Class 11 physics chapter 6 | Work,Energy and Power 03 | Work Energy Theorem IIT JEE NEET ||

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[@htga75]

Sir iska answer to bata dijiye plzz

[@ritu4888]

Sir iska ans Kya aa rha h

[@annanarang1783]

sir ji please make vedio on power

[@karan-zx4ws]

Bhiyaa Its (c) is coorect Naa Sir🙌

[@veyantsingh6003]

It is option c bt how to calculate the work done by friction ...

[@jadisrinija1338]

Better than coaching centers how many of u agree hit a like..

[@hec9lerwalker218]

Bilkul bhaii..💯Facts

[@alliswell7740]

Good teacher

[@lucifermorningstar1209]

I AGREE

[@anuragdwivedi5685]

Bilkul sai kai

[@venkateshchamalwar6372]

I agree

[@aryanpandey2929]

The answer to the last question is C Wf + Wg + WF + Wn =0 Wg = -mgh f = μmgcosθ Now s=l/cosθ Wf = -μMgl {as friction tries to retards its motion} Wn = 0 [as s is always perpendicular to N] by adding these WF= μmgl+ mgh THANK YOU SIR FOR THE TEACHINGS.

[@imranalam7060]

Love u bro u are great

[@anjulpriyadarshi5993]

Thanks

[@Muzann77]

@Mohit Malgaonkar it's a mountain so obviously there will be an angle ∅ N = mg cos∅ f=uN=umgcos∅

[@samimahmed8317]

Correct answer

[@bhoomirathi7894]

@Mohit Malgaonkar friction is u •Normal and here normal is mgcos€

[@Devilfish-r6u]

45:10 answer is whole root of 23/5

[@ShresthanshNayak]

Correct bhai

[@__Itz__Ravi__5452]

Thanks bro for answer I mistake last step 😊

[@Devilfish-r6u]

@@__Itz__Ravi__5452 you welcome 😊

[@PriyanshuMishra-u1b]

i thought my answer was wrong but its correct. thank you

[@Infintyworld015]

Yes that's right

[@amanagrawal6541]

Solution of last question is here by Amanagrawal :- Gradually means speed is constant so the change in kinetic energy is 0. So sum of work done by all four forces became 0. 1:- workdone by gravitational force=-mgh. 2:- workdone by normal reaction is 0 because theta is 90 between normal reaction and displacement. 3:-workdone by friction= -umgl It is because costheta = b/h so costheta= L/√L^2+h^2. Now friction (f)=u.N where N = mgcostheta. Displacement (s) = hypotenus =√L^2+h^2 Now workdone by friction=f.s.costheta and angle between friction and displacement is 180 so work done by friction= umg.L/√L^2+h^2 ×√L^2+h^2×cos180 =-umgl Now you can find work done by external force by equating the sum of all the work done with 0. Thank you

[@govindabhagat8767]

Well done

[@pragyanjena1054]

Why equating with zero

[@shubhamyadav-kj6pv]

Right bro good job

[@lockheedmartin286]

@@pragyanjena1054 because it's speed is constant (as given gradually ) and so , kinetic energy final = kinetic energy initial, and by this total work done will be ZERO

[@pragyanjena1054]

@@lockheedmartin286 yes.I understand after commenting

[@kingff3385]

Final velocity=initial velocity So that work done by all force is 0 Word done (by g)=f.s W=mg×h×Cos180 degree W=-mgh Word done (by Normal force)=f.s W=O(because theta is 90 degree and cos 90 degree is 0) Word done(by friction)=f.s ( Friction force=u×N F=u.mg) W=u.mg.l.cos180degree W=-umgl W(all force)=work done(by g)+work done(by Normal f)+work done(by friction)+work done by tangential force 0=-mgh+0-umgl+work done by tangential force Work done (tangential force)=umgl+mgh

[@Regina-hu6xk]

Work by Friction ke vakt s ko l kaise liya aapne

[@SurenderKumar-kl7dh]

But cos 180 is -1

[@f0rsaken296]

@@SurenderKumar-kl7dh he did write it minus look carefully

[@f0rsaken296]

@@Regina-hu6xk i nhi l liya hai

[@channelclosed9987]

Thanks

[@anujbhardwaj1729]

Answer of last q is c Wf+Wg+WF Wn=0 Wg=-mgh f=mu mgcos{teeta} Now s=l/cos{teeta} Wf=-Mu Mgl {as friction tries to retards its motion} Wn=0 [as s is always perpendicular to N] by adding these WF= mu mgl+ mgh

[@sourabhyadav1437]

Buddy can you plz tell me why Wg is taken as mgh instead of mg

[@sourabhyadav1437]

@UC2eog5QhTWyACuOTXfOAAEw thanks..now I got it

[@learner86765]

Thanks bro

[@rohanmishra8801]

Can you please tell me why it's my mgl instead of mu mgcos@

[@asiyapf7257]

@@rohanmishra8801 its bcz we need to find work done by frictional force and its fricional force multiply by distance here distance is l/costheta bcz costheta =l/hypotenus so hypote which is our distance comes out to be l/costheta so thaf work done by frictional force =mu mgcostheta multiplied by l/cos theta which comes out to be mu mgl hope u got it

[@rupamkarmakar7913]

4 yrs passed but still the value of this video can't be defined 🙌❤️

[@spider-man6932]

@@ajaysuryajacksparrow1618 is this correct ??

[@whiteglosss__]

Now it’s five years! ❤

[@Shinmen61]

Physics dies after 1 year or what?

[@riyanshsharma9541]

Bhai hme to w by all forces = Δk = m/2(v²-u²) kraya hai pr idhr sir to u²-v² le rhe hai ye clear krna

[@whiteglosss__]

@@riyanshsharma9541 jesa physics wallah padha rhe h vesa kerlo agar koi dikkat hoti to wo correction ker chuke hote khud!

