Sir please continue your series of jee mains extra syllabus topics Your 5 videos really helped me during my 1st attempt Please make more videos on semiconductors, communication system and optical instruments 🙏🙏🙏🙏
10:14, well i made a assumption while solving the question that if buoyancy force on ballon and normal force to the ladder at earth are sufficient to balance (M+m)g then no external force at on system so as man moves up ladder comes down , so when moves to top of ballon ladder descends (mh)/M+m so actually man moves height of (Mh)/M+m , Now applying WET ON man work done by internal force is equal to work done by gravity in magnitude since change in K.E. is zero (man moves slowly) so according to this ans should be mgMh/M+m.... If you dont consider above assumption and ladder doesnt descend then man simply moves h height so ans is then mgh Therefore there are 2 answer depending upon case
In the first case too answer will be mgh. If you take whole as system then displacement of COG/COM is zero so Wg=0 Now WET Wboun+Win=0 And Wbou=-(M+m)g×mh/(m+M)=-mgh So Win=mgh
10:20 is the answer of the question W= mg( h- mh/M ) As the man goes up by h height w.r.r the ladder but the ladder and balloon move down by mh/M distance so that the centre of mass does not move
Sir i tried an experiment to verify according to work theorem work done by internal forces should be same in different frame of reference So work by internal forces should change wrt plank I tried to find w by int but the answer comes 1/2m(v+(Mv/m))² Why sir it's coming wrong?
Sir what's wrong with writing total work done by man plus friction is equal to change in K.E. of boat and man. Now as total work by friction is zero work done by man equals to the total change in K.E. i.e work done by the man on boat plus man himself equals change in total K.E. now as the work done by man on boat is already found in option 1 we can calculate the work done by man on himself this makes option 4 also correct. Sir I am conflicted because I know what you have shown is correct but I am unable to find the mistake in the above logic.
Sir Please any motivation video if possible how to study physics property for jee aspirant which books. To be followed and how much time should we give to a chapter requested topic if you can make video On it 2022 aspirant here
@Innovative Learning true 👍 . That's why I avoid doing them. I might just provide suggested books and materials. Strategy is for students to prepare depending on their strengths or weakness
Sir can you plz help me with the hint for the problem no. 13 in mcq section in pathfinder kinematics of translational motion and also plz give hint for question 20 of same chapter and same book 🙂🙂 plz sir
I am planning to bring some questions from PATHFINDER kinematics, I just need some time to open that book. Once I start doing it , there will be no stopping it . I love the book that much👍
@@PHYSICSSIRJEE thank you so much sir... it's so relieving that I can just daily improve my concept and understanding of the subject and study more about the topic learnt in my coaching... 😄
Sir one question, if we are considering water offers zero resistance to the plank, so if a person want to jump from the plank, would never leave the last edge(or vertices) of plank and his left leg stick to the surface... So his left leg moving with velocity v toward left and body moves right with velocity Vm, quite lethal 🤓🤓🤓
Sir Can -(Work done by internal forces) of man be considered as change in potential energy of man ? I mean to ask if internal forces of man are conservative or non conservative ?
