Conservation of Energy

Some of the great tools in physics are so-called "conservation laws" that buttress the laws of motion with certain quantities that remain the same throughout time. Among these great laws is the conservation of energy which states that while energy can change forms, it cannot be created or destroyed. Here we'll explore the interconversion of kinetic energy and potential energy, the relationship between work and energy, and how energies can in some sense replace forces in our calculations.

For a person riding a bicycle, their kinetic energy \(E\) is equal to the amount of heat that would be dissipated from the moment they pull the brake, to the moment they come to rest. If they don't skid, most of this heat will go into heating the brake pad and the metal rim of the wheel. If they do skid, most of the heat will go to the rubber in the tire and to the road. Before they pull the break, this energy is observed as the ongoing motion of the parts of the bicycle, i.e. the net forward movement of the bicycle, and the spinning of the wheels.

There are other forms of energy that a particle can have such as gravitational potential energy, chemical potential energy, electrical potential energy, et cetera, but the kinetic energy is the portion of an object's energy that is due explicitly to its ongoing motion. For a point particle of mass \(\delta m\), moving at speed \(v\), the kinetic energy is given by the formula \(\frac12 \delta mv^2\), and the kinetic energy of any extended object can be built up from this. If a moving object collides with another object, and there is no dissipation, (no heat is given off, no chemical bonds are rearranged, etc.) then the total kinetic energy of the objects after the collision must equal the kinetic energy of the moving object before the collision.

In motion along inclined planes, we observed that when an object is accelerated down a plane by gravity, the product of the force of gravity along the plane, times the distance traveled, is equal to the quantity \(\frac12 mv_f^2 - \frac12 mv_i^2\). In fact, this quantity is the kinetic energy given to the block during its acceleration, and the mathematical form is the general expression for the kinetic energy of a point particle.

Translational kinetic energy is the kinetic energy due to an object's trajectory. In terms of a particle's mass and speed, the translational kinetic energy is given by \(\frac12 m\vec{v}^2\). In terms of its momentum \(\vec{p}=m\vec{v}\), the kinetic energy is \(\displaystyle \frac{\vec{p}^2}{2m}\).

Rotational kinetic energy is the kinetic energy associated with the rotation of an object about its center of mass: \(KE_\text{rotational} = \dfrac 1 2 I \omega^2 ,\) where \(I\) is the moment of inertia of the rotating body. A point particle cannot have rotational kinetic energy.

One thing that must be taken into consideration for engineers designing the Space Shuttle orbiter is the deceleration of the orbiter as it enters the earth's atmosphere. The orbital speed of the space shuttle is about \(8\text{ km/s},\) whereas when it finally comes to rest in the hangar, it is stationary. Now, if the weight of the orbiter is \(70,000\text{ kg},\) how much energy does the orbiter shed during its slowdown?

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The translational kinetic energy of the shuttle in orbit is equal to \[\begin{align}\textrm{KE}_\textrm{orbiter} &= \frac12 m_\textrm{orbiter}v_\textrm{orbiter}^2 \\ &= \frac12 \times 7\times10^4\textrm{ kg } \times \left(8\times 10^3\textrm{ m/s}\right)^2 \\ &= 2.2\times10^9\textrm{ kJ},\end{align}\] which is quite a bit. As it decelerates, this energy will go into many things such as pushing the gas of the atmosphere out of the way, forming shock waves, even the tires and ground upon landing. However, the vast majority of the energy is lost to the formation of a shock wave at the front of the flight path. The matter in the shock wave is made extremely hot by the high pressure, and some of this heat goes into heating the orbiter.

To compare, the orbital energy of the shuttle is about the same amount of energy that could be harvested from all the sunlight that shines on the entire state of Arizona in a day, and about a thirtieth as much energy as was released by the atomic bomb dropped on Hiroshima.

Keeping the orbiter cool as it comes to Earth is an important problem in keeping the crew safe.

On the motion along inclined planes page we pointed out a connection between the distance that an object travels down an inclined plane under the force of gravity, and the velocity that it has at the end. Because it is such a profitable example, we return to the calculation.

A coincidence for an open mind

Under constant acceleration, an object travels the distance \(d\) in the time \(t = \sqrt{\frac{2d}{a}}\). For the skier on the inclined plane, this time is given by \(\sqrt{\frac{2d}{g\sin\theta}}\). Starting from rest at the top of the incline, the skier has speed \(v = at = g\sin\theta \sqrt{\frac{2d}{g\sin\theta}} = \sqrt{2dg\sin\theta}\) at the bottom of the incline.

Consider the kinetic energy \(\frac12 mv^2\). At the top of the incline, this quantity is equal to zero, and at the bottom of the incline it's equal to \(dmg\sin\theta \).

Now, consider another quantity \(F\cdot s\), where \(F\) is the force of gravity along the incline, and \(s\) is the distance traveled down the incline. With \(F = mg\sin\theta\) and \(s=d\), we find \(F\cdot s = dmg\sin\theta\).

