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A chain of mass per unit length λ=2 kg/m is pulled up by a constant force F. Initially, the chain is lying on a rough surface and passes onto the smooth surface. The co-efficient of kinetic friction between chain and rough surface is μ=0.1. The length of the chain is L. Then, find the speed (in m/s) of the chain when x=L. Take g= 10 m/s2.


  1. FL
  2. F2L
  3. F4L
  4. FL2


Solution

The correct option is A FL
Here the friction force will be acting only on part of the chain in contact with rough surface i.e Lx


i.e Kinetic friction
fk=μmg=μ×λ(Lx)g

Net force on chain in horizontal direction =Ffk
Applying Newton's 2nd law on the chain gives
Fnet=ma=mvdvdx
Ffk=mvdvdx.....(1) 

Substituting fk in Eq.(1)
Fμλ(Lx)g=λLvdvdx

Multiplying by dx and integrating w.r.t x, considering the limits from x=0 to x=L and v=0 to v=v
FL0dxμλgL0(Lx)dx=λLv0vdv
FLμλg(L2L22)=λLv22 ...(2)

Putting the values of parameters λ=2 ,μ=0.1,g=10 in Eq.(2), we get
FL2L22=Lv2
FL=v2
v=FL  m/s 
 

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