\begin{align}
\mbox{System of Blocks} &= \mbox{[the child's room]} \\[1ex]
\mbox{Number of Blocks}= N_B = &
{\mbox{number of}\choose\mbox{blocks seen}} +
\frac{\mbox{(weight of box)}-\mbox{16 ounces}}{\mbox{(4 ounces)}} \\[1ex]
& + \frac{\mbox{(height of water)}-\mbox{(6 inches)}}{\mbox{(1/4 inch)}} + \cdots\\[3ex]
{\mbox{change in}\choose\mbox{number of blocks}}=\Delta N_B = & N_{B,\mbox{evening}}
- N_{B,\mbox{morning}} \qquad\Leftarrow\mbox{Conservation of Blocks}\\[1ex]
& \mbox{[Blocks enter and leave via the door or window.]}
\end{align}
in nearly complete analogy with what we do with energy
\begin{align}
\mbox{Energy System} &= \mbox{[well-defined but arbitrary and up to you]} \\[1ex]
\mbox{Energy} &= E = \tfrac12\,mv^2 + mgh + \tfrac12\,ks^2 + \cdots + U_{\mbox{int}}+\cdots \\[2ex]
\mbox{Work} &= \Delta E = E_\mbox{final}-E_\mbox{initial} \qquad\Leftarrow\mbox{Conservation of Energy}\\[1ex]
& \mbox{[Energy enters or leaves via work done by external agents.]}
\end{align}
The analogy is not exact because there are no blobs of
energy, they don't exist. There is no term analogous to "(number of
blocks seen)" in the context of energy. This means that energy is
actually a fiction, a made up thing, energy is a construct of the
human mind.
You will have correct insight into exactly what energy is ... if you listen to and
understand the Feynman explanation above.
The Joule
The SI unit of energy is the Joule (named after one of the early inventors of energy):
\[ \fbox{$ 1\,\mbox{Joule} = 1\,\mbox{J} = 1\,\mbox{kg (m/s)$^2$} = 1\,\mbox{N m} $} \]
Energy System
The energy system is a region of space defined by an imaginary boundary. The energy system
should be defined carefully so that it includes
what we want it to include. [The energy system is analogous to the child's room.]
Kinetic Energy
The kinetic energy of a point mass $m$ moving with velocity
$\vec{v}$ is defined to be
\[ \fbox{${K} = \tfrac12\,m\,v^2 $}\]
which depends only on the object's mass and its speed. This is energy of motion.
If there are several masses, all moving, the total kinetic energy is the
algebraic sum of the individual kinetic energies.
We invent that formula consistent with other facts.
It is used by looking in the
energy system, measuring system properties (e.g., masses and velocities), and then computing
a number we call kinetic energy.
[This is analogous to one term in the mom's formula, like the toy box term.]
Suppose at a given instant, five masses are observed to be moving as follows:
At this point, we need not consider any forces between these
masses.
We can
define the energy system by including some masses and [possibly]
not others:
Here, I define the energy system to include only particles 1,
3, and 5. As the masses move, this energy system boundary must
adjust as needed to always include only these three masses.
An energy system always has internal agents (the
included masses) and external agents (the excluded
masses).
At this early stage, we only know about kinetic energy. The total
kinetic energy of this defined energy system is
\[ K=\tfrac12\,m_1\,v_1^2 +\tfrac12\,m_3\,v_3^2 +\tfrac12\,m_5\,v_5^2 \]
which does not include the kinetic energy of the excluded masses.
Notice that there is no actual physically tangible boundary,
the energy system boundary is an imaginary boundary enclosing
the system of interest. It is a construction of the human mind.
Also notice that the formula for the kinetic energy will never give a
negative number, but either zero (if no objects in the energy system are moving) or positive
(if any of the masses in the energy system are moving).
A more complicated example will involve other things. Suppose we
drop a ball on a spring (diving board).
The first picture is an initial situation, just after the mass is released:
The second picture shows the mass in the lowest position, bending the
spring and momentarily at rest:
Between the two events shown above, the mass simply moved straight down,
first accelerating downward, then accelerating upward as it comes to a stop.
