A block on a rough inclined plane..... Question attached.
@Mashy
@shubhamsrg @DLS
Its easy.
What the heck is that ? Sry i only understand english
Ignore the hindi part. Its just the translation of the question.
I ve got d ans.
For the first question?
yes
Its a. Might me some minor mistake.
Here Suppose the block has moved a distance x
then at position x balance the forces i.e. mgsin(-) - umgcos(-) = 0 Integrate this from 0 to x
u get the answer on simplification.
Reply when done
Well i get this sin(-)=mu*x^2/2*cos(-) Doesnt give me the required answer? @lordcyborg
there is no m term on RHS
and u missed an 'x' term on LHS
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 ?
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
Oh. But you cant balance forces for the body to stop right? You have to conserve the work done?
When you balance forces, it means there is no acceleration. It doesnt necessarily mean that the body is at rest? And a man.
lol.........Dont go too deep...........i'm balancing forces when body is at rest
I dont think you can balance force for the body to be at rest.
ya we cannot all times.........bt we can
Why doesnt conservation of energy give me the answer?
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
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.
C'mon man............I gave u the perfect(i suppose) reasons.........now go and sleep it 2:15 a.m.
Yeah you did. Thank you.
|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
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