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yrelhan4
 3 years ago
A block on a rough inclined plane.....
Question attached.
yrelhan4
 3 years ago
A block on a rough inclined plane..... Question attached.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What the heck is that ? Sry i only understand english

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Ignore the hindi part. Its just the translation of the question.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0For the first question?

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Its a. Might me some minor mistake.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here Suppose the block has moved a distance x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then at position x balance the forces i.e. mgsin()  umgcos() = 0 Integrate this from 0 to x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u get the answer on simplification.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Well i get this sin()=mu*x^2/2*cos() Doesnt give me the required answer? @lordcyborg

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there is no m term on RHS

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and u missed an 'x' term on LHS

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0There is. You see the coefficient of friction given is mu=mu(not)*x. Ah, so we are not balancing forces, we are conserving the work. I get it. But that still doesnt give me a ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey man/woman................. on integrating u get mgsin(theta)*x=u.mgcos(theta)*(x^2)/2 Cut mg and 'x' on both sides.......... U get the answer........now try it fast

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Oh. But you cant balance forces for the body to stop right? You have to conserve the work done?

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0When you balance forces, it means there is no acceleration. It doesnt necessarily mean that the body is at rest? And a man.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol.........Dont go too deep...........i'm balancing forces when body is at rest

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0I dont think you can balance force for the body to be at rest.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ya we cannot all times.........bt we can

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Why doesnt conservation of energy give me the answer?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0coz u can apply dt when there are conservative forces.......in this case friction is not acting as a conservative force..........Now dont make me bore

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0You can leave if you are getting bored. You chose to answer my question. I didnt tell you to come here and help me. Thank you.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0C'mon man............I gave u the perfect(i suppose) reasons.........now go and sleep it 2:15 a.m.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah you did. Thank you.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364764688058:dw a = gsin@  u0 x gcos@ vdv = (gsin@  u0 xgcos@)dx => 0 = gsin@ x  u0 gcos@ x^2/2 x=0 and x = 2tan@/u0

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0for 2nd part, limit of v will be from 0 to v, of x will be 0 to tan@/u0 ^_^

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0:O mere dimaag me kabhi nhi aata ye, 2nd part konsa? Heat produced?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0oh sorry, nahi 3rd part me ye. heat produced me work energy lagayenge (SHAYAD) :/

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm. Equation likh do 2nd part ke liye. :/

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364765247915:dw mgh + integral (mg sin@  u0x mgcos@) dx = change in kinetic energy + heat change in KE = 0 hai x ka limit 0 to max tak jayega.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0check kar lena, maine bola hai to high probability hai ki galat hoga. 3 baj gaya! :O gtg bye

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Acha ruk.. Sahi hoga. :/ Ke 0 kyun hai? x to lamit 0 se x/2 tak nhi jayega?

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0@VincentLyon.Fr We got the first and the last part. For the second part what would be the work energy equation?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Answers to all 3 questions are no 1. I solved 1st question by dynamics 3rd question is just substituting cos and sin. 3rd question is computing the decrease in mechanical energy.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.01st one by 1=v*dv/dx right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For the first question, N's 2nd law gives : \(\ddot x+(\mu_og\cos\theta) x=g\sin\theta\) If you use \(\omega ^2=\mu_og\cos\theta\) the solution is \(x(t)=\Large\frac{\tan\theta}{\mu_o}\normalsize (1\cos\omega t)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll try solving it by workenergy theorem tomorrow, but now I have to go to bed.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah i am looking for that solution. Alright. Thank You.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0At the top, \(\cos \omega t=1\) At the bottom, \(\cos \omega t=1\) Halfway, \(\cos \omega t=0\) The rest is easy.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Right. I'll try and get back. And please see if we can do it by work energy theorem too.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if i use work energy theorem i get some weird answer.. Net work done = change in K.E since Initially its at rest and finally also at rest.. change in K.E = zero hence net work done = zero now there are two forces doing work, gravity Wg = mgcos(theta) x and friction Wf =  mgsin(theta)ux^2 Wg + Wf = zero but solving that gives a weird answer for x .. none of them listed :P..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry i flipped sin and cosine terms... so the answer should be 4 like someone said earlier.. :/..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you cannot balance forces.. why!? cause when forces are balanced.. the body will not stop.. it ll only stop ACCELERATING.. after that point it will start to decelerate .. and then ulimately stop :P.. thats how i see it..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and the answer can't be a) for the first part cause if it is.. then in second part none of them would satisfy the answer..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahh.. there is definitely something wrong with what i did.. :/... the answer should be ONE itself.. for the first question :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here is the solution using workenergy theorem. It is nicer than the one using N's 2nd law, because time does not appear in the equations. But you would need N's laws if one of the questions had been: "How long does it take the body to stop?" \(KE=1/2 \,m\,v^2\) \(PE=mgh=mg(x\sin\theta )\) Elementary work done by friction: \(dW=f\,dx=\mu \,mg\cos\theta\,dx=(\mu_o\,x) \,mg\cos\theta\,dx\) Total work done by friction: \(W=m\,(\mu_o\,g\,\cos\theta)\,\int{x\,dx}\) Let's use: \(\mu_o\,g\,\cos\theta=\omega^2\) \(W=m\,\omega^2\,x^2 /2 \) Now workenergy theorem goes: (KE + PE)  (0 + 0) = W leading to : \(\color{red} {v^2=x(2g\sin\theta \omega^2\,x)}\) So v = 0 for x = 0 and \(x=2\tan\theta /\mu_o\) The rest follows from new value \(x_1=\tan\theta /\mu_o\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0damn.. so stupid.. thats where i went wrong.. :/.. i forgot to consider the work done by friction as a variable.. and integrate

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By the way, "balancing forces" will not lead anywhere, as forces are only balanced at one point where the body is halfway down the plane; and you can find out that only in the end.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea logically i got that.. i was just wondering where i went wrong with my work energy theorem :P.. thanks man.. !!
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