There's another easy method.. This can be done by finding the maximum height using the speed eqn. (v²=2as+u²) when the body is going up and by putting the variables we will get the distance travelled. Then we will use the same speed equation when the body is coming down the slope and in this equation we will put the value of distance(s) and we will be getting the relation for friction coefficient..
interestingly that's essentially the same method as the kinetic energy being 1/2mv^2 can be derived from v^2=u^2+2as for constant acceleration (or with calculus in the general case). Thank you for the comment!
Try to bring more and more tough question..I am former jee adv aspirant but i still love to watch your videos!!And help me prep for physics olympiads..!!
@@zhelyo_physics it is ! thinking of it, I spend a lot of time watching your videos, and the remaining time thinking about them, :D !!! you are my favourite !!
Sir my approach to this method is as follows. Apply work energy theorem to the particle when it is travelling up and comes to rest. By taking zero potential level as the base of the inclined plane , we get , - f x d = 0 - (1/2)*m*(v^2) + mgh - 0 (f is the friction force) Similarly , applying work energy theorem , when it comes down along the incline plane , -f x d = (1/2)*m *(v/2)^2 - 0 + 0 - mgh By removing kinetic energy terms from the above 2 equations , we get 5 x f x d = 3 × mgh . Si
absolutely, always filming more problems - I put the problems here : th-cam.com/play/PLSygKZqfTjPCKus_q7vWTaRClR2hi6XFe.html&si=1ivlbBWb7r1nMFGm. Thanks for your comment!
Can you plz explain why U 238 can't participate in the fission process , I think becuz it need a neutron with more kinetic energy Correct me if iam wrong
The binding energy released by U 238 as it undergoes the process of absorbing a thermal neutron is less than the critical energy. It cannot sustain a chain reaction because the neutrons don't have another energy to cause further fission. Uranium 238 is fissionable but not fissile. So yes it would need higher energy neutrons
Hello ZPhysics! I was just wondering if you could suggest a book on errors and all for experimental part of physics olympiads. I think this topic is underrated and i think it is equally important. I wonder if you could help. Thank you!
Excellent question, let me research so I can give an appropriate answer. I did learn how to do those properly at university using lecture notes, but I am sure there would be a great book out there.
Hi! I'm in 8th Grade (IGCSE Curriculum), I started with University Physics for concepts,i solve the book problems, but what to do when i can't solve problems, stick to that problem or something else? And one more thing how to increase problem solving skill
can u make a playlist for CIE AS and A Level , like put all in one playlist instead of everyone finding seperately i would appreciate it. and many others would btw luv the vids !
hm, weird, I tried this problem and I got 1/7 tan(alpha) the thing is, I expressed the distance and then plugged that into a formula for the friction coefficient, which I then had to rearrange, which could have lead to mistakes along the way, i certainly didnt make an expression for v^2, thats for sure Update: expressing v^2 really is the key to this, otherwise the math gets too complicated
Sir my approach to this method is as follows. Apply work energy theorem to the particle when it is travelling up and comes to rest. By taking zero potential level as the base of the inclined plane , we get , - f x d = 0 - (1/2)*m*(v^2) + mgh - 0 (f is the friction force) Similarly , applying work energy theorem , when it comes down along the incline plane , -f x d = (1/2)*m *(v/2)^2 - 0 + 0 - mgh By removing kinetic energy terms from the above 2 equations , we get 5 x f x d = 3 × mgh . Since d = h tan(alpha) , we get the same exact answer for coefficient of friction.
Check out my Physics Olympiad Playlist here: th-cam.com/play/PLSygKZqfTjPCKus_q7vWTaRClR2hi6XFe.html
Once try a book named "Pathfinder for physics Olympiad" a very nice book for challenging problems in high school physics, you'll surely love it sir 😊
thank you for the suggestion!
There's another easy method..