[@smartenough722]

His previous videos are still better than pace series (physics and chemistry)

[@tushar5374]

Waha ki maths achhi hai... Physics aur chemistry ke lecture mai alakh sir ke hi dekhta hoon

[@studentlifestyle9673]

Option c is correct

[@jeekayvines1568]

@@tushar5374 ma bhi maths pace series se padta hu aur physics chemistry alakh sir se

[@jeekayvines1568]

@@studentlifestyle9673 yes mera bhi C aaya

[@harchan6274]

@@tushar5374 is maths there in pace series? if yes thx for the info

[@prernajasti6720]

answer for last question work done by all forces = change in kinetic energy WF=work done by force applied Wf=work done by frictional force Wn= work done by normal reaction Wg=work done by gravitational force WF+Wf+Wn+Wg=0 so, Wg= -mgh ( bcoz it is conservative)..............................(1) Wn= 0 (as normal reaction and displacement are perpendicular ,cos90=0)........................(2) n = mgcos theta ( cos component of gravitational force) frictional force (f)= mu (mg.cos theta) now lets assume the angle of repose to be theta cos theta = adjacent/ hypo ; so cos theta = l/s (where s is the hypo) ; s= l/cos theta ; so the displacement for frictional force is l/cos theta ; work done by frictional force = - mu mgcos theta x l/cos theta (where cos theta gets cancelled , negetive sign bcoz opposite direction ) so, Wf= -mu mgl.......................(3) WF= ?..................................(4) adding (1),(2),(3) and (4) WF+0-mgh-(mu mgl) = 0 WF = mgh +mu mgl correct answer option (c)

[@riyaa.s.p1600]

Frictional force is (u mg cos(theeta) ryt?

[@FaheemKhan-ef2wd]

Wo sab to thik ha lekin ham friction ki force ke liye work done nikaal rahe hai aur friction ki force non conservative hai aur iska work done path par depend karta hai to ham uske liye Roughen surface ko hata kar uske jagah "s= l/cos(theta)" kaise use kar skte hai Plzz batanaaa

[@prernajasti6720]

@@FaheemKhan-ef2wd i think sir ne rough surface isiliye banaya so that we know ki friction hoga .....its rough at a microscopic level toh isiliye usko straight hi consider karke s=l/cos(theta) lenge ...but friction hoga . i hope that helps😃

[@FaheemKhan-ef2wd]

@@prernajasti6720 oooo😂😂😂😂😂 DUE TO CONFUSION MENE YE QUESTION E NAI NOTE KARA

[@deehwat8033]

Thanks yaar plz send me diagram

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[@vaibhaviprakash9409]

Learn to use commas pls or you might as well ask sir to take some english lectures as well 😂

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Same here but I saw resistance (series and parallel) of class 10. And I just amazed and then I just saw all the concept of10 th and now 11th class

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Yui yoshioka why do you care if she uses punctuation marks or not? I can easily comprehend what she was trying to convey so there must be a problem with your pea-sized brain. You're an asshole.

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@@tiredcat6653 lol 🤣

[@tiredcat6653]

Ankit Mishra 😩

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I'm totally dependent on these lectures now for neet 2024. I have to crack it anyhow. Wish me luck everyone. So that I can enjoy this hard but beautiful journey.❤

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Same here! Wish you best of luck yrr… make god bless you! ❤

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@@whiteglosss__ best of luck to you too.❤️

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Bhai tum paper me tabahi macha dena

[@SatwikSinha-sf7kr]

@@Jdjdidhfb Thanks bhai. Best wishes for you too.

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4 year ago but this video is helping the students always

[@soniaghosh601]

Really

[@keshavsharma_8030]

Better than our offline classes

[@rajinikallu2689]

@@keshavsharma_8030 yes

[@SUSHanTM01]

yes

[@eimaan.3842]

Right..

[@notjustnerd]

45:21 answer v=√23/√5

[@SeemaYadav-cw5yt]

Ya same answer here

[@pacificscienceclasses2218]

Wrong hai apka

[@pacificscienceclasses2218]

Soryy shi h

[@adsportslover6714]

Ya same answer

[@swatipatra8303]

it is root of 46/10

[@aastik3565]

W by normal reaction =0 W by gravitation=-mgh friction force= µN here, normal reaction N =mgcostheta displacement =√(l^2+h^2 ) W by friction=- µmgcostheta √(l^2+h^2 ) So, w by external force - mhg- µmgcostheta √(l^2+h^2 ) =0 Costheta= b/h=l^2/√(l^2+h^2 ) By solving , work done by external force=mgh+µmgl so, answer is (c)

[@everywhere...9388]

Right 🔥 option is c

[@soumyaranjanpanda1227]

Thanks brother...I was wondering the value of costheta

[@somnathmukherjee585]

Thank u bro I was little bit confused of cos theta

[@pranavvyas7218]

Why normal reaction is equal to mgcostheta the block is moving upwards . Pls reply

[@prajwal7867]

@@pranavvyas7218 Because we take components of mg

[@anandmani1504]

45: 56 ​​my answer is umgl + mgh Work done by all forces = change in kinetic energy W.D by gravi. + W.D by friction + W.D by given force = 0. (Delta K.E=0) -mgh - umgcostheta(√h²+l²) + W.D by given force = 0 W.D by given force = mgh +umgcostheta(√h²+l²) (Note:costheta= base/hypotenuse= l/√h²+l²) W.D= mgh+ umg(√h²+l²)l/(√h²+l²) W.D=mgh+ umgl

[@abhigyan0140]

i appreciate your audacity to answer a question which was asked by sir 6 years back!!!!!

@@abhigyan0140 agar video abhi dekh rha hai toh answer bhi abhi hi karega na ab main agar abhi dekh rhi hu toh usme meri kya galti ki sir ne qs tab pucha tha when I was 5th.

[@spar-hr8du]

so?

[@Mohammedzehdanzehdan]

Semma

[@carguy7480]

Timestamps: 1:00 kinetic energy frame of reference 4:00 work energy theorem 10:00 derivation 25:40 imp 30:00 imp 38:00 q 43:00 q 45:30 hw q

[@tunesera2947]

😂👍🏼

[@carguy7480]

@@tunesera2947 In 11 some chemistry topics and in 12 a lot of chemistry topics are covered by awasthi sir on physics wala channel but I'm not understanding them🙄 can someone tell me if there is any other teachers on KZbin or unacademy?