The details of what happens is not important when we consider energy.
[This is analogous to the way the child can play with his blocks all day long,
but his mom looks in at only two times, and it doesn't matter how or when
the blocks got in the tub or toy box.]
The energy system will include everything relevant to the problem,
the dropped mass, the spring (diving board), the Velcro, and the
mass of Earth. It involves everything that exerts forces on each other,
or are moving (at some time in between the two events shown).
The purpose of the Velcro is to make the mass stick to the diving
board when it strikes the diving board.
The energy at each of the two events shown does not involve any
kinetic energy. This is a simplifying fact that makes these two
events special. There are other kinds of energy, but, because
there are no external agents, there is no work (see below).
Work
Work has a very specific definition in the context of physics.
It is the amount of energy added to or removed from the energy
system by an external agent, and it can do so only by exerting a
force on something moving within the energy system. We have no other way of
changing the energy of the energy system except via work done by an external agent
exerting a force on something moving within the energy system.
If the external force $\vec{F}$ is constant, and the displacement of the
thing moving that it acts on is also a straight line $\vec{d}$, the work
the external agent does is
\[ \mbox{Work} = \vec{F}\cdot \vec{d} = F\,d\,\cos(\theta) \]
where "$\cdot$" represents the "dot product" and $\theta$ is the angle
between $\vec{F}$ and $\vec{d}$.
The SI unit of work is the Joule, the same as that of energy. $1\,\mbox{J}=1\,\mbox{N m}$.
Encoded in this formula is the possibility that work is positive, negative, or zero.
$F$ and $d$ are never negative, but even if they are not zero, the work can be
zero if $\theta=90^\circ$, negative if $90^\circ\lt\theta\le 180^\circ$,
or positive if $0^\circ \le\theta \lt 90^\circ$.
If work is positive,
energy is added to the energy system. ($W=3\,\mbox{J}$ means three Joules of energy
are added to the energy system.)
If work is negative,
energy is removed from the energy system. ($W=-3\,\mbox{J}$ means three Joules of energy
are removed from the energy system.)
[In the analogy, the child's friend
bringing blocks over corresponds to positive work, throwing blocks out the
window corresponds to negative work.]
The next diagram gives, by example, situations where work is
negative, positive, or zero. Here $F$ is the external force and
$D$ is the displacement of the object that force acts on.
Quite often, introductory problems ask you to calculate the work
done without going through the process of defining an energy system.
In that case, if something does work, that something is assumed to be
external to the energy system (as it must be).
The problem is quite often also worded as, "how much work
does a force do?", but there is always an external agent
exerting that force and it is more pedantically correct to ask
"how much work does the external agent do?"
Keep these subtleties in mind as you do problems.
A 3-kg ball is dropped from rest through a distance of two meters. Disregard
effects due to the air, and assume $g=10\,\mbox{m/s}^2$.
Two ways to ask for the same task (work done on the ball):
Compute the work done by the Mass of the Earth.
Compute the work done by the force of gravity.
Here is the familiar diagram:
The displacement of the ball $\vec{d}$ is two meters straight down,
and the relevant force is the ball's weight,
also straight down and magnitude $30\,\mbox{N}$.
The work is computed by first drawing the displacement and force arrows
with their tails together (or slightly displaced if they point in the
same direction), then observing the angle and plugging into the formula,
\[ \mbox{Work} = W\,d\,\cos(\theta) = mg\,d\,\cos(0^\circ)=90\,\mbox{J} \]
where in this example, $\theta=0^\circ$.
A source of confusion is that work and weight both use the
letter $W$. To avoid that confusion, either write out the
word "Work" or use "$mg$" or "$F_g$" for weight, instead.
In this problem, consider a box of mass $M$ sliding down an 10-degree
incline through a distance of $D$. The coefficient of kinetic friction
between the bottom of the box and the incline surface is $\mu_k$.
Disregard effects due to the air and assume $g=10\,\mbox{m/s}^2$.
