me if I didn't make the Force applied arrow larger than the Force kenetic because if they are equally the force kenetic would have to be force static or force static max and by stating the friction is kenetic the applied force vector should be significantly larger than the force kenetic... Overall great video Thx :)
I don't think your physics teacher would torture you. The velocity is constant but the acceleration is 0. Since the acceleration is 0 the ma term goes to zero and force applied equals force of friction. Be careful not to confuse velocity and acceleration.
i am a great worshipper of science particularly physicsand i dont believe in god which our society shows to us. i believe in physics because for me physics is god and always shows us truth .i want your guidance. i want to physics in nature the way you teach . please provide me some books lectures or anything that continue the fire of learning physics
Outstanding! Awesome explanation of these often confusing topics, well done Mr. P!
Glad you enjoyed it!
I love you flipping physics
Love you too mateymax
Well done. I use your videos in my classes and I support you on Patreon. I hope enough others do that you are able to continue making videos for years to come. Yours are among the best available. You address the subtle details that many leave out or simply don't realize are there. Thanks
That is wonderful. Thank you so much for your support!
I LOVE YOUR VIDEOS!
Thanks for the love!
Loved the free body diagrams.......
great explanation sir.
5:26 why is the angle 180 and not 90 - a
Thank you so much for this video! It was intimidating at first but the students' explanations were very clear and and diagrams helped a lot. I will be watching this video many times, haha.
Great video! Peace out
THANK YOU SIR
YOU ARE VERY WELCOME
I have a question about the displacement we used for W_b. I might be wrong but isn't the displacement in the case be d and not circumference/2. Can someone please help me figure out what I'm missing out on? Thank you!
Some version of this question has been posed by several people on this video. Here is my best attempt at answering it, however, please realize a TH-cam comment is a very difficult way to answer these questions. There is a reason I make TH-cam videos with visuals, equations, demonstrations, etc.
The equation for a work done by a changing force (and if the force is changing direction, then the force is changing because force is a vector) is work equals the integral of force with respect to position (and only the force in the direction of the change in position). This means that, even though the work equation has displacement in it, because work equals the integral of force with respect to position, that displacement is for each infinitesimally small change in position, which means, as we move around the semicircle, the displacement is in the direction of the force for each infinitesimally small step around the semicircle, and the displacement used in the work equation is the distance around the semicircle. I hope that helps! I have a video about work as an integral here: www.flippingphysics.com/integral-work.html
@@FlippingPhysics I was about to ask that question! Thank you Mr. P for a succinct answer. Just want to ask how answering such questions are not so approachable on TH-cam platform?
You really have my respect sir,
Thanks for providing such a great explanation..!!!
You are most welcome
2:52 Shouldn’t the force experienced by the body on an incline be mg(sinα) and not just mg? Can anyone please clarify this.
Sir why potential energy is not defined for non conservative force??
Because the work done by a non-conservative force is path dependent. It matters how you got there. If you drag a box to the top of a mountain, friction will do more negative work when you take a zig zag path, than when you take a path straight up the mountain. The work done by gravity will be equal to mg(change in height) regardless of the path taken.
nice video
PLEASE MAKE VIDEOS ON PHYSICAL OPTICS
Someday.
Not soon.
Sorry.
Please help me understand why the displacement for the last demonstrations is not equal to 0. The started and ended at the same location.
Some version of this question has been posed by several people on this video. Here is my best attempt at answering it, however, please realize a TH-cam comment is a very difficult way to answer these questions. There is a reason I make TH-cam videos with visuals, equations, demonstrations, etc.
The equation for a work done by a changing force (and if the force is changing direction, then the force is changing because force is a vector) is work equals the integral of force with respect to position (and only the force in the direction of the change in position). This means that, even though the work equation has displacement in it, because work equals the integral of force with respect to position, that displacement is for each infinitesimally small change in position, which means, as we move around the semicircle, the displacement is in the direction of the force for each infinitesimally small step around the semicircle, and the displacement used in the work equation is the distance around the semicircle. I hope that helps! I have a video about work as an integral here: www.flippingphysics.com/integral-work.html
So this is a part of Dynamics chapter in ap physics1
Nope. It's part of the Work and Energy topics in AP Physics 1.
@@FlippingPhysics ok thanks mr p
I have a doubt, Work done by a force is the scalar product of The force vector and displacement vector. While calculating friction you took it as Distance and not displacement. Am i wrong, I am new to this concept so could ya help out
Some version of this question has been posed by several people on this video. Here is my best attempt at answering it, however, please realize a TH-cam comment is a very difficult way to answer these questions. There is a reason I make TH-cam videos with visuals, equations, demonstrations, etc.
The equation for a work done by a changing force (and if the force is changing direction, then the force is changing because force is a vector) is work equals the integral of force with respect to position (and only the force in the direction of the change in position). This means that, even though the work equation has displacement in it, because work equals the integral of force with respect to position, that displacement is for each infinitesimally small change in position, which means, as we move around the semicircle, the displacement is in the direction of the force for each infinitesimally small step around the semicircle, and the displacement used in the work equation is the distance around the semicircle. I hope that helps! I have a video about work as an integral here: www.flippingphysics.com/integral-work.html
How displacement equals the curve path of the semi circle, not the diameter ?
Please help me out?
Some version of this question has been posed by several people on this video. Here is my best attempt at answering it, however, please realize a TH-cam comment is a very difficult way to answer these questions. There is a reason I make TH-cam videos with visuals, equations, demonstrations, etc.
The equation for a work done by a changing force (and if the force is changing direction, then the force is changing because force is a vector) is work equals the integral of force with respect to position (and only the force in the direction of the change in position). This means that, even though the work equation has displacement in it, because work equals the integral of force with respect to position, that displacement is for each infinitesimally small change in position, which means, as we move around the semicircle, the displacement is in the direction of the force for each infinitesimally small step around the semicircle, and the displacement used in the work equation is the distance around the semicircle. I hope that helps! I have a video about work as an integral here: www.flippingphysics.com/integral-work.html
sir i watched your video which was uploaded 5 year from now ,Those video was more realistic because u do expermient in class with ur homies and that was intresting 😀😀.this video i did not find that much intresting
Two mistakes I found in this vedio I could be wrong btw:
3:54 the W= -(-mgh) wouldn't that just make is mgh and not -mgh the negative shouldn't be there in front of the prenthesis
9:49 I understand the point you are trying to make here but my physics teacher would torture me if I didn't make the Force applied arrow larger than the Force kenetic because if they are equally the force kenetic would have to be force static or force static max and by stating the friction is kenetic the applied force vector should be significantly larger than the force kenetic...
Overall great video Thx :)
I don't think your physics teacher would torture you. The velocity is constant but the acceleration is 0. Since the acceleration is 0 the ma term goes to zero and force applied equals force of friction. Be careful not to confuse velocity and acceleration.
Most conceptual videos
i am a great worshipper of science particularly physicsand i dont believe in god which our society shows to us. i believe in physics because for me physics is god and always shows us truth .i want your guidance. i want to physics in nature the way you teach . please provide me some books lectures or anything that continue the fire of learning physics