[@tunesera2947]

@@carguy7480 Unacademy ka lecture attain kar sakte ho. Wahn pe achhe lecturer hote hain

[@carguy7480]

@@tunesera2947 bhai koi ache chemistry educators ka nam btao pls

[@Godgifted9540]

@@carguy7480 arvind arora

[@Chinnu-je7hr]

42:00 Moving gradually means velocity remains constant, change in kinetic energy is zero, work done by all forces is zero.

[@gamingwithnoobinsaan6314]

Option c is correct

[@shailamanishgupta3168]

Not velocity but speed

[@vishnuteja99]

No

[@vishnuteja99]

Work done is zero

[@NitinSingh-nx5tc]

Sir u r the best teacher ever I have seen in my life..

[@SanjayKumar-ds8xv]

Answer of 45:21 is √23/5 m/s

[@adityakadam4977]

correct hai mera bhi woi ara hai

[@AyanMujawar-u1c]

My Ans is also under root 23/5

[@Bipralab]

Are bhai par vo Force to conservative hai na

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Sir I m in kota... And after watching u I think to drop my coaching.. And only watch ur magical videos..... Thnx a lot sir ... 💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎💎

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Same case is with me bro... I am also in kota but I find sir's videos more useful than 6 hrs coaching.

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Chirag Garg 🖕

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Chirag Garg even I'm in kota, living in hostel ... seeing his videos I have the same thought ... so I'll be doing skooling in 12th class

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Same case here but I m not in Kota, I m going to career launcher

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Bokachoda

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I Tell you all people a 'Fact ' / 'Obvious Fact ' That he is world's best Teacher For physics. And it's Incredible that he is Teaching for free on this platform. SALUTE SIR

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Who like physics wallah,'s coaching hit a Like

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[@BHarshaRajkumar]

yes correct

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probably we should hit a like to his videos and there should be no "if" cuz there isnt a single student who wouldnt love it you know that the dislikes are from the teachers of other coaching institutes who are getting paid

[@prishachouhan]

ANSWER: 47:09 ( last question) Work done by all force = ∆K.E. As mass is taken gradually so K.E. will become 0. Wmg + Wfriction + Wnormal + Wforce = 0 W mg = mg in downward direction and block is pushed upwards so it will be negative and W= FS cos theta Force is 'mg' and S is ' h '(only vertical displacement to be considered) Wmg => -mgh WNormal is zero (cos 90=0) Wfriction => negative as block is going upward and friction is opposing its motion ,friction force is uN Here normal is perpendicular to mg so N = mg and hence force of friction will become ' umg' Then, displacement is 'l'( horizontal displacement) All all forces -> Wmg + Wfriction + W force=0 => -mgh -umgl + Wforce=0 => Wforce = umgl + mgh Hence , option "C" is correct.

[@ankitmode8133]

Displacement by sir (here) =0 Therefore word done = 0 😂😂 Just joking thank you sir actually work done by you for us=infinity

[@harshitkumarrai8183]

No bro only the work done by conservative force is zero..😂😂😂

[@rajeshgupta4548]

No bro not in all case only for closed loop 😂😂

[@rajmukherjee234]

yaa

[@AnmolSingh-vt6wn]

LOL.. Sir ka hei gyaan chod rhe h log yaha😑🤣🤣🤣

[@AlokAlokTariyal]

@@rajeshgupta4548 bhai tu khud bhi krr raha hai 🤣🤣🤣🤣irony

[@coderYT1]

Sir! I am a student from allen and I need daily revision of topics from each chapter. Sir your videos are definitely enough for revising all those concept! I never need anything else.

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TheTeenTuber same here bro

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iSAP RaJ u in Allen

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[@AyushGupta-yj8jz]

Same here....which centre are u from?

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Ayush Gupta give ur phone no. We can talk

[@harshpachori4874]

10:19 at 🕺0. 5x

[@truelyindian518]

Gzb

[@kakartik8469]

What a sound

[@Only_for_study_]

47:23 option C If we taken gradually then W by all force =0 Wg=work done by gravity Wf=work done by friction force f=friction s=displacement F=force applied m=mass a= acceleration Theta =_ N=normal or mgsin_ g=gravity ú=mew Now, Wg+Wf=0 Wg=mgh f - mgsin_ - úmgcos_=ma we taken gradually then acceleration is 0 f=mg(sin_+úcos_) Wf=f.s Wf=mgs(sin_+úcos_) Wf=mgssin_+úmgscos_ Wf=mgh+úmgl Thank you 🙏😊

[@MurliD_T]

Bro tune theta kyu use kia given hi nhi h?

[@patrick2.3.4.5]

​@@MurliD_Thaa rakha bhi toh surface even thodi hai Har bar change hoga

[@MurliD_T]

@@patrick2.3.4.5 bhai bina theta ke ho gya

[@patrick2.3.4.5]

@@MurliD_T haan ho gaya

[@abhishek_kumarvelloreinsti1866]

Trust me As long as He is there.... He will keep revolutionizing this coaching mafia with his sincere efforts.......

[@kamleshvbbs7820]

1:00 kinetic energy frame of reference 4:00 work energy theorem 10:00 derivation 25:40 imp 30:00 imp 38:00 q 43:00 q 45:30 hw q

[@legendofallpro]

thanks

[@irtesaamatfi]

To all of those who didnt got how to calc W by frctnl force in the last qstn.. Start by taking a very small disp ds on tht hill such that the a small and prfct right angled triangle is formed..let the angle it made with the horizontal be theta and let the horizontal displacement be dl.. Now dW = - f.ds (since f is opp to ds) => dW = -(u×mgcos theta)ds..(where theta is the angle it made during tht small disp and nt the angle of tht curved hill) = -u×mg(dscos theta) = -u×mg(dl)..(check this frm rght triangle made) =>W= -u×mg×integral dl => W = -umgl..(since dl is horizontal disp which on integrating gives horizontal length l) Also..taking disp as √l²+h² is conceptually wrong as frctnl force is non conservative..i.e..workdne by it depends on path taken..hope evry1 got now.

