# Example: Using energy conservation with friction

## Understanding the situation

In our previous example, we showed how the conservation of mechanical energy worked with only kinetic and potential energies (no explicit work — though of course PEs correspond to work done by conservative forces). To see an example with work that cannot be described as a potential energy consider the situation shown in the figure below. Our description is the same as before, except now the region between B and C has friction. We can't treat friction with potential energies since it is not a conservative force.

A mass is held against a squeezed spring. When the spring is released, it pushes the mass along a frictionless track. After some distance, the track becomes a ramp and tips upward. The mass feels the force of the spring, the force of gravity, the normal force of the track, and, in the region between B and C, a frictional force. Let's analyze this using the combined energy-conservation work-energy theorem principle.

## Presenting a sample problem

Suppose we can approximate the track as frictionless except for the region between B and C and for the motions considered here we can model the spring with Hooke's law. B The moving object has a mass m = 0.2 kg. The spring is initially compressed a distance of 4 cm, and the spring constant has a value of 2.5 N/cm (see Example 1). When the mass is released, it slides up the ramp part way, at which point it turns around and slides back down. It slides back through the frictional region but comes to a stop at the point B. If L is 20 cm long, find the coefficient of friction of the frictional region, μ.

## Solving this problem

In this case let's start with the  "change in energy = work done by non-conservative forces" form of the work-energy theorem.

$$\Delta \bigg( {1 \over 2} mv^2 + U \bigg) = \int \overrightarrow{F}_{non-cons} \cdot d\overrightarrow{r}$$.

where $\overrightarrow{F}_{non-cons}$ represents the non-conservative forces -- those that can't be changed into a PE. (In this case, friction.) In general the work has to be written as an integral since the force might change with position so you would have to take small steps (small enough where you could treat the force as constant and then add up the $FΔr$terms).

In this case, since the frictional force is always constant and opposite to the displacement of the object, the integral can be simplified. It becomes just the negative of the force times the distance. Since friction is given by $μN$ where $N$ is the normal force squeezing the surfaces together, and since the friction part of the track is horizontal, $N$ is just equal to the object's weight, $mg$. This gives the result

$$\Delta \bigg( {1 \over 2} mv^2 + U \bigg) = -F_{fric} \Delta x$$

This is the principle we will use. Now let's "tell the story of the problem" and see how it is different from last time.

As before, the variables in the problem are: $v$,  the velocity of the mass, $Δl$,  the squeeze of the spring, and $h$, the height of the mass above the track. The parameters are the mass of the object, $m$, the gravitational field strength, $g$, and the spring constant $k$. Now we also have a coefficient of friction $μ$. The values of the variables in the initial state are labeled by i and in the final state by f.

The story is as follows: In our initial state the mass is at rest ($v_i = 0$), the spring is compressed by an amount $Δl = 4$ cm, and we'll take the initial height as $h = 0$. The catch is released and the spring force accelerates the mass, putting all of the energy stored in the spring into kinetic energy. The mass slides across the track at a constant speed (and KE) until it reaches the hill. As it starts up the hill, the gravitational force slows it, converting KE into gravitational PE until it reaches some unknown height. It won't go up as high as before since it loses some of the energy it had gotten from the spring passing through the frictional region. At that point it stops for a moment and then begins to slide down. It speeds up coming down, reaches the frictional region again, slows and comes to a stop at B.

This time, let's take the final state at B. Since it comes to a stop, we still know everything about the variables: the spring is uncompressed, the KE is zero, and the height it 0 cm. Let's start by putting in the new term -- the friction. Since the object passes through the frictional region twice, our distance is $2L$. And as usual, "change" means "final minus initial".

$$\Delta \bigg( {1 \over 2} mv^2 + U\bigg) = - \mu mg (2L)$$

$$\bigg( {1 \over 2} mv_f^2 + U_f\bigg) - \bigg( {1 \over 2} mv_i^2 + U_i\bigg) = -2\mu mg L$$

By our choices, both the initial and final KEs are 0. In this case, the initial PE is that of the spring and, since we are finally at the base level, the final PE is 0. This gives an equation we can solve for our unknown, μ, in terms of the other parameters of the problem.

$${1 \over 2}k(\Delta l)^2 = 2\mu mgL$$

$$\mu = \frac{k(\Delta l)^2}{4mgL}$$

Plugging in numbers we can get a value. (We are careful to put in all our units to make sure our numbers are consistent and that our equation is not obviously wrong.)

$$\mu = \frac{k(\Delta l)^2}{4mgL} = \frac{\mathrm{(2.5\;N/cm)(4\;cm)^2}}{4 \mathrm {(0.2\;kg)(10\;N/kg)(20\;cm)}} = \frac{40}{160} \frac{\mathrm{N\;cm^2\;kg}}{\mathrm{cm\;kg\;N\;cm}} = 0.25$$

All the units cancel correctly  as they should, since the coefficient of friction is unitless.