# Example: Energy conservation

## Understanding the situation

As an example of how the conservation of mechanical energy and its extension to a the work-energy theorem with potential energies, consider the situation shown in the figure below.

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 object we are focusing on is the mass, and it feels the force of the spring, the force of gravity, and the normal force of the track. Although we could analyze the motion of the mass using forces, since the mass is always traveling perpendicular to the normal force, the only forces affecting the kinetic energy of the mass is the force of the spring and the force of gravity. And both of these are conservative forces that can be described by the use of potential energies. {Recall from the work-energy theorem that a force perpendicular to an object's motion does no work and therefore does not change its KE.}

## Presenting a sample problem

Suppose we can approximate the track as frictionless and for the motions considered here we can model the spring with Hooke's law. The moving object has a mass $m$ = 0.2 kg. The spring is initially compressed a distance of $\Delta l$ =4 cm. When the mass is released, it slides up the ramp to a height $h$ = 10 cm, at which point it turns around and slides back down. Find the spring constant, $k$, of the spring.

## Solving this problem

Since we know that energy conservation will give us a relation that constrains the relevant variables of the problem — displacements and velocities, let's see what it gives us. We can use energy conservation either in the form "initial energy = final energy" or "change in energy = 0". Let's use the first:

$${1 \over 2} \; m v_i^2 + U_i = {1 \over 2} \; m v_f^2 + U_f$$

To begin with a problem of this sort (indeed, of almost any sort), it's best to "tell the story of the problem" — to describe what is happening step by step focusing on the features (in this case forms of energy) that we will be using to do our calculations.

We start with the situation shown in the figure. There are three kinds of energies we are going to have to pay attention to: the kinetic energy of the mass, the potential energy of the spring-mass system, and the potential energy of the earth-mass system. So we have a conservation of energy equation that looks like the following.

$${1 \over 2} m v_i^2 + {1 \over 2} k (\Delta l_i)^2 + mgh_i = {1 \over 2} m v_f^2 + {1 \over 2} k (\Delta l_f)^2+ mgh_f$$

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 $k$. 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 a height of $h = 10$cm. At that point it stops for a moment and then begins to slide down. Let's take the final state as the top so we know everything about the variables: the spring is uncompressed, the KE is zero, and the height it 10 cm. But just for good practice, let's put in the 0s we know and keep the values as symbols and solve for k

$$0 + {1 \over 2} k (\Delta l)^2 + 0 = 0 + 0 +mgh$$

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

$$k = \frac{2mgh}{(\Delta l)^2}$$

This shows that we can indeed find $k$ given the information we have, since we have values for all of the symbols on the right.

$$k = \frac{2 \times (0.2\;\mathrm{kg})\times (10\;\mathrm{N/kg}) \times (10\;\mathrm{cm})}{(4\;\mathrm{cm)^2}} = \frac{40}{16} \frac{\mathrm{kg\; N\; cm}}{\mathrm{kg\;cm^2}} = 2.5 \;\mathrm{N/cm}$$

I haven't forced the expression into all SI units (should be meters instead of centimeters) since for a spring a few cm long, compressions make more sense in terms of centimeters rather than meters.

Joe Redish 11/19/14

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