## yrelhan4 Group Title A block on a rough inclined plane..... Question attached. one year ago one year ago

1. yrelhan4 Group Title

2. yrelhan4 Group Title

@Mashy

3. yrelhan4 Group Title

@shubhamsrg @DLS

4. DLS Group Title

Its easy.

What the heck is that ? Sry i only understand english

6. yrelhan4 Group Title

Ignore the hindi part. Its just the translation of the question.

7. lordcyborg Group Title

I ve got d ans.

8. yrelhan4 Group Title

For the first question?

9. lordcyborg Group Title

yes

10. yrelhan4 Group Title

Its a. Might me some minor mistake.

11. lordcyborg Group Title

Here Suppose the block has moved a distance x

12. lordcyborg Group Title

then at position x balance the forces i.e. mgsin(-) - umgcos(-) = 0 Integrate this from 0 to x

13. lordcyborg Group Title

u get the answer on simplification.

14. lordcyborg Group Title

15. yrelhan4 Group Title

Well i get this sin(-)=mu*x^2/2*cos(-) Doesnt give me the required answer? @lordcyborg

16. lordcyborg Group Title

there is no m term on RHS

17. lordcyborg Group Title

and u missed an 'x' term on LHS

18. yrelhan4 Group Title

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 ?

19. lordcyborg Group Title

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

20. yrelhan4 Group Title

Oh. But you cant balance forces for the body to stop right? You have to conserve the work done?

21. yrelhan4 Group Title

When you balance forces, it means there is no acceleration. It doesnt necessarily mean that the body is at rest? And a man.

22. lordcyborg Group Title

lol.........Dont go too deep...........i'm balancing forces when body is at rest

23. yrelhan4 Group Title

I dont think you can balance force for the body to be at rest.

24. lordcyborg Group Title

ya we cannot all times.........bt we can

25. yrelhan4 Group Title

Why doesnt conservation of energy give me the answer?

26. lordcyborg Group Title

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

27. yrelhan4 Group Title

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.

28. lordcyborg Group Title

C'mon man............I gave u the perfect(i suppose) reasons.........now go and sleep it 2:15 a.m.

29. yrelhan4 Group Title

Yeah you did. Thank you.

30. shubhamsrg Group Title

|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

31. shubhamsrg Group Title

for 2nd part, limit of v will be from 0 to v, of x will be 0 to tan@/u0 ^_^

32. yrelhan4 Group Title

:O mere dimaag me kabhi nhi aata ye, 2nd part konsa? Heat produced?

33. shubhamsrg Group Title

oh sorry, nahi 3rd part me ye. heat produced me work energy lagayenge (SHAYAD) :/

34. yrelhan4 Group Title

Hmm. Equation likh do 2nd part ke liye. :/

35. shubhamsrg Group Title

|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.

36. shubhamsrg Group Title

(SHAYD) :/

37. shubhamsrg Group Title

check kar lena, maine bola hai to high probability hai ki galat hoga. 3 baj gaya! :O gtg bye

38. yrelhan4 Group Title

Acha ruk.. Sahi hoga. :/ Ke 0 kyun hai? x to lamit 0 se x/2 tak nhi jayega?

39. yrelhan4 Group Title

@Vincent-Lyon.Fr We got the first and the last part. For the second part what would be the work energy equation?

40. Vincent-Lyon.Fr Group Title

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.

41. yrelhan4 Group Title

1st one by 1=v*dv/dx right?

42. yrelhan4 Group Title

a*

43. Vincent-Lyon.Fr Group Title

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 $$x(t)=\Large\frac{\tan\theta}{\mu_o}\normalsize (1-\cos\omega t)$$

44. Vincent-Lyon.Fr Group Title

I'll try solving it by work-energy theorem tomorrow, but now I have to go to bed.

45. yrelhan4 Group Title

Yeah i am looking for that solution. Alright. Thank You.

46. Vincent-Lyon.Fr Group Title

At the top, $$\cos \omega t=1$$ At the bottom, $$\cos \omega t=-1$$ Halfway, $$\cos \omega t=0$$ The rest is easy.

47. yrelhan4 Group Title

Right. I'll try and get back. And please see if we can do it by work energy theorem too.

48. Mashy Group Title

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..

49. Mashy Group Title

sorry i flipped sin and cosine terms... so the answer should be 4 like someone said earlier.. :-/..

50. Mashy Group Title

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..

51. Mashy Group Title

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..

52. Mashy Group Title

ahh.. there is definitely something wrong with what i did.. :-/... the answer should be ONE itself.. for the first question :(

53. Vincent-Lyon.Fr Group Title

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 $$x_1=\tan\theta /\mu_o$$

54. Mashy Group Title

damn.. so stupid.. thats where i went wrong.. :-/.. i forgot to consider the work done by friction as a variable.. and integrate

55. Vincent-Lyon.Fr Group Title

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.

56. Mashy Group Title

yea logically i got that.. i was just wondering where i went wrong with my work energy theorem :P.. thanks man.. !!