A heavy chain with a mass per unit length $\rho$ is pulled by the constant force $F$ along a horizontal surface consisting of a smooth section and a 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 rough surface is $\mu$, determine the velocity of the chain when $x=L$.
I am applying work energy theorem. Work done by constant Force will be Force × displacement of centre of mass i.e $FL$ but not able to find work done by friction. The friction force at an instant when chain length $x$ lies on the rough surface should be $\mu\rho x g$. This force is continuously decreasing. I feel calculus is involved here but I am unable to apply it. Please help me.
Best Answer
The chain is initially at rest, so \begin{equation} KE_{o} = 0 \end{equation}
The force of friction is given by \begin{equation} f(x) = \mu \rho (L-x) g \end{equation}
The net force on the chain is \begin{equation} \sum F = F - \mu \rho (L-x) g \end{equation}
Work done on the chain is the integral of force over distance, so \begin{equation} W = \int_{0}^{L}F - \mu \rho (L-x) gdx \end{equation}
Integrate and get
\begin{equation} W = FL - \frac{1}{2} \mu \rho g L^{2} \end{equation}
Use Work-Energy Theorem \begin{equation} KE_{f} = KE_{o} + W \end{equation}
and final kinetic energy is \begin{equation} KE_{f} = FL - \frac{1}{2} \mu \rho g L^{2} \end{equation}
Kinetic Energy equation \begin{equation} KE_{f} = \frac{1}{2} m v_{f}^{2} = \frac{1}{2} (\rho L) v_{f}^{2} = FL - \frac{1}{2} \mu \rho g L^{2} \end{equation}
Solve for final velocity \begin{equation} v_{f} = \sqrt{\frac{2F}{\rho} - \mu gL} \end{equation}