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

- yrelhan4

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

- jamiebookeater

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- yrelhan4

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- DLS

Its easy.

- anonymous

What the heck is that ?
Sry i only understand english

- yrelhan4

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

- anonymous

I ve got d ans.

- yrelhan4

For the first question?

- anonymous

yes

- yrelhan4

Its a. Might me some minor mistake.

- anonymous

Here
Suppose the block has moved a distance x

- anonymous

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

- anonymous

u get the answer on simplification.

- anonymous

Reply when done

- yrelhan4

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

- anonymous

there is no m term on RHS

- anonymous

and u missed an 'x' term on LHS

- yrelhan4

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 ?

- anonymous

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

- yrelhan4

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

- yrelhan4

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

- anonymous

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

- yrelhan4

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

- anonymous

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

- yrelhan4

Why doesnt conservation of energy give me the answer?

- anonymous

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

- yrelhan4

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.

- anonymous

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

- yrelhan4

Yeah you did. Thank you.

- shubhamsrg

|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

- shubhamsrg

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

- yrelhan4

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

- shubhamsrg

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

- yrelhan4

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

- shubhamsrg

|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.

- shubhamsrg

(SHAYD)
:/

- shubhamsrg

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

- yrelhan4

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

- yrelhan4

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

- Vincent-Lyon.Fr

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.

- yrelhan4

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

- yrelhan4

a*

- Vincent-Lyon.Fr

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

- Vincent-Lyon.Fr

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

- yrelhan4

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

- Vincent-Lyon.Fr

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

- yrelhan4

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

- anonymous

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..

- anonymous

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

- anonymous

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..

- anonymous

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..

- anonymous

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

- Vincent-Lyon.Fr

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 \(x_1=\tan\theta /\mu_o\)

- anonymous

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

- Vincent-Lyon.Fr

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.

- anonymous

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

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