Here, we ask,
What is the work done by the frictional force (...by the incline)?
What is the work done by the force of gravity (...by the mass of Earth)?
Since there are only two things in the environment that exert a force on the
sliding box, these are the only two possibilities for work.
We need the free body diagram for the box.
To the right of the FBD is a force diagram showing the relationship
between forces, and to the right of that are the diagrams needed to compute the
works.
Since this problem doesn't specify acceleration, only that it moves through
a distance $D$, down the incline, we can't say anything about the relation
between the friction force $f_k$ and $W\sin(10^\circ)$.
However, since there is no acceleration away from the incline, we can
say that $N=mg\cos(10^\circ)$. This will help us compute the friction force
$f_k=\mu_kN=\mu_k\,mg\,\cos(10^\circ)$ in terms of things that are usually
given.
Putting these facts together,
\[\mbox{Work done by friction} = f_k\, D \,\cos(180^\circ) = -\mu_k\,\left[mg\,(0.985)\right]\,D\]
\[\mbox{Work done by gravity} = W\, D \,\cos(80^\circ) = +mg\,D\,(0.174) \]
With numerical values for $\mu_k$, $m$, and $D$, it would be possible to
compute final numerical answers.
Note how the friction does a negative amount of work, decreasing the energy
of the energy system.
Note how the gravity does a positive amount of work, increasing the energy of the
energy system.
To be pedantic, the work is really done by the incline and the mass of Earth,
but I'll be imprecise like everyone else.
Also, to be pedantic, I can actually define the energy system to be everything except
the incline and the mass of Earth, so it really only includes the box. This
way, I can speak of external agents (incline and ME) actually doing
work. But, again, introductory problems on work usually disregard this subtlety.
Energy
There are many types of energy besides kinetic energy.
To determine how much energy is in the system at a given time,
one looks in the energy system at the system properties, and
plugs those system properties into formulas to tally up the
total energy.
[In the analogy, the child's mom enters the room, measures the
weight of the toy box, depth of the bath water, and so on; then
plugs those results into the formula to determine the total
number of blocks in the room.]
Things do change as time marches on, but often we are not
concerned about the details of those changes, we are only
concerned about the situation at a given instant, how much
energy is in the system at a given event.
The formulas for the energy are defined in the context of
this construct, not derived. In the end, they must provide
answers to questions consistent with Newton's laws.
A summary of the types of energy and the generic formulas we encounter in this chapter is:
kinetic energy: Energy of motion (here only linear, non-rotating motion),
\[K=\tfrac12\,mv^2 \]
where $m$ is the mass and $v$ is its speed;
gravitational potential energy: Energy related to the force of gravity and
the relative position of mass near the ground,
\[ U_g=mgh \]
where $m$ is the mass, $g$ is the gravitational acceleration, and $h$ is some elevation
above a reference elevation;
elastic potential energy: Energy related to the elastic force of springs and
their deflection.
\[ U_s = \tfrac12\,k\,(\Delta x)^2 \]
where $k$ is the spring constant (how many Newtons does it take to stretch or compress
the spring by one meter), and $\Delta x$ is the amount the spring is deflected from
equilibrium. (The subscript s represents spring.)
thermal energy (heat): Energy that remains in the system as thermal
energy because of the kinetic friction force (rubbing) between two surfaces.
\[ U_\mbox{int} = \mbox{internal energy} \]
where there is no simple formula. We usually consider only the change in internal
energy in real problems,
\[ \Delta U_\mbox{int} = U_\mbox{int}\mbox{(final)} - U_\mbox{int}\mbox{(initial)} \]
Given the above limited set of possible energies, at a given instant,
the total energy in the system is the sum,
\[ E_\mbox{tot} = K + U_g + U_s + U_\mbox{int} \]
Work-Energy Bar Charts
At a given instant, or event, energy computations turn into a sort
of accounting problem (like counting the blocks in the child's
room). To develop a visual picture, we find that a [qualitative]
bar chart is quite useful to represent the various energies in a
system:
In that diagram, hints of where the bars can be placed are given.
