# How to calculate work done by variable kinetic friction force?

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.

• 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

## 2 Answers

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}

Great question!! We will use the work-energy theorem for solving this.

$$KE_i$$ = 0

Let the final velocity be v.

Therefore, $$KE_f$$ = $$\frac{1}{2} m v_{f}^{2}$$

$$KE_f - KE_i = W_F + W_{fric}$$

As you said, $$W_F = FL$$

Force of friction: $$\begin{equation} f(x) = \mu \rho (L-x) g \end{equation}$$ $$\begin{equation}W_{fric} = -\int_{0}^{L}\mu \rho (L-x) gdx = -\mu \rho g \frac{L^2}{2} \end{equation}$$ (The work is negative because friction and displacement are in opposite direction)

$$\begin{equation}KE_f - KE_i = FL -\mu \rho g \frac{L^2}{2}\end{equation}$$

$$\frac{\rho Lv^2}{2}$$ = $$FL -\mu \rho g \frac{L^2}{2}$$

Solving for v,

$$\begin{equation} v_{f} = \sqrt{\frac{2F}{\rho} - \mu gL} \end{equation}$$

This could have even been done with Forces but using Work-Energy theorem makes it much more easier.

Note: This is just my second answer on this platform and I would love to get your responses and suggestions. Thank you!!