@@PHYSICSSIRJEE sir to fir aapne change in velocity sirf cm ko kyon liya vcm-V aayega na change in kinetic energy qand also initial kinetic energy is zero i know but final ke =1/2m(vcm-V)^2 therefore final ke = change in ke @6:14
@@thermmech4835 as mentioned in video , most of man's system moves with Vcm(ignoring his size) except for feet points (very less mass) . Hence KE is considered in such Problems as KEof CM . We ignored Rotational KE
Sir I am in 10th icse board my friends have started jee prep now I am also starting sir i read mechanics of university physics till now Should I go with gc agarwal and anurag mishra grb publication or anurag mishra balaji or physics galaxy or tmh sbt I am confused sir please help me out sir
Sir here's a problem that has been bugging me for an year now, probably because none has been able to answer it satisfactorily. The T=I alpha eqn=rxF, has been derived from newton's second law. Therefore whenever we apply newton's 2nd law in a problem, we shouldn't need to use the t=ia as it shouldnt give us extra information as it has been derived from f=ma only. But we see in countless examples that t=ia eqn gives information which is apart from information gathered from f=ma. Why does this happen. In theory we should be able to solve all dynamics problems using f=ma. Like say for example let's take a simple case of a rigid rod lying on a horizontal plane with it's one end pivoted by a friction less hinge. A tangential force of magnitude constant f starts acting on the body at t=0. We need to obtain omega and alpha at given time t. Using only f=ma Not even the concepts of work, com etc. My approach to this was 1. Given rod is rigid velocity of any dm element I will write as xw, where s is distance from the pivot. 2. I will assume 4 forces acting on the dm element, 2 tangential forces due to elemts on the left and right side of dm, and 2 radial forces due to left and right side of dm 3 write it's acceleration and use newton's 2nd law in the 2 directions 4. Using the boundary conditions that tangential force at x=l is equal to our applied force and radial force at the end is zero. But sir doing this doesnt give me the ans. Which indicates there is something missing. But that missing thing can't be the torque eqn as it is derived from f=ma only. Pls clarify. There is only one person on the stack exchange that has asked this, and none of the answers are good enough.. I hope you understand what I want to say because it has happened many times that the other person is unable to understand what I'm trying to say.
Actually a rigid body has a large number of particles, so to solve a dynamics problem in 2 dimensions, we will need to write two F = ma equations for each particle. Assuming n particles, this will give us 2n equations. Then, we will have 2(n - 1) equations from the condition that the body is rigid and distance between any pair of particles is fixed (constraint equations basically). Solving these equations simultaneously will reduce the math to 3 equations, with new variables defined in terms of the old ones. Depending on how you choose these new variables, you can get different versions of these 3 equations. The most common version being: 1. 2 F = ma equations for the CM 2. τ = I α relative to the CM
![[JEE ADVANCED] WORK DONE BY INTERNAL FORCES | MAN LADDER BALLOON | SCHOOL PHYSICS SIR JEE QUICKIES](http://i.ytimg.com/vi/yYNCbGxsAVc/mqdefault.jpg)
Oh You Are Alive ❤️❤️🙏🙏
😂😂
@@Fighter_Believer_Achiever no, I guess
Sir please continue your series of jee mains extra syllabus topics
Your 5 videos really helped me during my 1st attempt
Please make more videos on semiconductors, communication system and optical instruments
🙏🙏🙏🙏
Great conceptual question
Only intuitive man in you tube who really have knowledge.... Can handle any problem.
Sir this exactly same ques i solved from allen grs sheet today and was unconvinced by the soln given
but now I'm fully satisfied 😌🌟🏆
Sir please take online live classes also cuz I cannot come to Hyderabad but really wish to study from you. Plzzz sir any solution for us
I would have definitely , if a day had 40 hrs 😁. But, on a serious note , I will think about something to connect to more students
@@PHYSICSSIRJEE sir are u willing to join unacademy in the future as jee advanced is over now
Thanks sir. Cleared a lot of concepts
10:14, well i made a assumption while solving the question that if buoyancy force on ballon and normal force to the ladder at earth are sufficient to balance (M+m)g then no external force at on system so as man moves up ladder comes down , so when moves to top of ballon ladder descends (mh)/M+m so actually man moves height of (Mh)/M+m ,
Now applying WET ON man work done by internal force is equal to work done by gravity in magnitude since change in K.E. is zero (man moves slowly) so according to this ans should be mgMh/M+m....
If you dont consider above assumption and ladder doesnt descend then man simply moves h height so ans is then mgh
Therefore there are 2 answer depending upon case
In the first case too answer will be mgh.
If you take whole as system then displacement of COG/COM is zero so Wg=0
Now WET Wboun+Win=0
And Wbou=-(M+m)g×mh/(m+M)=-mgh
So Win=mgh
nice concept sir, thank you
Sir if there is no slipping then wouldn't the displacement of the point of application of force be 0,then sir how friction is doing work?
