1
Abhisek
·2011-03-24 23:39:46
I think the Work-Energy Theorem is correct.
dW = Fdx
W = 0∫2πr (Fdx)
= F 0∫2πr (dx)
= F [x]02Ï€r
= F(2Ï€r)
=2Ï€rF......................................this is a positive value.
So Work done is not zero...(please verify)
1
Vinay Arya
·2011-03-25 00:40:53
oho.You have taken distance travelled.You have to take net dispacement.Which is equal to zero.
1
kunl
·2011-03-25 01:32:26
let me make it pretty clear...
answer urself these two questions which force and which diosplacement are u tlking about n u will get ur answer!
btw both the theories are correct!it's just tht u r messing up with identification of "which work" are u calculating!
its ok if u realize what slipped in ur thought process...if not after sometime i will post a long reply regarding this "APPARENT PARADOX"
1
kunl
·2011-03-25 01:41:54
just a hint
non-uniform motion so work done by tangential force is ∫F.dl and note u can't take F out of integration here as u were thinking tht ∫dl=0 but the force itself is varying in direction(and if by ur grace maybe even in magnitude [3]...just kidding)
1
Abhisek
·2011-03-25 02:29:11
@ Kunl : Can u post the *LONG REPLY* plz ?
1
kunl
·2011-03-25 02:37:25
did u not read the hint?
yaar long wala will take some time i already had spent time i allocated for long reply with rahul :P
1
Vinay Arya
·2011-03-25 03:39:26
Sorry,the direction of the net force is changing.I took F.dr but it is F vector.dr vector.
Thank you