[@sanjaykumarchoudhary4945]

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[@AIG-FFYT]

Thanks !!

[@chittaranjanhait2689]

THANKSSSS

[@harshbhamre5087]

thanks!!!

[@ayushpandey9994]

thanks

[@galacticwarrior5278]

Homework answer is option (c) Am watching these lectures again after 5 years for IIT Jam exam So trust me dhyan se saare 11th 12 th ke lectures dekho chaye board exam hee nikalna ha tab bhi.. Nahi toh baad me firse dekhne pdenge and write everything make proper notes .. Lifetime kaam aaenge

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sir you are the best teacher in the world for us ,do u agree guys

[@bibo8231]

At least during online classes

[@barsharanisethy5631]

No

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Mere channel ko check karlo aacha lage to subscribe and like nahi to dislike new channel need support

[@khtmttabyebye3929]

@@barsharanisethy5631 toh app yeha pr kiu ati h..jab ap sir ki rspct nhi kar pati...!?!

[@vinodkanwar1541]

Ya he is bestest teacher

[@rijulsinghal4729]

Solution work done by friction is -umgcos(angle)×√l^2+h^2 where cos (angle) =l/√l^2+h^2 Work done by normal is zero as displacement is perpendicular to normal. Force And gravitational force is conservative force so following it from base to height path net gravitational force -mgh so acc. To equation ∆K.E.=0 Wf+wgravity+Wfriction+Wnormal =0 Wf=umgl+mgh So right ans is c consider that rough surface as a straight line than the problem will be too easy

[@msakhtarsk6660]

Thanx a lot friends ☺😍👍🏻... Your solution very helpful for me bro , Good job

[@adityabaghel1270]

I too got the same answer.. just the way was different 😉

[@kamyakumari9979]

Please reply how displacement is perpendicular to normal reaction....😥

[@prabjotsingh2834]

Thanks a lot

[@studybooster7]

@@kamyakumari9979 see diagram

[@phythematics2188]

10:10 😂 🤣😂 🤣😂 🤣 when he says ''potential energy kaha gaya?'' was too dramatic .... love you sir .. love you from bangladesh

[@yogeshraghuwanshi1595]

😂 haha

[@uknome2754]

4 real

[@ranjankr_yt]

😂😂😂

[@STUDIOUSBRO]

In last question frictional force is not constant then how we can consider the Of as umgl

[@lazy_bug4246]

@Sarvesh Tiwari dimaag mein gobar hai naa?

[@BabaBoee5198]

3:00 for those wondering, baat thori incomplete h kyuki, energy can’t be destroyed so there is a conservation of energy, and to prove this, we know that B has kinetic energy, from frame of reference C, okay, now A ke frame se B ke Pas nahi h kinetic energy, it’s okay also, butttt A ke frame of reference se C ke Pas ulti direction me kinetic energy hai( negative K.E) , matlab ki kinetic energy toh hai hi, bas baat h yeh direction ka , and we know ki direction is totally dependent open frame of reference, so no single frame I’d reference can be correct only, every frame is correct,

[@souvikmandal1919]

W by Normal roaction is Zero because Normal is always perpendicular to instantaneous displacement W by gravity is -mgh W by friction is -µmgl because friction is always perpendicular to instantaneous change in height but friction is opposite to horizontal displacemant (which is 'l' ) And total W is Zero Hence W by F is mgh + µmgl

[@Nimisreeni123]

but here N is not equal to mg. mg makes an angle with Normal reaction

[@narichan9399]

silpa shaji it doesn’t matter because angle between normal reaction and the surface along which it is moved is 90°

[@Nimisreeni123]

@@narichan9399 hey that`s not my doubt. work done by friction=/\/\N S. where N= mg sin thetta. then how it become N=mg.What happens to thetta?

[@narichan9399]

silpa shaji oh sorry I misread it 🙈 can someone help 🙏😪 idk either

[@gdgqdhd]

W by frictional force will be - uN. S (u=mew) N = mgcos theta and s=√(l^2+h^2) ... Cos theta = b/h which is equal to l/√(l^2+h^2) sabki value put kardo... Ans aa jyega

[@arunthorat2100]

22:07 hame tumse pyaar kitna ye tum nhi jaannte mgr jii nhi skte tumhare bina........ Papa😂😆

[@uchihasauske5634]

He is a great singer too 😂😂

[@akshay_chaudhary]

dont u know that i think his father was no more with him until his childhood so can u remove those laughing emojis!

[@Ingenuity_0]

@@akshay_chaudhary bruh he didn't even say papa, he said tn..tn

[@Sc4r69]

@@akshay_chaudhary His father is still alive.

[@dhritiranjan9453]

Anyone who has never made a mistake has never tried anything new. impossible is not a word, it's just a reason for someone not to try.⭐⭐

[@shraddhabisht9377]

Right

[@ganeshgalaxygg2549]

Good explanation sir, For,,W-E theorem ,K.E &P.E derivation in easy way .click:--)kzbin.info/www/bejne/rILFgKqjeJt6q8U Like share and subscribe my channel😊😊

[@sibanisahu4962]

Sir mera next year 12th hai aap unke topics bhi explain kijiyega aur sir sach mei itne selflessly koi kisike liye ni karta salute to you😳😳😳😳😘😘😘 🙏🙏🙏🙏🙏🙏😊😊😊😄😄😄😶😶😶😶😶😮😮😮😄😄😄😄😃😃😃😃😃😃😃😃😃😃😃😳😳😳😳😳😍😍😍😍😍😍😍😘🙏🙏🙏🙏

[@somaruprasad7804]

Sir my also

[@unhuman537]

Duniya me har tarah me log h

[@diwangaming6640]

@@unhuman537 ap kis tarah ke ho

[@bokesh2215]

@@somaruprasad7804 sir sabki hai akela tumara nahi 🤩🤩😎😎😎😎😎😎

[@norablossoms9157]

@@diwangaming6640 time waste karne wale mai se hai wo banda xDDD

[@akancha8717]