The kinetic energy $K$ is never negative and the gravitational potential
energy $U_g$ can be zero, positive, or negative.
If there is some energy of the various kinds at a given
instant, the bar chart might look like:
Here, the total energy is computed to be $E=(3)+(-2)+(4)+(0)=5\,\mbox{J}$,
but it may simply be a qualitative representation showing relative
amounts and signs, or if something is zero.
Potential Energy and Conservative Forces
If the force is conservative, we can introduce a potential
energy. A force is conservative if a video recording of some
process looks natural when played in reverse. Two examples:
The gravitational force is conservative: If I video record a
ball tossed in the air, and play it backward, it looks
natural. To the gravitational force, we can associate a
gravitational potential energy, $U_g$.
The kinetic friction force is not conservative: If I video record a
box sliding across the floor, slowing to a stop because of
friction, and then play it backwards, it will not look natural
as the box spontaneously slides back with increasing speed. To
the friction force, we cannot associate a potential
energy. Instead, since the surfaces rub together, they warm
slightly and the energy goes into thermal energy,
$U_\mbox{int}$. (Ultimately, thermal energy is a form of
kinetic energy because of the increased kinetic energy
of the atoms and molecules in the heated material.)
The relationship between conservative forces and potential energy is
the reason why potential energy is sometimes invented ($U_g$ and $U_s$)
and sometimes not ($U_\mbox{int}$).
Conservation of Energy
The reason the energy construct is so useful is that the energy
of a closed system will not change. Energy is conserved.
The only way the energy of the system can change is if an
external agent does work, and the only way that happens is
if it reaches in through the energy boundary and exerts a
force on something moving within the energy system.
This means that we can look at the energy of an
energy system at two different times, and then write
\[\fbox{$\displaystyle \mbox{Initial Energy} + \mbox{Work} = \mbox{Final Energy} + \mbox{Internal Energy Change} $}\]
Note that $\Delta U_\mbox{int}$ (Internal Energy Change) is put on the right-hand side of
the equation. This change in internal energy represents the
"loss" of mechanical energy, usually to thermal energy, but also
to other non-mechanical forms of energy like sound, light, or
chemical energy.
This equation for energy conservation represents a very
abstract construct and I find that the bar chart can be a very
useful visual representation.
Work-Energy Bar Charts and Conservation of Energy
The bar chart can be used to represent the amount of energy
at two different times, but we must also include the possibility that
the energy will increase or decrease by including work in the menu of
possible items.
There are some hints in the bar chart. For example,
kinetic energy and elastic potential energy are never
negative, but gravitational potential energy, work, and
internal energy change can be negative, zero, or positive.
An important feature of the bar chart is that it allows
one to set up the problem, realizing which terms are
zero or non-zero, and the relative amounts of each energy.
As we fill up the bar chart with bars, we must be careful
to balance the bars on both sides (taking into account sign).
To see how it all works, let's consider a familiar example.
I will also do this problem twice by using two different
energy systems.
First consider that we only know about kinetic energy and work,
and we haven't learned anything about potential energy, and, as
usual, we will disregard any effects due to the air and assume
$g=10.0\,\mbox{m/s}^2$. We also assume here that we know all of
the kinematic quantities.
We drop a 3.00-kg ball from a height of $h=2.00\,\mbox{m}$. The
two events are (initial) just after the ball is released and
(final) just before it strikes the ground.
In this first pass, the energy system includes only
the ball. The Earth is an external agent that will do work.