Plank is moving back, so displacement of point of application is backward
@@PHYSICSSIRJEE thank you sir
10:20 is the answer of the question
W= mg( h- mh/M )
As the man goes up by h height w.r.r the ladder but the ladder and balloon move down by
mh/M distance so that the centre of mass does not move
No 🙂
Good to see you sir!
Sir i tried an experiment to verify according to work theorem work done by internal forces should be same in different frame of reference
So work by internal forces should change wrt plank
I tried to find w by int but the answer comes 1/2m(v+(Mv/m))²
Why sir it's coming wrong?
Sir what's wrong with writing total work done by man plus friction is equal to change in K.E. of boat and man. Now as total work by friction is zero work done by man equals to the total change in K.E. i.e work done by the man on boat plus man himself equals change in total K.E. now as the work done by man on boat is already found in option 1 we can calculate the work done by man on himself this makes option 4 also correct. Sir I am conflicted because I know what you have shown is correct but I am unable to find the mistake in the above logic.
Always enjoyed the video throughout the solution🔥OP
Yes
God teacher after 3 days is back 💖💖💖
10:14 is answer W_int=mgh ?
Sir Please any motivation video if possible how to study physics property for jee aspirant which books. To be followed and how much time should we give to a chapter requested topic if you can make video On it 2022 aspirant here
I will plan something 👍. Give me some time. Say 2 Weeks
@Innovative Learning true 👍 . That's why I avoid doing them. I might just provide suggested books and materials. Strategy is for students to prepare depending on their strengths or weakness
Sir can you plz help me with the hint for the problem no. 13 in mcq section in pathfinder kinematics of translational motion and also plz give hint for question 20 of same chapter and same book 🙂🙂 plz sir
I am planning to bring some questions from PATHFINDER kinematics, I just need some time to open that book. Once I start doing it , there will be no stopping it . I love the book that much👍
@@PHYSICSSIRJEE yes sir we want pathfinder solutions at top priority plz sir maja aayega
@@PHYSICSSIRJEE Sir what do you do when you are unable to do a question??
Even after multiple attempts
@@Fighter_Believer_Achiever leave it for the time being and come back to it later with a fresh mind
Amazing concepts just clearly visualised all the saying in the video thankyou sir
10:11 sir, is the answer "mgh" ?
Yes , I have made two videos on this. Please go through them, you will definitely enjoy
@@PHYSICSSIRJEE thank you so much sir... it's so relieving that I can just daily improve my concept and understanding of the subject and study more about the topic learnt in my coaching... 😄
A man moving on plank from one end to other end sir what is work done by man on boat sir if he stopped suddenly and if he stopped gradually
Sir one question, if we are considering water offers zero resistance to the plank, so if a person want to jump from the plank, would never leave the last edge(or vertices) of plank and his left leg stick to the surface... So his left leg moving with velocity v toward left and body moves right with velocity Vm, quite lethal 🤓🤓🤓
10.24: (mg-(bouyancy on man))h please reply sir is it correct
Thankyou sir...❤️
Really a conceptual problem
Sir can you please tell source of this question ?
The DPP I give to my students while teaching . Btw, many standard coaching centers have similar problems in their material , I guess
@@PHYSICSSIRJEE Ok sir thanks for replying
sir is ans of homewrok question 2mMgh/M-m
Sir Can -(Work done by internal forces) of man be considered as change in potential energy of man ? I mean to ask if internal forces of man are conservative or non conservative ?
Non-conservative. You cannot possibly store and retrieve that work
also sir it is wrong to say i think that friction is doing negative workdone since friction will be doing positive instead of negative
On plank/boat positive ,.... On person negative.... Total sum is zero
@@PHYSICSSIRJEE sir to fir aapne change in velocity sirf cm ko kyon liya vcm-V aayega na change in kinetic energy qand also initial kinetic energy is zero i know but final ke =1/2m(vcm-V)^2 therefore final ke = change in ke @6:14
@@thermmech4835 as mentioned in video , most of man's system moves with Vcm(ignoring his size) except for feet points (very less mass) . Hence KE is considered in such Problems as KEof CM . We ignored Rotational KE
@@PHYSICSSIRJEE ok sir thankyou sir aapne mere do doubt ek saath clear kr diye especially the microscopic one thankyou sir..