This is curious. In descending the slope, the skier gained an amount of kinetic energy that is equal to the product of the force that acted on the particle, and the distance over which the particle traveled. One might be tempted to hypothesize that forces acting through a distance give objects kinetic energy.

Mathematically,

\[\vec{F}\cdot\vec{d} = \Delta \textrm{KE} = \frac12mv_f^2-\frac12mv_i^2.\]

This suggests that the cumulative pull of the skier by gravity has given rise to the kinetic energy of the particle. This relationship would be quite useful if it holds in general, but as yet, it is just a nice coincidence we've noticed in our calculation. Next we look to Newton's laws for a way to put it on firm ground.

Newton's laws: A basis for work

According to the second law, we have the following relationship between changes in velocity and a net applied force (for simplicity we work in one dimension, though the result easily generalizes)

\[\begin{align}F_\textrm{net}&=ma\\ &=m\frac{\Delta v}{\Delta t}\end{align}\]

Rearranging, we have

\[\begin{align} F_\textrm{net} \Delta t &= m \Delta v \\ &= \Delta p\end{align}\]

Because \(\Delta x = v \Delta t\), we can write \(F_\textrm{net} \Delta x / v = m \Delta v\), or

\[\boxed{F_\textrm{net} \Delta x = m v\Delta v}.\]

This relation shows that if the object is traveling with velocity \(v\) and it is pushed through some small distance \(\Delta x\) parallel to the force \(F_\textrm{net}\), it will pick up the additional velocity \(\Delta v = \frac{F_\textrm{net}\Delta x}{mv}\).

In three dimensions, our result is \(\vec{F}_\textrm{net}\cdot\Delta\vec{x} = m\vec{v}\cdot\Delta\vec{v}\).

This relation provides the basis for what we suspected above, that forces can do work to endow particles with kinetic energy. We'll now exploit the relation to prove the work-kinetic energy theorem.

Work-kinetic energy theorem

The work done by the net force on an object is equal to its change in kinetic energy.

If we add up all of the incremental pushes \(F_\textrm{net}\Delta x\) that the particle receives over the distance \(d\), we get \(F_\textrm{net}\sum \Delta x = F_\textrm{net}d\).

However, we showed that \(F_\textrm{net}\Delta x = mv\Delta v\), so that we also have \(F_\textrm{net}d = m\sum v\Delta v\), a sum over the incremental increases in velocity.

Let us now perform the sum of the \(mv\Delta v\).

To start, when \(v\) is zero, \(mv\) is zero; at the end, it is equal to \(mv_f \). If we divide the increases up into \(n\) little pieces, the velocity increases \(\Delta v\) are each given by \(\frac{1}{n}v_f\), we can pull them out of the sum so that the sum becomes \(\frac{v_f}{n}\sum v\).

Now, the sum of \(v\) from \(0\) to \(v_f\) in \(n\) equally sized chunks is simply \(v\) times the average value of \(v\) over the range: \[\sum\limits_{v=0}^{v_f} v= n\frac{v_f}{2}\]

Therefore, \(\sum mv\Delta v\) is equal to \(\displaystyle \frac{mv_f}{n}\sum\limits_{v=0}^{v_f}\Delta v = \frac{mv_f}{n}n\frac{v_f}{2} = \frac12 mv_f^2\)

And we have \(F_\textrm{net}d=\frac12 mv_f^2\)

This proves that if we act on an object of mass \(m\), with a force \(F_\textrm{net}\), over a distance \(d\) it ends up with a kinetic energy \(\frac12 mv_f^2\), where the velocity \(v_f\) is given by \(\sqrt{2F_\textrm{net}d/m}\).

Hence, we have shown that a force acting on an object through a distance transfers kinetic energy to the object, and we call this quantity, \(W_\textrm{net} = \vec{F}\cdot \vec{d}\), the work. In a frictionless system, the work is equal to the change in kinetic energy caused by the force: \[W_\textrm{net} = \Delta \left(\frac12 mv^2\right)\]

Here's another way of proving it:

We know \[W = \int \vec F \cdot d\vec s.\] Since \(\vec F =m \vec a=m\dfrac{d\vec v}{dt},\) it follows that \[\begin{align} W &= \int \vec F \cdot d\vec s \\ &=\int m\dfrac{d\vec v}{dt}\cdot d\vec s =\int\limits_{\vec {v_0}}^{\vec{v_f}}m\dfrac{d\vec s}{dt} d\vec v \\ &=m\int\limits_{\vec {v_0}}^{\vec{v_f}} \vec v \cdot d\vec v =\Delta \textrm{KE}, \end{align}\] which proves \(\Delta \textrm{KE} = W.\)

We just showed that a net force can perform net work on an object to increase its kinetic energy. This happens all the time, e.g. when a steam engine drives tugboat paddles, or when a bicyclist turns the cranks to speed up. Sometimes, we want to do work now so that we can spend energy later, e.g. it would be convenient if the electrical energy made through water flowing over a dam could be stored until the customers want to use it, and it is crucial that the energy in jet fuel is not released until it's ignited in the engine. When work is accumulated, the stored energy is called potential energy.

