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yrelhan4 Group Title

A block on a rough inclined plane..... Question attached.

  • one year ago
  • one year ago

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  1. yrelhan4 Group Title
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    • one year ago
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  2. yrelhan4 Group Title
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    @Mashy

    • one year ago
  3. yrelhan4 Group Title
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    @shubhamsrg @DLS

    • one year ago
  4. DLS Group Title
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    Its easy.

    • one year ago
  5. Eyad Group Title
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    What the heck is that ? Sry i only understand english

    • one year ago
  6. yrelhan4 Group Title
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    Ignore the hindi part. Its just the translation of the question.

    • one year ago
  7. lordcyborg Group Title
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    I ve got d ans.

    • one year ago
  8. yrelhan4 Group Title
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    For the first question?

    • one year ago
  9. lordcyborg Group Title
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    yes

    • one year ago
  10. yrelhan4 Group Title
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    Its a. Might me some minor mistake.

    • one year ago
  11. lordcyborg Group Title
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    Here Suppose the block has moved a distance x

    • one year ago
  12. lordcyborg Group Title
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    then at position x balance the forces i.e. mgsin(-) - umgcos(-) = 0 Integrate this from 0 to x

    • one year ago
  13. lordcyborg Group Title
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    u get the answer on simplification.

    • one year ago
  14. lordcyborg Group Title
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    Reply when done

    • one year ago
  15. yrelhan4 Group Title
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    Well i get this sin(-)=mu*x^2/2*cos(-) Doesnt give me the required answer? @lordcyborg

    • one year ago
  16. lordcyborg Group Title
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    there is no m term on RHS

    • one year ago
  17. lordcyborg Group Title
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    and u missed an 'x' term on LHS

    • one year ago
  18. yrelhan4 Group Title
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    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
  19. lordcyborg Group Title
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    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
  20. yrelhan4 Group Title
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    Oh. But you cant balance forces for the body to stop right? You have to conserve the work done?

    • one year ago
  21. yrelhan4 Group Title
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    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
  22. lordcyborg Group Title
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    lol.........Dont go too deep...........i'm balancing forces when body is at rest

    • one year ago
  23. yrelhan4 Group Title
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    I dont think you can balance force for the body to be at rest.

    • one year ago
  24. lordcyborg Group Title
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    ya we cannot all times.........bt we can

    • one year ago
  25. yrelhan4 Group Title
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    Why doesnt conservation of energy give me the answer?

    • one year ago
  26. lordcyborg Group Title
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    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
  27. yrelhan4 Group Title
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    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
  28. lordcyborg Group Title
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    C'mon man............I gave u the perfect(i suppose) reasons.........now go and sleep it 2:15 a.m.

    • one year ago
  29. yrelhan4 Group Title
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    Yeah you did. Thank you.

    • one year ago
  30. shubhamsrg Group Title
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    |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
  31. shubhamsrg Group Title
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    for 2nd part, limit of v will be from 0 to v, of x will be 0 to tan@/u0 ^_^

    • one year ago
  32. yrelhan4 Group Title
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    :O mere dimaag me kabhi nhi aata ye, 2nd part konsa? Heat produced?

    • one year ago
  33. shubhamsrg Group Title
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    oh sorry, nahi 3rd part me ye. heat produced me work energy lagayenge (SHAYAD) :/

    • one year ago
  34. yrelhan4 Group Title
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    Hmm. Equation likh do 2nd part ke liye. :/

    • one year ago
  35. shubhamsrg Group Title
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    |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
  36. shubhamsrg Group Title
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    (SHAYD) :/

    • one year ago
  37. shubhamsrg Group Title
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    check kar lena, maine bola hai to high probability hai ki galat hoga. 3 baj gaya! :O gtg bye

    • one year ago
  38. yrelhan4 Group Title
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    Acha ruk.. Sahi hoga. :/ Ke 0 kyun hai? x to lamit 0 se x/2 tak nhi jayega?

    • one year ago
  39. yrelhan4 Group Title
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    @Vincent-Lyon.Fr We got the first and the last part. For the second part what would be the work energy equation?

    • one year ago
  40. Vincent-Lyon.Fr Group Title
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    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
  41. yrelhan4 Group Title
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    1st one by 1=v*dv/dx right?

    • one year ago
  42. yrelhan4 Group Title
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    a*

    • one year ago
  43. Vincent-Lyon.Fr Group Title
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    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
  44. Vincent-Lyon.Fr Group Title
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    I'll try solving it by work-energy theorem tomorrow, but now I have to go to bed.

    • one year ago
  45. yrelhan4 Group Title
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    Yeah i am looking for that solution. Alright. Thank You.

    • one year ago
  46. Vincent-Lyon.Fr Group Title
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    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
  47. yrelhan4 Group Title
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    Right. I'll try and get back. And please see if we can do it by work energy theorem too.

    • one year ago
  48. Mashy Group Title
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    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
  49. Mashy Group Title
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    sorry i flipped sin and cosine terms... so the answer should be 4 like someone said earlier.. :-/..

    • one year ago
  50. Mashy Group Title
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    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
  51. Mashy Group Title
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    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
  52. Mashy Group Title
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    ahh.. there is definitely something wrong with what i did.. :-/... the answer should be ONE itself.. for the first question :(

    • one year ago
  53. Vincent-Lyon.Fr Group Title
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    Here is the solution using work-energy 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 work-energy 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
  54. Mashy Group Title
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    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
  55. Vincent-Lyon.Fr Group Title
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    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
  56. Mashy Group Title
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    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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