Sir aapki ankhe nind se bhari hui hai 😢phir bhi aap hmre liye video banate Ho wo bhi raat K 1 ,2 bje tk. Aap bbut mehnti person Ho sir. Aap apni khushi Na dekh kr hmare problems ko apne mind me rkhte hue ki hmlog lock down me kaise parhai kr payenge. Or aap it a sacrifice karte Ho. 😞 sir aapko shri narayan apko bhut bari bulandiyo pe phuchaye😊aap bhut ache Ho sir. May god blessings be with you always Sir.😊 Maine apne trf ki sare coaching se parh K dekh liya but apke jaisa koi Nai itna concept clearance krwa Pate hai. Aap hm apne bacho jaisa treat karte hai. 😍 Hme jha samjh Nai ata hai aapko bina kuch bole pta chal jata hi 😊you're are the best teacher of my world 😍

[@pramanandjimaharaj66]

Right

[@somethingcreative4893]

Hello

[@yritik50]

c): is right option

[@rahullubana4338]

Urs

[@harsh26may]

Kese

[@KaranKumar-in8cv]

Yes c is correct

[@Darkgamer-wl1iy]

Ya

[@aravindmr3262]

@@Darkgamer-wl1iy pls send the solution

[@rajmishra6566]

The video of alakh sir is always help the students no one can compit him in the world

[@manankhan4506]

@@satyarthkhajuria0309 bhaii jee Kai liyae kitna helpful hai aur phr agr aik vedio oneshot daekhuga for jee

[@rajnish1629]

Hello everyone.. Here is the solution of last question.. .. assume angle of inclination to be theta.. Let displacement is D So D cos theta = L So D = L/cos theta.. Now W by friction= - Mu×N×D =- Mu×mg cos theta ×L/cos theta = -MU mgl Now Work done by gravity =-mgh and work done by Normal reaction =0 As Total work done is 0 Work done by external force is =Mu mgl+mgh Option c

[@Ashutosh-qm8gu]

You had took component of gravity in the normal reaction of friction then why you had not took the component of g along the plane in gravitation work done

[@ajaymahawar2862]

@@Ashutosh-qm8gu because Gravity is a conservative Force and Sir already said we only take Vertical displacement of it.

[@Ashutosh-qm8gu]

@@ajaymahawar2862Hmm

[@qwerty1885]

Are bhai only u likh dete me kabse paresan ho raha tha ki M ka kya use matlab friction ka formula (u × N) hota he tumne to dara hi dya tha

[@_Charon_IV]

Thanks a lot bro❤️..I was confused in the w.d. by friction part.

[@krishnakumarkushwaha85]

Sir- baccho smjh gye jo mne samjaya ajj Bacche- yes sirrrrrrr (Mann mein) ghr ja ke alakh sir ka video dekhenge tb jake kch ayega 😂😂😂😂😂😂

[@radheyshyamsharma4411]

I am suffering from this situation

[@reemabansal9594]

Pay attention to coaching and then understand you are completely relying on Alakh sir which is not right because in the FUTURE you will always tend to rely on someone

[@athletesplayercr7]

Top.pr sir key jgh school teacher liktte toh jyada better hota...😀😊😊

[@clashtm8210]

@@reemabansal9594 you didn't state that this is your opinion and hence it you are trying to say that the once in coaching can't fetch good marks. Aankhon ki patti hateke dekhlo duniya alag hai

[@paramitabiswal7110]

Same here. I also think the same thing....

[@anjalistudylover8423]

Sir apka selection IIT m isliye ni hua qki apki qismat m usse bi bda kuch likha h...... Well done sir

[@ananyagupta6075]

The video of alakh sir is always help the students no one can ever compete with him....tysm sir

[@imprldragon1938]

Kaash alakh sir english bhi padhate to tumhari english acchi hoti 😅

[@noxen.8818]

Cute anime dp 🤩

[@kunalu02]

@@imprldragon1938 ghoul 😰😬

[@darkager2256]

@@noxen.8818 Nhi pategi rhen do aur padh lo

[@vineetameta9944]

The answer to the last question is C Wf + Wg + WF + Wn =0 Wg = -mgh f = μmgcosθ Now s=l/cosθ Wf = -μMgl {as friction tries to retards its motion} Wn = 0 [as s is always perpendicular to N] by adding these WF= μmgl+ mgh

[@dhanveersinghshekhawat7210]

Yaha prr s=l/costheta kese aaya

[@ranbirkaur2079]

Could you plz tell ki s=1/cos theta kese aya

[@rajkashyap204]

Wow copied answer

[@hfhfjjfjf-z7z]

@@dhanveersinghshekhawat7210 cos theeta = opposite side / hypotenuse

[@mannelanithin5463]

Can u plz tell why sum of all forces is 0.

[@alishkasingh657]

These video's are far better than the recent video's. Don't wanna say bt i think alakh sir teaching method quite changed now. And this is the reason u all are here.

[@rajshekhar308]

U R ✅

[@Xoranium]

frrr, i totally agree on this point. He just teaches like a professional and industrial teacher now whereas this dude used to teach like he's someone from my own fam, miss his old style :(

[@abhinavmharnur6901]

Old is gold

[@abhinavmharnur6901]

Sir is now In excited state

[@alishkasingh657]

If possible try to tell sir that we want the old alakh sir.

[@VishalSingh-eo9zc]

25:37 sir aapke is expression ko dekh kr muje TONY STARK ki yaad aa jati h

[@Yashkumar-pv5tj]

@@dipeshhhhh who is Tony stark

[@asmitsingh2093]

@@AtulKumar-rd2wc who is Tony Stark?

