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

Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, dynamic Exercise and much more on Physicswallah App.

LAKSHYA Batch(2020-21)

Registration Open!!!!

What will you get in the Lakshya Batch?

1) Complete Class 12th + JEE Mains/ NEET syllabus - Targeting 95% in Board Exams and Selection in JEE MAINS / NEET with a Strong Score under Direct Guidance of Alakh Pandey.

2)Live Classes and recorded Video Lectures (New, different from those on YouTube)

3)PDF Notes of each class.

4)DPP: Daily Practice Problems with each class having 10 questions based on the class of JEE Mains/NEET level.

5)Syllabus Completion by end of January, 2021 with topicwise discussion of Last 10 Years Problems in Boards, JEE Mains/NEET within Lecture.

6)The Complete Course (Video Lectures, PDF Notes, any other Study Material) will be accessible to all the students untill JEE Mains & NEET 2021 (nearly May 2021)

7)In case you missed a live class, you can see its recording.

8)You can view the videos any number of times.

9)Each chapter will be discussed in detail with all concepts and numericals

10)Chapterwise Approach towards JEE Mains/ NEET & Board Exams.

****Test Series for XI & XII****

We provide you the best test series for Class XI,XII, JEE, NEET chapterwise, which will be scheduled for whole year.
The test series follows very logical sequence of Basic to Advance questions.&
Evaluation of Test and Solution to all the questions at the end of the test.

class 11 physics chapter 6 | Work, Energy and Power 01 | Introduction | Formulae for Work IIT JEE
Class 11 physics chapter 6 | Work,Energy and Power 02 | Conservative and Non Conservative Forces|
Class 11 physics chapter 6 | Work,Energy and Power 03 | Work Energy Theorem IIT JEE NEET ||
Class 11 physics chapter 6 || Work,Energy and Power 04 | Potential Energy IIT JEE NEET
Class 11 physics chapter 6 | Work,Energy and Power 05 | Equilibrium - Stable , Unstable , Neutral |
Class 11 physics chapter 6 | Work,Energy and Power 06 || Conservation Of Mechanical Energy 1 IIT JEE
Class 11 physics chapter 6 | Work,Energy and Power 07 | Chain Problems | Conservation of Energy 2 |

[PhysicsWallah]

Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, dynamic Exercise and much more on Physicswallah App.



LAKSHYA Batch(2020-21)

Registration Open!!!!

What will you get in the Lakshya Batch?

1) Complete Class 12th + JEE Mains/ NEET syllabus - Targeting 95% in Board Exams and Selection in JEE MAINS / NEET with a Strong Score under Direct Guidance of Alakh Pandey.

2)Live Classes and recorded Video Lectures (New, different from those on YouTube)

3)PDF Notes of each class.

4)DPP: Daily Practice Problems with each class having 10 questions based on the class of JEE Mains/NEET level.

5)Syllabus Completion by end of January, 2021 with topicwise discussion of Last 10 Years Problems in Boards, JEE Mains/NEET within Lecture.

6)The Complete Course (Video Lectures, PDF Notes, any other Study Material) will be accessible to all the students untill JEE Mains & NEET 2021 (nearly May 2021)

7)In case you missed a live class, you can see its recording.

8)You can view the videos any number of times.

9)Each chapter will be discussed in detail with all concepts and numericals

10)Chapterwise Approach towards JEE Mains/ NEET & Board Exams.


****Test Series for XI & XII****

We provide you the best test series for Class XI, XII, JEE, NEET chapterwise, which will be scheduled for whole year.
The test series follows very logical sequence of Basic to Advance questions.&
Evaluation of Test and Solution to all the questions at the end of the test.

class 11 physics chapter 6 | Work, Energy and Power 01 | Introduction | Formulae for Work IIT JEE
Class 11 physics chapter 6 | Work, Energy and Power 02 | Conservative and Non Conservative Forces|
Class 11 physics chapter 6 | Work, Energy and Power 03 | Work Energy Theorem IIT JEE NEET ||
Class 11 physics chapter 6 || Work, Energy and Power 04 | Potential Energy IIT JEE NEET
Class 11 physics chapter 6 | Work, Energy and Power 05 | Equilibrium - Stable, Unstable, Neutral |
Class 11 physics chapter 6 | Work, Energy and Power 06 || Conservation Of Mechanical Energy 1 IIT JEE
Class 11 physics chapter 6 | Work, Energy and Power 07 | Chain Problems | Conservation of Energy 2 |

[jadisrinija]

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

[hemlatakhatri]

4 year ago but this video is helping the students always

[rupamkarmakar]

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

[smartenough]

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

[SatwikSinha-sfkr]

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[manosmitanath]

Sir im following..ur vdos since so long..first time i saw ur chem vdos of cls .10 ..
I was starstucked..watching..such contents over youtube but the viewers that time were less..but I knew..that one day..ur great effort..clear concepts..excellent teaching..methods would reach to..millions..Sir...& that's on the way...Wish you Good luck sir..
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[aryanpandey]

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.

[anandmani]

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

[goalsuccess]

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

[NitinSingh-nxtc]

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

[prernajasti]

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
Wn= 0 (as normal reaction and displacement are perpendicular
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
WF=
adding (1), (2), (3) and (4)
WF+0-mgh-(mu mgl) = 0
WF = mgh +mu mgl

correct answer option (c)

[atharvabhalekarixb]

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[anujbhardwaj]

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

[amanagrawal]

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

[rajmishra]

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[galacticwarrior]

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

[kingff]

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

[NavjotSingh-quii]

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[ananyagupta]

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