Sir I am in 10th icse board my friends have started jee prep now
I am also starting sir i read mechanics of university physics till now
Should I go with gc agarwal and anurag mishra grb publication or anurag mishra balaji or physics galaxy or tmh sbt
I am confused sir please help me out sir
If you have to choose one book among your options , go for Anurag Mishra, Balaji
@@PHYSICSSIRJEE thanks sir any comment on gc agarwal and anurag mishra series sir I am worried sir many told grb is better but no review.confused sir
@@senthil7002 I found Anurag Mishra Balaji better
@@PHYSICSSIRJEE thank you sir for replying me 💯
Following your problems eventhough couldn't understand
Sir plzz use hinglish language ( hindi+english ) ...
Plzzz sirr
Sir here's a problem that has been bugging me for an year now, probably because none has been able to answer it satisfactorily. The T=I alpha eqn=rxF, has been derived from newton's second law. Therefore whenever we apply newton's 2nd law in a problem, we shouldn't need to use the t=ia as it shouldnt give us extra information as it has been derived from f=ma only. But we see in countless examples that t=ia eqn gives information which is apart from information gathered from f=ma. Why does this happen. In theory we should be able to solve all dynamics problems using f=ma. Like say for example let's take a simple case of a rigid rod lying on a horizontal plane with it's one end pivoted by a friction less hinge. A tangential force of magnitude constant f starts acting on the body at t=0. We need to obtain omega and alpha at given time t. Using only f=ma
Not even the concepts of work, com etc.
My approach to this was
1. Given rod is rigid velocity of any dm element I will write as xw, where s is distance from the pivot.
2. I will assume 4 forces acting on the dm element, 2 tangential forces due to elemts on the left and right side of dm, and 2 radial forces due to left and right side of dm
3 write it's acceleration and use newton's 2nd law in the 2 directions
4. Using the boundary conditions that tangential force at x=l is equal to our applied force and radial force at the end is zero.
But sir doing this doesnt give me the ans.
Which indicates there is something missing. But that missing thing can't be the torque eqn as it is derived from f=ma only. Pls clarify. There is only one person on the stack exchange that has asked this, and none of the answers are good enough.. I hope you understand what I want to say because it has happened many times that the other person is unable to understand what I'm trying to say.
Actually a rigid body has a large number of particles, so to solve a dynamics problem in 2 dimensions, we will need to write two F = ma equations for each particle. Assuming n particles, this will give us 2n equations. Then, we will have 2(n - 1) equations from the condition that the body is rigid and distance between any pair of particles is fixed (constraint equations basically). Solving these equations simultaneously will reduce the math to 3 equations, with new variables defined in terms of the old ones. Depending on how you choose these new variables, you can get different versions of these 3 equations. The most common version being:
1. 2 F = ma equations for the CM
2. τ = I α relative to the CM
Wint = (M-m)gh Sir, is this correct?
No , if M=m , is the work=0? . Please think again 👍
@@PHYSICSSIRJEE Sir is it 2Mmgh/(M+m)
@@girirajgupta5438 I don't think so, I don't remember the answer. Please wait for my Solution video in a few days time 🙂
@@PHYSICSSIRJEE Ok sir, I will be waiting for the video
Sir W int=mgh
🙌🙌
Sir wouldn’t it be better to say the work done by man *by* himself as he is only providing kinetic energy to both himself and the boat
Yes , if we take both both man and boat as system .👍 I wanted to take individual systems to make students realise the intricate points involved
🔥🔥🙂
Answer to the practice problem is W=(2Mmhg)/(M+m). Please verify my answer.
Nope . Please try again
Ok Sir
@@PHYSICSSIRJEE I tried again and the answer is mgh.
Previously I did not take buoyant force into consideration.
@@swatiprusty5256 perfect 👍