Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
First we need some clarification. What is $x$ precisely? And What is $L$? How long is the rough section, and how long is the smooth section? A picture will sure help. – Rody Oldenhuis Aug 20 '12 at 8:22
"x is an assumed length of chain at any instant" Do you mean by "instant" "moment"? In that case I though chain has a fixed length. – Yrogirg Aug 20 '12 at 13:51
"I feel calculus is involved here but I am unable to apply it" Do you know how to solve differential equations? – Yrogirg Aug 20 '12 at 13:55

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}

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.