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Consider a box of mass m moving along a rough, level surface. The box is subject only to horizontal applied forces, gravity, normal force and kinetic friction as it moves.

Part A

First, suppose the box is moved in one dimension from position x_i directly to position x_f as shown in the figure below.

Assuming that the coefficient of kinetic friction is a constant μ_k, find an expression for the work done by friction in the course of this movement.

Solution

A free body diagram for the box will look like:

The friction force will be:

\begin{large}\[ F_{f} = \mu_{k}N\]\end{large}

where N is the normal force on the box from the floor. However, since the box is only moving horizontally (meaning a_y is zero) the free body diagram implies we can write Newton's 2nd Law for the y direction as:

\begin{large}\[ N-mg = ma_{y} = 0\]\end{large}

This tells us that the normal force is equal in magnitude to the box's weight and so the force of kinetic friction has the constant magnitude:

\begin{large}\[ F_{f} =\mu_{k}mg\]\end{large}

Further, since kinetic friction is always directed opposite to the motion, and the motion is always in the + x direction, we can write:

\begin{large}\[ \vec{F}_{f} = -\mu_{k}mg\hat{x}\]\end{large}

and:

\begin{large}\[ d\vec{r} = dx\hat{x}\]\end{large}

Thus, our path integral is reduced to a one-dimensional integral of the form:

\begin{large}\[ W = \int_{x_{i}}^{x_{f}} (-\mu_{k} mg)\;dx = -\mu_{k}mg(x_{f}-x_{i})\]\end{large}

which depends only upon the endpoints...

Part B

Now consider the alternate path between the same endpoints x_i and x_f shown here:

What is the work done by kinetic friction on the box in the course of this movement?

Solution

This path has the box first making the trip from the position x_i to a position x_m between the initial and final positions of the box, then returning to the initial position x_i and then completing the trip to the final position x_f. Just as in Part A, the box is assumed to be subject to purely horizontal applied forces so that the friction force has a constant magnitude of:

\begin{large}\[F_{f} = \mu_{k}mg\]\end{large}

To evaluate the path integral for the work, we must break the path up into three parts which consist of motion in one direction only. The parts coincide with the stages labeled 1 through 3 in the figure above. In each stage, we must determine the vector form of the friction force.

\begin{large}\[ \mbox{Stage 1:   }\vec{F}_{f} = - \mu_{k}mg\hat{x}\]
\[ \mbox{Stage 2:   }\vec{F}_{f} = + \mu_{k}mg\hat{x}\]
\[ \mbox{Stage 3:   }\vec{F}_{f} = - \mu_{k}mg\hat{x}\]\end{large}

We can now write the path integral for the work, using the fact that all motion is in the x direction:

\begin{large}\[  W = \int_{x_{i}}^{x_{m}} (-\mu_{k} mg)\;dx + \int_{x_{m}}^{x_{i}} \mu_{k}mg\;dx + \int_{x_{i}}^{x_{f}}(-\mu_{k}mg)\;dx\]\end{large}

The result of the integrations is:

\begin{large}\[ W = -\mu_{k}mg (x_{m}-x_{i} + x_{m} - x_{i} + x_{f} - x_{i}) = -\mu_{k}mg(2(x_{m}-x_{i}) + (x_{f}-x_{i})) = -\mu_{k}mgd \]\end{large}

where d is the total distance (not displacement) traveled by the box. Since distance depends on the path (it is not a function of the endpoints only) we see that the exact path traveled by the box is important.