[@asmitsingh2093]

@@AtulKumar-rd2wc iron man Usse toh kon nhi janta 😁😁

[@kunalsharma5107]

@@dipeshhhhh tony star film mein naam h asli mein uska naam robert downey jr. H lol🤣

[@deepikadwivedi2089]

@@Yashkumar-pv5tj he is iron man💪

[@rajshekhar308]

45:25 Underoot 23/5 ♡●♡●♡●♡

[@_sx7293]

Yaa right it's √23/5

[@sakshamsingh2125]

AUR AAGE BADANA HAI KYA ISKO

[@sakshamsingh2125]

It's the ans itself

[@SmitaNaik-gs7ej]

​@@sakshamsingh2125it depends upon what options are given

[@nirvikpaul9603]

sir woh 20:00 par diya gaya question f.s se bhi ho gaya . s=2t square -2t+10 ko double differentiate karoge toh acceleration 4 ayega . ma= force = 4N aa gya (mass =1kg) aur displacement 0 se 2 tak 4 agaya . hence work done is 16 . f and s are in same direction

[@Logitang]

Sir, please don't delete these videos... I study from these and my friends too... You are our emotion as well as Guru

[@tomandjerryhindi5493]

Last question is v.v.v. conceptual and tricky Thanks alakh sir for boosting our mind

[@anoop_ix]

Hey guys in last question While finding work done by friction we took total displacement as hypotenuse but friction is non conservative then why not we take total distance of mountain as it will be obviously greater than hypotenuse because it has pits that will increase distances

[@bcdplayz2884]

@@anoop_ix m is taken gradually so force is conservative....

[@suyashnegi8344]

@@bcdplayz2884 kuch bhi bakar bakar karni hai matlab. Aata jaata kuch ni chala aaya

[@dangerouslybad58]

@@suyashnegi8344 😂😂

[@yashraj5080]

@@suyashnegi8344 reason?

[@Rohittpall]

Currently you are 11th class and watch this video hit like 👍👍 This video value can't be defined

[@devanshtyagi25]

I don't understand that how a teacher can be so good at teaching.you make all difficult concepts easy as bread and butter.i will try my best to gain as much as I can from u.lovr u sir🙂

[@shobha1302]

45:22 answer is coming as (23/5)^1/2

[@92patelaayushi68]

Yes

[@ganeshgalaxygg2549]

Good explanation sir, For,,W-E theorem ,K.E &P.E derivation in easy way .click:--)kzbin.info/www/bejne/rILFgKqjeJt6q8U Like share and subscribe my channel😊😊😊

[@bechanger_007]

How bro

[@onedreammanygoal4747]

East or West Physics wallah (Alakh sir) is the best.. How many of u agree hit like👍

[@dhruvaetoor1553]

North or South too😕

[@shortsongsguy8350]

@@dhruvaetoor1553😂

[@Thecannibalfrombeyond]

@@dhruvaetoor1553 no south

[@ganeshgalaxygg2549]

Good explanation sir, For,,W-E theorem ,K.E &P.E derivation in easy way .click:--)kzbin.info/www/bejne/rILFgKqjeJt6q8U Like share and subscribe my channel😊

[@rushilajaria4307]

Hit his face

[@raghav5832]

Alternate method for this question 21:09 differentiating the equation v=4t-2 again to get a=4 Here,the acceleration came out to be constant.Hence the applied force is also constant. F=ma=1×4=4N Displacement at t=0 is 10m Displacement at t=2 is 14m (From the equation in the question) displacement covered=14-10=4m Since force is constant,we can use W=F.s W=4×4 W=+16J

[@raaaaa7081]

1'Another cleue 👍 take angle@ and now, cos@ =l/√h^2 + l^2 therefore, N (normal reaction) mgcos@, u can put cos@ value, 2' take displacement as √h^2+ l^2 and multiply as they are opposite, u get - umgl - work of friction 👍👍👍👍

[@vishalmishra025]

Accha laga toh like jarur karna kzbin.info/www/bejne/noOleIyimdeHjsk☺☺☺

[@ustaad6637]

42:24 wale sawal ka jawab √23/5 m/s hai kya?? Kya maine sahi jawab diya hai???

[@devchauhan2203]

Yup...√4.7 same..

[@sillicon8227]

Bhai Mera toh √(29/5) a raha he

[@arshdeep011]

MAY YOU got 500k subs. in *Dec 2018*

[@nishkrsh_]

for sure itill december he gonna get 750k subscribe rs

[@YashChaphekar07]

1 million till next year

[@pranjayjain]

Nov may he ho gaye 500k ☺️

[@sameermaneri8426]

@@nishkrsh_ kay bhavishavani he bro nice

[@RandomGuy-vq3gd]

I am from the future and sir has achieved 1million subs by febuary 2019.

[@ParmodKumar-ct7ij]

2:46 yes, kinetic energy depends upon frame of reference

[@vivekkumarprajapati1366]

W by gravity =-mgh, W by normal reaction = 0 becoz it's perpendicularly acting. W by friction = mu mgcos(theeta)×l/cos(theeta) = mu mgl (Becoz we have taken theeta as angle between inclined plane and base). Now, Workdone by all forces = ∆KE Wnet=0 Wg + Wf + Wnormal + Wapplied force = 0 -mgh - mu mgl + 0 + Wapplied force = 0 Wapplied force = mgh + mu mgl (Hope you will understand it:)

[@abhisheksinha9452]

Bhai work done by friction nhi samaj aaya please samjha do 👐

[@vivekkumarprajapati1366]

@@abhisheksinha9452 bhai friction = Normal reaction * frictional coefficient (mu) hota hai.

[@abhisheksinha9452]

@@vivekkumarprajapati1366 thanks bro 😇

[@Shirtskiduniya124]

Bhai w by g kaise aya bata do please...

[@vivekkumarprajapati1366]

@@Shirtskiduniya124 bhai W by g= force* component of displacement in the direction of force. Toh force downward lg rhi thi (mg) aur displacement upward ho rha tha (h) which means that displacement force ke opposite Ho rhi thi isiliye W by g = -mgh Hope you have understood it:)

[@thekidmagician7920]

Homework Answer - (μmgl + mgh) Option C. Thank You very much Sir.

[@nehakumari6111]

SIR JEE I AM FROM KENDRIYA VIDYALAYA KANKARBAGH PATNA I HAD VISITED YOUR CHANNEL FIRST TIME IN CLASS 10 I GOT 91% IN BOARD AND NOW I AM IN CLASS 11 AND I GOT 85% IN MY FIRST PERIODIC THIS IS JUST BECAUSE OF U SIR . I CAN NOT IMAGINE MYSELF WITHOUT U. I NEVER EXPECTED THAT THERE IS A TEACHER LIKE U IN THE WHOLE WORLD . AFTER MY FAMILY ONLY U MATTERS FOR ME

[@mrdontknow8347]

Qualified Advanced?

