  Question

# 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. √F−L√F−2L√F−4L√F−L2

Solution

## The correct option is A √F−LHere the friction force will be acting only on part of the chain in contact with rough surface i.e L−x i.e Kinetic friction fk=μmg=μ×λ(L−x)g Net force on chain in horizontal direction =F−fk ∴ Applying Newton's 2nd law on the chain gives Fnet=ma=mvdvdx F−fk=mvdvdx.....(1)  Substituting fk in Eq.(1) F−μλ(L−x)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 F∫L0dx−μλg∫L0(L−x)dx=λL∫v0vdv ⇒FL−μλg(L2−L22)=λLv22 ...(2) Putting the values of parameters λ=2 ,μ=0.1,g=10 in Eq.(2), we get FL−2L22=Lv2 ⇒F−L=v2 ⇒v=√F−L  m/s  Suggest corrections   