Question

A heavy chain with a mass per unit length $p$ is pulled by the constant force $F$ along a horizontal surface consisting of a smooth section and rough section. The chain is initially at rest on the rough surface with $x=0$. If the coefficient of kinetic friction between the chain and the rough surface is ${\mu }_k,$ determine the velocity $v$ of the chain when $x=L$. The force $F$ is greater than ${\mu }_k,$ $pgL$ in order to initiate the motion.

• A. $\sqrt{\frac{2F}{P}-{\mu }_{k}gL}$
• B. $\sqrt{\frac{F}{P}-{\mu }_{k}gL}$
• C. $\sqrt{\frac{2F}{P}+{\mu }_{k}gL}$
• D. $\sqrt{\frac{4F}{P}-{\mu }_{k}gL}$

Answer 1

The correct option is A $\sqrt{\frac{2F}{P}-{\mu }_{k}gL}$

Using work energy theorem

Work done by $F+$work done by friction$=$change in kinetic energy of chain

$FL-{\int }_0^L{\mu }_{k}N\left(x\right)dx=\frac{1}{2}m{v}^2$...(I)

$N\left(x\right)$ reaction force of the rough region which is equal to weight of chain in rough region

$N\left(x\right)=P(L-x)g$

$m$ is weight of total chain

$m=PL$

Substituting values of m and N in equation (I)

$FL-\frac{1}{2}{\mu }_{k}P{L}^{2}g=\frac{1}{2}PL{v}^2$

$v=\sqrt{\frac{2F}{P}-{\mu }_{k}Lg}$