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EyadBest ResponseYou've already chosen the best response.0
What the heck is that ? Sry i only understand english
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
Ignore the hindi part. Its just the translation of the question.
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
For the first question?
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
Its a. Might me some minor mistake.
 one year ago

lordcyborgBest ResponseYou've already chosen the best response.0
Here Suppose the block has moved a distance x
 one year ago

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

lordcyborgBest ResponseYou've already chosen the best response.0
u get the answer on simplification.
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
Well i get this sin()=mu*x^2/2*cos() Doesnt give me the required answer? @lordcyborg
 one year ago

lordcyborgBest ResponseYou've already chosen the best response.0
there is no m term on RHS
 one year ago

lordcyborgBest ResponseYou've already chosen the best response.0
and u missed an 'x' term on LHS
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
There 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 ?
 one year ago

lordcyborgBest ResponseYou've already chosen the best response.0
hey 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
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
Oh. But you cant balance forces for the body to stop right? You have to conserve the work done?
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
When you balance forces, it means there is no acceleration. It doesnt necessarily mean that the body is at rest? And a man.
 one year ago

lordcyborgBest ResponseYou've already chosen the best response.0
lol.........Dont go too deep...........i'm balancing forces when body is at rest
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
I dont think you can balance force for the body to be at rest.
 one year ago

lordcyborgBest ResponseYou've already chosen the best response.0
ya we cannot all times.........bt we can
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
Why doesnt conservation of energy give me the answer?
 one year ago

lordcyborgBest ResponseYou've already chosen the best response.0
coz 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
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
You 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.
 one year ago

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

yrelhan4Best ResponseYou've already chosen the best response.0
Yeah you did. Thank you.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
dw: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
 one year ago

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

yrelhan4Best ResponseYou've already chosen the best response.0
:O mere dimaag me kabhi nhi aata ye, 2nd part konsa? Heat produced?
 one year ago

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

yrelhan4Best ResponseYou've already chosen the best response.0
Hmm. Equation likh do 2nd part ke liye. :/
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
dw: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.
 one year ago

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

yrelhan4Best ResponseYou've already chosen the best response.0
Acha ruk.. Sahi hoga. :/ Ke 0 kyun hai? x to lamit 0 se x/2 tak nhi jayega?
 one year ago

yrelhan4Best 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?
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.2
Answers 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.
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
1st one by 1=v*dv/dx right?
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.2
For 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 \(x(t)=\Large\frac{\tan\theta}{\mu_o}\normalsize (1\cos\omega t)\)
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.2
I'll try solving it by workenergy theorem tomorrow, but now I have to go to bed.
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
Yeah i am looking for that solution. Alright. Thank You.
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.2
At the top, \(\cos \omega t=1\) At the bottom, \(\cos \omega t=1\) Halfway, \(\cos \omega t=0\) The rest is easy.
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
Right. I'll try and get back. And please see if we can do it by work energy theorem too.
 one year ago

MashyBest ResponseYou've already chosen the best response.0
if 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..
 one year ago

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

MashyBest ResponseYou've already chosen the best response.0
you 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..
 one year ago

MashyBest ResponseYou've already chosen the best response.0
and 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..
 one year ago

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

VincentLyon.FrBest ResponseYou've already chosen the best response.2
Here 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 \(x_1=\tan\theta /\mu_o\)
 one year ago

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

VincentLyon.FrBest ResponseYou've already chosen the best response.2
By 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.
 one year ago

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