This can be done by finding the maximum height using the speed eqn. (v²=2as+u²) when the body is going up and by putting the variables we will get the distance travelled. Then we will use the same speed equation when the body is coming down the slope and in this equation we will put the value of distance(s) and we will be getting the relation for friction coefficient..
interestingly that's essentially the same method as the kinetic energy being 1/2mv^2 can be derived from v^2=u^2+2as for constant acceleration (or with calculus in the general case). Thank you for the comment!
@@zhelyo_physics Okay Thank U sir...
Please bring some questions of Indian Physics Olympiad (INPHO) as it is considered very hard in India...
Try to bring more and more tough question..I am former jee adv aspirant but i still love to watch your videos!!And help me prep for physics olympiads..!!
Thanks for the comment and the suggestion!
I was just watching another one of your videos when this notification dropped - more physics time!!
wohooo, thank you for watching! Hope this is helpful!
@@zhelyo_physics it is ! thinking of it, I spend a lot of time watching your videos, and the remaining time thinking about them, :D !!! you are my favourite !!
Sir my approach to this method is as follows.
Apply work energy theorem to the particle when it is travelling up and comes to rest. By taking zero potential level as the base of the inclined plane , we get ,
- f x d = 0 - (1/2)*m*(v^2) + mgh - 0
(f is the friction force)
Similarly , applying work energy theorem , when it comes down along the incline plane ,
-f x d = (1/2)*m *(v/2)^2 - 0 + 0 - mgh
By removing kinetic energy terms from the above 2 equations , we get 5 x f x d = 3 × mgh . Si
Make an IPHO problem solving series please!
absolutely, always filming more problems - I put the problems here : th-cam.com/play/PLSygKZqfTjPCKus_q7vWTaRClR2hi6XFe.html&si=1ivlbBWb7r1nMFGm. Thanks for your comment!
Well! It's a good question but take more than this tough question
I agree : ) Glad you enjoyed it!
Can you plz explain why U 238 can't participate in the fission process , I think becuz it need a neutron with more kinetic energy
Correct me if iam wrong
The binding energy released by U 238 as it undergoes the process of absorbing a thermal neutron is less than the critical energy. It cannot sustain a chain reaction because the neutrons don't have another energy to cause further fission. Uranium 238 is fissionable but not fissile.
So yes it would need higher energy neutrons
Hello ZPhysics! I was just wondering if you could suggest a book on errors and all for experimental part of physics olympiads. I think this topic is underrated and i think it is equally important. I wonder if you could help. Thank you!
Excellent question, let me research so I can give an appropriate answer. I did learn how to do those properly at university using lecture notes, but I am sure there would be a great book out there.
Hi! I'm in 8th Grade (IGCSE Curriculum), I started with University Physics for concepts,i solve the book problems, but what to do when i can't solve problems, stick to that problem or something else? And one more thing how to increase problem solving skill
can u make a playlist for CIE AS and A Level , like put all in one playlist instead of everyone finding seperately
i would appreciate it. and many others would
btw luv the vids !
excellent idea! I will try to get it done before exams. Thank you.
hm, weird, I tried this problem and I got 1/7 tan(alpha)
the thing is, I expressed the distance and then plugged that into a formula for the friction coefficient, which I then had to rearrange, which could have lead to mistakes along the way, i certainly didnt make an expression for v^2, thats for sure
Update: expressing v^2 really is the key to this, otherwise the math gets too complicated
You're the best!!
thank you so much!
Sir my approach to this method is as follows.
Apply work energy theorem to the particle when it is travelling up and comes to rest. By taking zero potential level as the base of the inclined plane , we get ,
- f x d = 0 - (1/2)*m*(v^2) + mgh - 0
(f is the friction force)
Similarly , applying work energy theorem , when it comes down along the incline plane ,
-f x d = (1/2)*m *(v/2)^2 - 0 + 0 - mgh
By removing kinetic energy terms from the above 2 equations , we get 5 x f x d = 3 × mgh . Since d = h tan(alpha) , we get the same exact answer for coefficient of friction.
first!
thanks for the comment!