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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Jan 2008 00:09:45 IST
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Hey karthik for the first one:
I have solved it completely and found that the answer given
is wrong.
For getting answer sa V = 1.7 m/s
I think a change has to be made in the question.
K = 5mg/L
Now try it Karthik you will get the answer.
Hope it is useful.
Rate if useful.
Cheers!!!!! @@@ !!!!!! :)
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4 Jan 2008 00:15:32 IST
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For the 1st one, I"m getting the distance as
am /  (M2 - m2)
Karthik tell me how are u getting the nearby answer..give the equations...
Rate if this is right!!
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4 Jan 2008 00:18:00 IST
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Hey tell me how u got v=1.7 m/s
Explain it plz
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Hello.
The solution for fig.
On drawing the daigram join the starting and end points it
forms a triangle.
Now we know height and base.
Let @ be the angle with horizontal.
So,
Cos@ = L
h2 + L2
Frictional Force = (mgCos@)K
= Kmg * L
 h2+L2
= KmgL
h2+L2
Work done by Force = (net work done) + W.D against
frictional Force.
= mgh + KmgL * h2+L2
h2+L2
= mgh + KmgL
= mg(h+KL)
Hope you find it useful.
Rate if useful.
Cheers !!!!!! @@ !!!!!!!! 
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4 Jan 2008 00:19:53 IST
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first one is quite simple...if the mass m goes down by a distance h(which we have to find) then mass M will go up by root(a2+h2)-a....now by conserving energy...Mg( root(a2+h2)-a)=mgh now rearranging and squaring...M2(a2+h2)=m2h2+M2a2+2Mmah or h2(M2-m2)=2Mmah or h= 2Mmah/(M2-m2)
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4 Jan 2008 00:22:50 IST
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Dude,
The first question is wrong.
Check ur book again.if K = 5mg/L only then answer comes out
to be V = 1.7 m/s
Cheers !!!!!!!!!!!! :)
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4 Jan 2008 00:24:27 IST
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Hey Rohit
Thats a rocking answer..Cheers to his brain!!
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4 Jan 2008 00:25:26 IST
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Its K =5mg/l
See the question again
Now plz provide explanation
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4 Jan 2008 00:40:40 IST
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Here is the solution:
Let "B" move down "x" distance before the string breaks off
So "A" will move towards right by "x".
At the moment of breaking say spring makes an angle @ with
the vertical so the new stretched length will be say "L1"
L1 = L2+x2
The change in length will be:
dL = L2+x2 - (L)
As per law of conservation of energy we have:
Loss in PE of B
= Gain in KE of B + gain in KE of A + gain in PE of spring
Therefore we get:
mgx = (1/2)mv2 + (1/2)mv2 + (1/2)(K)(dL)2
On solving above equation and putting K = 5gm/L
we get:
v2 = gx - (5g/2L) ( L2+x2 - (L) )2 ........ (1)
Now if we draw an FBD of A we have:
TCos@ = mg ........ (2)
And if we draw FBD of B we will have:
T = K(dL) .........(3)
Equationg (2) and (3)
solving we will get:
x = 3L/4
Put the value of x = 3L/4 in equation(1) and solve:
finally you will get:
V2 = 19gL/32
Substitute the values in above expression we get:
V = 1.7 m/s
Hope it is useful.
Rate if useful.
Cheers!!!!!!!!@@@!!!!!!!!!
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4 Jan 2008 00:51:31 IST
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Solution to Q.4)
Let "T" be the tension at the top poitn of the hanging part
and let x be the length of the chain inside the tube at any
instant.
Now we have Equation of motion as:
mgh - T = (mh)a ........... (1)
m = mass per unit length of the chain.
Equation of motion of the horizontal part:
T = (mx)a
We know:
a = dv/dt = (dv/dx)(dx/dt) = v.(dv/dx)
T = (mx)v.(dv/dx) .......... (2)
Now adding (1) and (2) we get:
mgh = mv.(dv/dx)(h+x)
(gh/h+x)dx = v.dv
Now integrating both sides.
gh [0] [L-h] (1/h+x)dx = [0][v] v.dv
On solving above we get:
v2/2 = (gh)Loge(h+x) [ limit is from 0 to (L-h) ]
v2 = (2gh)Loge(L/h)
v = (2gh)Loge(L/h)
Hope you find it useful.
Rate if useful.
Cheers !!!!!! @@@ !!!!!!!
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Always available for help !
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4 Jan 2008 01:14:00 IST
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Hey dude,
I have got the answer to your Q.8)Ans:0.09J
Change in length due to stretching
dL = LSec@ - L = L(Sec@-1)
From the figure provided by me Below:
TSin@ = uR ...........(1)
TCos@ + R = mg
R = mg - TCos@ ......(2)
Substituting the value of "R" in eq.(2) into eq.(1)
we get:
Tsin@ = u(mg-TCos@)
T = umg
(Sin@ + uCos@)
If say "K" is spring constant,then T = K(dL)
Therefore,
K(dL) = umg
(Sin@+uCos@)
K = umg
L(Sec@-1)(Sin@+uCos@)
Now,
PE of Spring = Work done against friction
W = (1/2)K(dL)2
W = umg (L2(Sec@-1)2
2 * L(Sec@-1) * (Sin@ + uCos@)
W = umgL (1-Cos@)
2Cos@ * (Sin@ + uCos@)
Now given @ = 300,L=0.4m,m=1,u=0.2
Substituting the values we get:
W = 0.09Joule
Hope you find it useful.
Rate if useful.
Cheers!!!!!!!@@@@!!!!!!!!!
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4 Jan 2008 02:53:02 IST
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