We are fortunate to have already solved this problem, so it
should be clear that the initial and final kinetic energies
are:
\[K_i = \tfrac12\,m\,v_0^2 = 0\,\mbox{J}\]
\[K_f = \tfrac12\,m\,v_1^2 =\tfrac12\,(3.00)(6.32)^2= 60.0\,\mbox{J}\]
We also know the mass of the Earth is reaching in through the
system boundary with the gravitational force
$F=W=mg=30.0\,\mbox{N}$, pulling down on the ball as it moves
through a distance of $h=2.00\,\mbox{m}$. The work that $M_E$
does is
\[ \mbox{Work} = F\,h\,\cos(0^\circ) = mg\,h\,(+1) = (3.00)(10.0)(2.00)=60.0\,\mbox{J}\]
Using these facts, and using only kinetic energy and work, the bar chart becomes:
The work is the amount by which the initial [kinetic] energy changes to yield the
final [kinetic] energy. The formula is
\[ K_i + \mbox{Work} = K_f \]
\[ (0) + mg\, h\, \cos(0) = \tfrac12\,m\,v_1^2 \]
\[ (0) + (60) = (60) \]
If we didn't know the numbers, we would merely know that
$K_i=0$ and $K_f>0$ and $W>0$, and so $W=K_f$,
which would allow us to construct a qualitative
bar chart, and (after dividing by $m$):
\[ g\,h = \tfrac12\,v_1^2 \]
If we are given two numerical values, we could find the remaining one.
For example, if $g=10\,\mbox{m/s}^2$ and $D=2\,\mbox{m}$, then we could
compute \[v_1=\sqrt{2gh}=\sqrt{(40)}=6.32\,\mbox{m/s}\]
Now, repeat the problem, but this time assume we know
gravitational potential energy $U_g=mgh$ for when the ball is a
given height or elevation $h$ above or below some reference level
(we'll use the release elevation, two meters above the floor). We
also know about the other forms of energy, but since there is no
spring or friction, those will be zero.
In this second pass, the mass of the Earth is included in the
energy system, so it is no longer doing work.
Even though the mass of the Earth is still pulling down on the
ball as before, the amount of work the Earth does is zero since
it (mass of Earth) is now in the energy system. [I know I wrote
it twice, but it is worth repeating.] We account for the
corresponding energy using $U_g$. Now the work-energy bar
chart looks like:
Note how I carefully annotate the bar chart when an energy is zero, giving
a brief explanation. The purpose of this is to make sure you don't forget
to account for one of the energies, which is easy to do with blank entries.
[In the analogy, the child's mom has to look in all the places
where there might be blocks, she must not forget to consider the
toy box, the tub, etc., and she checks off each place to be sure
she doesn't forget.]
In the initial event, the height relative to the reference level was $h=0\,\mbox{m}$
and in the final situation it is at the floor where $h=2\,\mbox{m}$, but it is down from
that reference level. We can incorporate the minus sign (from the bar chart) in the
term for $mgh$ on for the final situation.
The formula now becomes:
\[ K_i + U_{gi} = K_f + U_{gf} \]
\[ \tfrac12\,m\,(0)^2 + mg\,(0) = \tfrac12\,m\,v_1^2 - mg\,h \]
\[ g\,h = \tfrac12\,v_1^2 \]
and we have the same relationship as in the previous example ... because it was
the same physics problem. The only difference: do we include or not include
the mass of Earth in the energy system?
Recall from kinematics (one-dimensional, constant acceleration),
\begin{align} v_1^2 &= v_0^2 + 2\,a \cdot\left(x_1-x_0\right)^2 \\[1ex]
v_1^2 &= (0)^2 + 2\,g \, h^2 \\[1ex]
g\,h & = \tfrac12\,v_1^2
\end{align}
which is the same result.
Power
is the rate at which energy is converted from one for to another.
The SI unit of energy is the Watt, which is a Joule per second.
\[ P = \frac{\Delta E}{\Delta t} \qquad \qquad [P] = \frac{\mbox{J}}{{s}}=\mbox{W} \]
In all problems related to power, we must compute the rate at which
energy is converted from one form into another.
Efficiency of an energy conversion is related to whether or
not the final forms of energy are desired or not. If a lamp converts
100 Joules of electrical potential energy into 80 Joules of desirable
visible light energy and 20 Joules of undesirable thermal energy, then
the efficiency of the conversion (electricity to light) is
80%. We can also use the language of power: The 100-Watt lightbulb
is 80% efficient. If we now consider the heat (thermal energy) given
of by the bulb to be useful to warm a room, its efficiency would
become 100%.