Thought of in this way, potential energy is a currency that can be used to purchase processes (e.g. pulling an elevator to the top floor), transformations (e.g. melting ice to water), and other physical phenomena. When people live paycheck to paycheck, such that any money they make is immediately dispensed to pay for their immediate needs, they're quite limited in the things they can do. However, when someone can put money aside each paycheck toward a savings account, new possibilities open up to them, e.g. vacations, buying a car, a subscription to Brilliant\(^2\), etc.

Similarly, building up potential energy over time, rather than immediately converting it to kinetic energy, enables all kinds of physical phenomena that would not otherwise be possible. Some examples include volcanic eruptions (quick releases of accumulated pressure), heartbeats (fast relaxation of ionic gradients across cells), and survival during starvation (liberation of energy from chemical bonds in glycogen, which cells use to store excess glucose).

Thus, we can hypothesize a generalization to the work-kinetic energy theorem in which work can result in either motion (kinetic energy) or future motion (potential energy).

Work-energy theorem

Work done on a particle (in the absence of friction) results in kinetic energy, and/or stored potential energy. \[\Delta W = \Delta \textrm{K.E.} + \Delta \text{P.E.}\]

Gravitational potential energy

A mass-less elevator carries two passengers from the first floor to the twentieth floor of the building. Each floor is \(d=2\) m high, and each person weighs \(m_p = 100\) kg. What is the net work done on the people? What is their increase in potential energy?

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The external force of gravity pulls the passengers down with force \(\vec{F}_g = -2m_pg\hat{z}\). To lift them to the top floor at constant velocity, the elevator pushes them up with equal and opposite force \(\vec{F}_e = -\vec{F}_g = 2m_pg\hat{z}\).

Hence, while the elevator performs \(2\times m_p g \times 19 d\) J of work, gravity performs \(-2\times m_p g \times 19 d\) J.

If we were to consider the whole universe as our system, there is no net work performed on the passengers, because the force from the elevator balances the force of gravity. However, the elevator has done work on the people against the external force of gravity. The idea is that as soon as the internal force stops acting (the elevator ceases to support their weight), the external gravity of the Earth will do work on the people so that they obtain kinetic energy equal in magnitude to the work done by the elevator to bring them to the twentieth floor.

Thus, if we restrict our system to include the people and the elevator, but not Earth, then for all intents and purposes, it is truly as if the elevator gives them kinetic energy in the future. This store of energy is called potential energy.

As the last example shows, we can partition the universe into our system of interest \(\Gamma_S\), and the rest of the universe \(\Gamma_R\), and we can identify work by objects in \(\Gamma_S\) against external forces from \(\Gamma_R\) as creating potential energy in \(\Gamma_S\).

We can see that the gain in potential energy is independent of how the passengers made it up to the twentieth floor, i.e. whether they walked up a staircase, were lifted by the elevator, or flew there with a jetpack, they still end up with the same amount of gravitational potential energy \(mg\Delta h\) which is now available to spend. Moreover, we see that any object held at height \(\Delta h\) above Earth has the same potential energy per unit weight.

In this way, we can envision a series of levels above the Earth between which objects can move. If an object descends from a higher level to a lower level, it spends some potential energy for kinetic energy. Likewise, to go up a level, work must be performed to buy the extra potential energy. To move an object within the same level (at constant velocity), i.e. move laterally without changing our height above the ground, that can be done without spending any energy.

A skydiver is brought up to the red ring in the diagram above by a helicopter. They jump out (with zero velocity) and don't pull their chute until they cross the yellow ring. If \(h_0 = 4.25\) km, what is their velocity when they pull the chute? Assume no friction with the atmosphere.

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On the red ring, the skydiver's potential energy is \(3mgh_0\) and on the yellow ring it is equal to \(mgh_0\). Thus, in transit, the skydiver's potential energy is reduced by \(2mgh_0\).

If we consider Earth outside of our system \(\Gamma_S\), we have \(\Delta \textrm{K.E.} = -\Delta\textrm{P.E.}\), hence \(\frac12 m v_f^2 = 2mgh_0\),

\[\begin{align} v_f &= \sqrt{4gh_0} \\ &= \sqrt{4\times10\textrm{ m/s}^2\times4.25\times10^3\textrm{ m}} \\ & \approx 408\textrm{ m/s} \end{align}\]

If we consider Earth inside of our system, we have \(\Delta W = \Delta \frac12 mv^2 = Fd = mg\Delta h\), yielding the same answer. Thus, we see that potential energy amounts to a bookkeeping device for the skydiver's position in the gravitational field. Whether we calculate from the perspective that the skydiver has potential energy, or from the perspective that the Earth does work on him in descent, we get to the same answer. It is often much simpler to work from the energy perspective.

Note, this speed is much faster than the terminal velocity of a human in Earth's atmosphere, but is the speed one would have in the absence of atmosphere. For example, Felix Baumgartner achieved a top speed of 377.1 m/s during his skydive from the edge of space.