1
Manmay kumar Mohanty
·2010-04-23 21:43:24
1) work done in the whole process is equal to Wa,0 + Wa,a
now dWa,0 = -K(a\hat{i}+0\hat{j}).dr \Rightarrow W = -K\int(a\hat{i}+0\hat{j}){dr}\hat{i}=-Ka^{2}/2
similarly for 2nd part
hence net work done is -Ka2/2 + ( - Ka2/2 ) = - Ka2
1
Manmay kumar Mohanty
·2010-04-23 22:01:12
2) let length of the chain is L
since chain is allowed to free fall velocity acquired v = √2gL
weight = mg ( downward )
thrust force is given by Ft = vr dmdt where vr = relative velocity = v,
and dmdt = λv ( λ = mass per unit length = mL )
hence Ft = λv2 = mL x 2gL = 2mg
hence net force = Ft + W = 2mg + mg = 3mg
hence required answer is 3mg . So option (c) is correct [1]
11
SANDIPAN CHAKRABORTY
·2010-04-23 22:04:24
@manmay thats what i had also done...but the solution says..
"In the first part F=-K(xi+0j) = -Kxi dr = dx i ∫F.dr = -Ka2/2 . Similarly for the 2nd part......"
i couldnt get what the solution is trying to tell....
1
Manmay kumar Mohanty
·2010-04-23 22:29:55
3) since sine wave has a peroid of 2Ï€ so force will be repeated after every 2Ï€ cycle
hence we will take force for one cycle
force per unit length ( F/l ) hence = B i ( 2Ï€) ,,,,,,,,,,,,, which is (c)