mg - 200 = ma a = (10m-200)/(50+m) 5= -----#------- 250+5m = 10m - 200 450= 5m m = 450/5 = **90kg** However sir thank you so much for your service to this topic. This was my toughest topic
Nice video sir.. But I have a question. If we apply two different forces both on the upper and lower block simultaneously in the same direction and if friction exists between the blocks as well as in the ground, then when will the relative sliding between the blocks occur??
Sir last wale question mai aap jab mg-T=m × ac aur dusri equation jo ki T -200 =50× ac ko eliminate Kar rahe the toh aapne equation likhi elimination kain baad main (10m-200)/50 =ac par aapne firstequation for main mac bhi likha tha wo toh 50ac kain sang add hi nahi kara
Your way of teaching is perfect. No extra talk. Keeping things simple. Thank you sir
That's true bro
This is definitely a most systematic way of explaining things.
Sir i have watched your past year neet problems trust me sir bahut accha explanation ka tarika hai aapka hats off🙏 sir
Thank you sir.... crystal clear explanation 👌
Thnx, thnx, thnx a lot and lot sir
The best explanation and even good examples to understand the concept
Super sir..... awesome explanation 👌
Brilliant explaination sir!!
Thank you so much
Sir last problem me (50+m) hoga na sir.And answer 90 kg ayega.
Very very systematic explanation...Great sir...Lot to learn from phalke sir....
Thank you very much sir
mg - 200 = ma
a = (10m-200)/(50+m)
5= -----#-------
250+5m = 10m - 200
450= 5m
m = 450/5 = **90kg**
However sir thank you so much for your service to this topic. This was my toughest topic
Watch my latest video on this topic, in Standard Results series. Hope you will like it.
Same doubt sir..?????????
Thank you so much for the wonderful explaination air
Sir in the first two cases at 1:00,
Is it the coefficient of static friction or dynamic friction?
Sir last case me (50+m) ac hoga
Sir plzz upload previous videos on this topoc
Direction to tell for upper block
Thank you very much sir, excellent explanation
sir 10:38 pe (mu)*mg hoga na likin aapne tho keval (mu)*g krra h
Nice video sir.. But I have a question. If we apply two different forces both on the upper and lower block simultaneously in the same direction and if friction exists between the blocks as well as in the ground, then when will the relative sliding between the blocks occur??
Very helpful and wonderful lecture !
Sir last wale question mai aap jab mg-T=m × ac aur dusri equation jo ki T -200 =50× ac ko eliminate Kar rahe the toh aapne equation likhi elimination kain baad main (10m-200)/50 =ac par aapne firstequation for main mac bhi likha tha wo toh 50ac kain sang add hi nahi kara
nice explanation
Sir last question please check once final answer iam getting 90kg.
Sir pls make jee oriented videos.
Woow sir what an explanation
21:17 (50 + m)ac hoga na sir
Aur answer 90kg aayega
Awesome lecture sir :)
Thank you so much sir
Sir, second case me upar vali block pe 20N friction kaise aaya? Friction =uN=umg= 0.5*10*10=50N ???
Action reaction f = ma
👉❤❤loved this video
😊😊😊😊🙏🙏🙏nice
Sir plz upload sequence plzzs sir
Ans-90kg hoga last vale m kya?
Yes 90kg hoga, sir ne (50+m)lena tha 50 le liya
a(m+50)
sir aapko allen ne nikaal diya na
Ty sir
god bless you and your family sir
Sir, your teaching way is very systemetic but you had not teach block on block problem completely
50+m hoga
Oiie