[@aaditbhargava6874]

@@mrdontknow8347 xd

[@vishnuteja99]

@@mrdontknow8347 🤣🤣😂

[@priyanzhu]

@@mrdontknow8347 bitches use brute force technique bro! Pseudo working is their root of scoring 90 above!

[@hemantpradhan5043]

Kankarbagh Patna me kahan?

[@artesart1398]

Watching in 2024... just speechless the way u teach.... 34:58 those points just stunned me sir... clear cut explanation!!! Thank u so much sir❤

[@madhulikasinha975]

In the question at 40:00 I found the work done by air friction by another method. First I used the 3rd law of motion to find the final velocity which should be possesed by the body and that came out as 20/s. Then using ∆K.E (1/2*5*(10) ^2-1/2*5*(20) ^2) I got -750J. Is this method correct?

[@Shhhh463]

no

[@vsijahsjsjshsf599]

No

[@bhanu.p5120]

Neverrrrrr everrrrrr

[@hahhaharsh]

47:12 The change in kinetic energy will be zero beacuase the mass moves gradually with which we assume that velocity is constant .. Now W(Total Force) = 0. All the forces accting in the illustrations are Gravitational force , Friction and The given force.. So the Work done( Force) = - [ Work(G) + Work(f) ] .. ; W(G) = -mgh & W(f) = -umgl .. The W(Force) = umgl + mgh Option(c)

[@imsumitbhade]

mgh + umgl hoga bhai

[@hahhaharsh]

@@imsumitbhade Thank you for correcting the mistake bro.. I took height as l and distance as h 😅😅

[@294adityayadav7]

The given force F is tangential to the surface so the work must be zero right?

[@rasla322]

@@294adityayadav7 the force F is along the distance . How can it become zero???

[@shivamchouhan5077]

How is friction force -umgl, friction force is non conservative so it depends on path.

[@abhaytomar9236]

You teaches virtually this much good really learning from you will be excellent who wants sir to start own class plz like

[@AbhinavKushwahaB]

Best teacher for clearing basics of physics

[@adambhai7384]

MOTHER : MODI JI MERA BACHHA TUTION NAHI KARTA KOACHING NAHI KARTA MODI : YE PHYSICS WALLAH HAI?

[@aryashah1579]

Ye Physics wallah wallah hai kya xD

[@Aman-dh9wz]

Mention. Bas youtube chalta rahata hai.😁

[@vasundharavydyula9957]

Copied comment. It was mine.

[@AryanSingh-mu8pu]

@@vasundharavydyula9957 tumne kis video me ye comment kiya tha bhai....?

[@eternalsoul1764]

@@vasundharavydyula9957 aree bhai chill yaar

[@zargarinaam403]

Statement of Work energy theorm: Work done by all the forces on a body is equal to change in kinetic energy of the body.

[@nsfactian2422]

Sir I am a student from government school and i am very thankful to you because you are giving highly content in free of cost now my physics is very improved after watching your videos

[@priyu8312]

Anyone here in ending of 2023

[@opera1235vt]

Yeah sure 😊

[@opera1235vt]

Are u preparing for Jee

[@priyu8312]

@@opera1235vt Yes

[@SmitaKirtania-xq2bf]

Me😊😊

[@Jayram.2211]

Hii

[@arthyvenkatesh8579]

Sir answer is option (c) umgl+mgh I am your huge fan sir. Your videos make fiitjee and other institutes look like dust. Keep up the good work sir. Love you.

[@NavjotSingh-qu6ii]

Sir bhut acha smjhaya apne. Very nice teaching by you sir well done.

[@happybhadoria9583]

Since friction is a non conservative force. So it depends upon the path then how we can consider the displacement as hypotenuse

[@madhubalarajhans9854]

Ha bhai sahi bola or baki sb displacement ko hypotenuse consider kr rha h

[@Physicskiyatra-Rishabhsoni]

Ha bhai Shi h....lekin iska answer kse aayga fir

[@handauu2005]

Yes sir ...hame aap se pyar Kitna ye aap nehi jante, Magar concept clear nehi Hota aap ke Bina ☺️🙂..22:10

[@PreetiTiwari-hk8br]

At 10:13 what a movement by the sir

[@realnaveenkumar-vj6kr]

Answer is -> W(F) = μmgl + mgh Solution you all know very well I guess.... because Sir ne bahut hi achhi tarike se samjhaya hai 😊

[@naziahussain1036]

When you teach, it seems that physics is so easy...!!

[@edwinjudo5917]

last question: work done by friction=uN=umgcostheta=umgxl/(l/h^2+l^2)==Wfriction=-umgl Wg=-mg(as W horizontally=0 as Wtotal=0 WF+Wfriction+Wgravity=0 therefore WF=umgl+mgh ------ option c)

[@fanofabdevillersandmathslo5960]

Answer to last question is Option C (umgl+mgh) Thanks for your extraordinary lecture sir

[@Tea_lover_vishnu_jani]

Ryt

[@muzamilbhat4757]

How?can u help

[@shivamvishwakarma941]

@@muzamilbhat4757 Work done by normal 0 kyunki displacement ke perpendicular hai Work done by friction ke liye f=uN normal ko mgcostheta balance karta hai costheta =l/√(l^2+h^2) aur friction non conservative hai to uske path ke hisaab se displacement √(l^2+h^2) hojayega W by friction= - √(l^2+h^2) umgl/ √(l^2+h^2) negative kyunki force ke opposite direction mai hai Work by gravity = -mgh vertical displacement dekhenge isme kyunki horizontal mai 0 rahega aur conservative hai path se farak nahi padhta W by F dhundna hai W by all= ∆KE=0 gradually diya hai isliye 0 W by normal+ W by Friction + Work by mg + W by F=0 W by F=umgl+mgh aayega I hope samjha hoga

[@sanjuthreddy6435]

Hi sir Like who are happy with phy wallah sir

[@socialstatus7554]

Go through my channel and subscribe it

[@STUDIOUSBRO]

In last question frictional force is not constant then how we can consider the Of as umgl

[@STUDIOUSBRO]

@@socialstatus7554 Come to my channel , and put your hand on my °°°°° channels subscribe butoon Then open your -- finger and press your boo-- , I mean like button This is what u will get by begging , but I explained it in nice way otherwise it can be more harsh

[@ganeshgalaxygg2549]

Good explanation sir, For,,W-E theorem ,K.E &P.E derivation in easy way .click:--)kzbin.info/www/bejne/rILFgKqjeJt6q8U Like share and subscribe my channel😊😊

[@anupamaanu371]

How do you know the exact doubt in students' mind..You are really a great teacher❣️❣️❣️

[@itachu.]

experience.

