Question

A chain of mass per unit length $\lambda =2\text{kg/m}$ 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 $\mu =0.1$. The length of the chain is $L$. Then, find the speed ($\text{in m/s}$) of the chain when $x=L$. Take $g=10{\text{m/s}}^2$.

• A. $\sqrt{F-L}$
• B. $\sqrt{F-2L}$
• C. $\sqrt{F-4L}$
• D. $\sqrt{F-\frac{L}{2}}$

Answer 1

The correct option is A $\sqrt{F-L}$

Here 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
$f_k=\mu mg=\mu \times \lambda (L-x)g$

Net force on chain in horizontal direction $=F-f_k$
$\therefore$ Applying Newton's 2nd law on the chain gives
$F_{net}=ma=mv\frac{dv}{dx}$
$F-f_k=mv\frac{dv}{dx}$.....(1)

Substituting $f_k$ in Eq.(1)
$F-\mu \lambda (L-x)g=\lambda Lv\frac{dv}{dx}$

Multiplying by $dx$ and integrating w.r.t $x$, considering the limits from $x=0\text{to}x=L$ and $v=0$ to $v=v$
$F{\int }_0^{L}dx-\mu \lambda g{\int }_0^L(L-x)dx=\lambda L{\int }_0^{v}vdv$
$\Rightarrow FL-\mu \lambda g(L^2-\frac{L^2}{2})=\frac{\lambda L{v}^2}{2}$ ...(2)

Putting the values of parameters $\lambda =2,\mu =0.1,g=10$ in Eq.(2), we get
$FL-\frac{2L^2}{2}=L{v}^2$
$\Rightarrow F-L=v^2$
$\Rightarrow v=\sqrt{F-L}\text{m/s}$