The law of conservation of energy states that within a closed system the total amount of energy is always conserved.

Another way of this is saying this is that energy can be neither created of destroyed, it can only be converted from one form to another.

There are, however, many different forms of energy.

For mechanics problems it is useful to think about a more restricted law, which considers mechanical energy to be conserved.

Conservation of mechanical energy can be assumed whenever the forces which apply to a system are entirely conservative.

We first define an energy associated with motion, which is called Kinetic Energy. This should be a scalar quantity, it should not depend on the direction of motion, but it should depend on the mass of the moving object, a heavier object moving with a given speed should have a higher kinetic energy than one a lighter one.

$K=\frac{1}{2}mv^{2}$

The unit for energy is the joule, $1\,\mathrm{J}=1\mathrm{kg\,m^{2}s^{-2}}=1\,\mathrm{N\,m}$.

The kinetic energy will depend on the reference frame of the motion.

When certain forces act on objects it can be useful to consider the concept of potential energy. These are forces where the change in the energy of an object due to the action of that force depends only on the position on the object. We call these forces conservative forces and define a change in potential energy $U$ as

$\Delta U=U_{f}-U_{i}=-\int_{x_{i}}^{x_{f}}F_{x}dx$

for motion in one direction or more generally

$\Delta U=U_{f}-U_{i}=-\int_{r_{f}}^{r_{i}} \vec{F}\cdot d\vec{r}$

We will return to this second equation in more detail in our next class.

$\Delta U_{G}=\int_{y_{1}}^{y_{2}}mg\,dy=mg(y_{2}-y_{1})$

The change in potential energy depends only the change in height.

We've seen already that potential energy due to gravity should have the form $mg(y_{2}-y_{1})$. We can measure potential energy relative any position that makes sense to us, the most sensible place to measure from will depend on the problem. We can then say that the gravitational potential energy of an object is given by the height $h$ of that object above that reference point is $mgh$. An object below the reference point will have negative gravitational potential energy.

As we saw previously the force due to gravity on an object of mass $m$ due to an object of mass $M$ at distance $r$ is $-\frac{GmM}{r^2}\hat{r}$

If we raise an object from the surface of a planet to a distance $h$ above it the change on potential energy is

$\Delta U = \int_{R_{E}}^{R_{E}+h}\frac{GmM}{r^2}\,dr=-\frac{GmM}{R_{E}+h}+\frac{GmM}{R_{E}}$

We should consider where a suitable zero of potential energy is

Choosing $r=\infty$ as our reference position is attractive as then the potential energy change in bringing an object from infinity to a position $r$

$\Delta U = \int_{\infty}^{R}\frac{GmM}{r^2}\,dr=-\frac{GmM}{r}$

is the same as the potential energy, so generally we say the gravitational potential energy at point $r$ from the center of a mass $M$.

$U = -\frac{GmM}{r}$

Another example of a conservative force is the force provided by a spring

$F_{s}=-kx$

where $x$ is the distance of the end of the spring from the equilibrium end position of the spring. The change in potential energy associated with compressing or extending a spring is

$\Delta U = -\int_{x_{i}}^{x_{f}}F_{s}\,dx=\int_{x_{i}}^{x_{f}}kx\,dx=\frac{1}{2}k(x_{f}^2-x_{i}^2)$

If only conservative forces are active within an isolated system then

$\Delta K + \Delta U = 0 $

We call the sum of the Kinetic Energy and Potential Energy the Total Mechanical Energy E

$E=K+U$

and we can see that under the condition of only conservative forces acting this is a conserved quantity.

Which track should give a ball a faster velocity at the end of the track?

A. Track A B. Track B or C. Neither, the velocity should be the same.

On which track should a ball arrive first if they are released simultaneously?

A. Track A B. Track B or C. Neither, the time taken should be the same.

We can use the conservation of mechanical energy to know that if the change in height and therefore potential energy is the same in both cases, the kinetic energy gained should be the same in both cases, and thus the velocity at the end of both tracks should be the same.

We can't, however, use this to tell the time taken, this depends on the path taken. We can understand the counter-intuitive result that the longer path takes less time based on the fact that throughout the middle section of the path the ball is always moving faster than it is on the straight path.

In reality there is a limit to how far we can bend the path and still have the ball take less time. The fastest path is the one closest to the Cycloid.

$\frac{dv}{dt}=\frac{-GM_{E}}{(R_{E}+x)^{2}}$

$\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{-GM_{E}}{(R_{E}+x)^{2}}$

$\int^{v}_{v_{0}}v\,dv=\int^{x}_{0}\frac{-GM_{E}}{(R_{E}+x)^{2}}\,dx$

$\frac{1}{2}(v^{2}-v_{0}^2)=\frac{GM_{E}}{R_{E}+x}-\frac{GM_{E}}{R_{E}}$

$v^{2}=v_{0}^2+2(\frac{GM_{E}}{R_{E}+x}-\frac{GM_{E}}{R_{E}})$

as $x\to\infty$ the object will escape if

$0\geq v_{0}^{2}-\frac{2GM_{E}}{R_{E}}$

For the object to escape

$v_{0}\geq \sqrt{\frac{2GM_{E}}{R_{E}}}$

At the Earth's surface the total mechanical energy of a rocket is

$\frac{1}{2}mv^{2}-\frac{GmM_{E}}{R_{E}}$

The condition for escape is that at infinity the velocity equals zero (just avoids being negative). As at infinity the gravitational potential energy is equal to zero the condition for the escape velocity is

$\frac{1}{2}mv_{escape}^{2}-\frac{GmM_{E}}{R_{E}}=0$

and

$v_{escape}=\sqrt{\frac{2GM_{E}}{R_{E}}}$