[@rajshekhar308]

✅

[@almondespa]

Option (C) Acc to question block is taken gradually therefore its speed=constsnt, Therefore Delta{K}=0, Wg+Wf+Wn+WF=0, Taking the longer part of mountain with length(l) and height (h) as total s vector. Wg along l=0,Wg along h=-mgh{angle b/w mg and h is 180} Now friciton f = umg, Now the friction is between the surface so friction in h is 0. friction along l = -umgl{180degree} Wn=0{Perpendicular} WF=umgl+mgh

[@RejinaParbin-w4g]

Displacement=h kese bro Can you tell me plzz

[@goal2success586]

4 years and this video is very helpful to stu yet thanks alakah sir

[@teamindian3990]

Anybody watching in 2024 😅😅😁

[@Gameonrexff]

Me

[@rajeshsingh5660]

🙋🏻

[@NANI-u6e]

Me

[@puspamanjarimishra4603]

Me 😅

[@BhavanaBurli]

✋🏻

[@yashverma-yb7qc]

What a brilliant lecture Sir really you are a God of physics and chemistry 😇😇

[@kaivalya8]

The answer to the last question is C Wf + Wg + WF + Wn =0 Wg = -mgh f = µmgcosθ Wf = -µMgl Now s=l/cose {as friction tries to retard its motion} Wn = 0 as S is always perpendicular to N] By adding these WF= µmgl+mgh

[@ly_lols]

17:00 I took out d accn by d slope then used F=ma n multiplied it with s(area...1/2×10×2) n got d same ans

[@omkhype4665]

Same !!!! lol

[@kshmasingh8746]

Kya mai isko neet 2021 ke liye pdh skti hu shi hai ya nhi plz ans

[@ly_lols]

@@kshmasingh8746 yes u can. N best of luck 👍💯

[@kshmasingh8746]

@@ly_lols thanku

[@aayushi1978]

Duniya ke sabse best physics ke teacher ❤ Thank you sir🤘🤘👨‍✈️

[@blitzTheAVR16U2Enthusiast]

The last question is to be solved under the assumption that the 'bumps' are very small, because if they are not then length of the hypotenuse won't be equal to the length of the path along which the body traverses and as a result, none of the options would be correct.

[@rohh_gh]

not required because for work we consider displacement so agar bumps chahe kitne bhi bade ho displacement to wahi hogi

[@blitzTheAVR16U2Enthusiast]

@@rohh_gh It is clearly mentioned that force acts tangentially to the surface, this means that the direction of force keeps on changing with respect to the surface, so saying that W = F x 'hypotenuse' will not be valid. To find the work done you will have to find the length of the slope and then multiply it with the magnitude of the force, this would be the actual work done. And of course, here I am assuming that the bumps are very huge.

[@dhruvinvekariya975]

Tumane iss concept ka sahi feel liya hai Agar jyada bade bump rakhenge to path taken by body hypo se jyada ho jayega aur -w by friction badh jayega Sahi hai na

[@blitzTheAVR16U2Enthusiast]

@@dhruvinvekariya975 Yes eekdam sahi bole.

[@veenasingh5945]

@@blitzTheAVR16U2Enthusiast very true People dont get that

[@KGVines418]

Work(by gravitational force)= -mgh Work by friction= -uN times under root l square+ h square Where, N= mgcos theta and cos theta= l/under root l square+ h square So friction force= - umgl Hence answer is option C

[@eklavyajoshi8053]

Anyone here in december 2020

[@seemasinghal2976]

45:28 is it under root23/5

[@bhushanwanjare2039]

I'm here

[@naindichandu1]

Im

[@nanditaghose3055]

Me

[@Aman-_-XD]

Yup

[@dakshjethva1490]

Sir 17:24 can we find work by finding area under graph and slope of graph.

[@alivlogs151]

जैसे चांद से आगे कोई बस्ती, नहीं ऐसे physics wala se बड़ी कोई हस्ती नहीं ❤❤❤❤🎉🎉🎉

[@vajjasureshkumar2844]

45:36 I think answer for last question is mg(h+ul)

[@yashrustogi2156]

correct!

[@aspirants8627]

Haa sahi hai

[@anugrahdeolia1186]

yeh toh option hi nhi hai bhai ....according to options mgh+umgl sahi hai ...but u have taken common...but yeah anyways answer is correct

[@AYUSH-fm8mi]

U ki value kaise nikali plese btao🙏🙏🙏

[@rishabhmajumdar50]

@@AYUSH-fm8mi u ka value nikaali nahi hai bas variable form mein li hai

[@prarthanabidare4280]

Someone give him an award for his wonderful teaching!!!!! :)

[@albartros8257]

Chalo deta h

[@PrashantYadav-dw6vg]

@@albartros8257 haa bhaii too agaa agaa chall hamm picha pichaa ataanhaa

[@PrashantYadav-dw6vg]

Dill ki bart khaahaa dii

[@manabsaha2733]

And for singing also :P Sir is awesome 🔥🔥🔥

[@vedantpatel8532]

🏆🏆🏆🏆🏆🏆🏆🏅🥇🥈🎖🏆🏅🎖🏆🏅🏆🎖🏆🏅🏆🏆🏆🏅🏆🏆🏆🏆🏆🎖🎖🏆🏆🏆🏆