Example 1: Dropping stone
What is the average rate at which energy is converted from
gravitational potential energy into kinetic energy when we drop a
one-kilogram stone through a distance of one meter? (Disregard
effects due to the air.)
The total change in energy is the change in gravitational potential
energy of the stone:
\[ \Delta E = U_g = mgh = (1.00\,\mbox{kg})(9.80\,\mbox{m/s}^2)(1.00\,\mbox{m})
= 9.80\,\mbox{J} \]
Because there is no friction, all of that energy is converted into kinetic
energy: It is 100% efficient.
The time it takes to fall is
\[ \Delta t = \sqrt{\frac{ 2h}{g}} = \sqrt{\frac{2\,(1.00\,\mbox{m})}{(9.80\,\mbox{m/s}^2)}} = 0.452\,\mbox{s} \]
where I used kinematics, $x=x_0+v_0t+\tfrac12 a t^2\qquad\Rightarrow\qquad
2h=g\cdot(\Delta t)^2$.
Putting these result together, the power is
\[ P = \frac{\Delta E}{\Delta t} = \frac{(9.80\,\mbox{J})}{(0.452\,\mbox{s})} = 21.7\,\mbox{W} \]
Answer: The average rate at which gravitational potential energy is converted into kinetic
energy is 21.7 Joules.
Note: We need the adjective "average" since the rate will change as
the ball falls. To see this, allow the stone to continue falling for
another meter, the change in energy will be the same as above, but the
time it takes will be less. The power would thus be greater in the second
meter. In general, the power increases as the stone falls.
Example 2: Climbing stairs
Suppose I walk up a flight of stairs (vertical height of
$h=2.50\,\mbox{m}$) and it takes me $\Delta t=7.40\,\mbox{s}$. My mass
is $m=80.0\,\mbox{kg}$. At what rate am I converting chemical
potential energy (via biological metabolism) into gravitational
potential energy.
This process is less than 100% efficient: Not all of the energy I
convert from stored chemical potential energy is converted into
gravitational potential energy while climbing the stairs. Compare the
calculated power required to climb the stairs to my basal metabolic
rate (wikipedia) of about 1700 Calories per day, about
$82\,\mbox{W}$. $(1\,\mbox{Calorie}=1000\,\mbox{calories})$
The total change in energy is the change in gravitational potential
energy of me:
\[ \Delta E = mgh = (80.0\,\mbox{kg})(9.80\,\mbox{m/s}^2)(2.50\,\mbox{m})
1960\,\mbox{J} \]
The time it took is given, so the power is,
\[ P = \frac{(1960\,\mbox{J})}{(7.40\,\mbox{s})} = 265\,\mbox{W} \]
Compared to my basal metabolic rate, this is 3.2 times bigger.
Answer: In climbing the stairs, I convert chemical potential energy
into gravitational potential energy at a rate of 265 Watts. This is 3.2
times bigger than my basal metabolic rate of 82 Watts.
Example 3: Charging cell phone
Suppose we charge a cell phone via a typical 12-Watt USB supply
(5VDC/2.4A) for thirty minutes. During that time, how much electrical
potential energy is converted into chemical potential energy (battery "charge")
and thermal energy (the battery gets a little warm). This charging process
is certainly not 100% efficient.
If that same amount of energy were used to lift a one-kilogram mass
($1\,\mbox{liter}$ of water) instead, how high would it be lifted
assuming 100% efficiency?
Note: This problem is a little unrealistic since the actual power
decreases as the battery approaches "full charge".
The total amount of electrical potential energy converted is
\[ \Delta E = P\cdot \Delta t = (12\,\mbox{W})(1800\,\mbox{s})=2.16\times 10^4\,\mbox{J} \]
If that energy is used to lift a one-kilogram mass, it would be
lifted a vertical height of
\[ \Delta E = mgh \qquad\Rightarrow\qquad h=\frac{\Delta E}{mg}
= \frac{(2.16\times 10^4\,\mbox{J})}{(1.00\,\mbox{kg})(9.80\,\mbox{m/s}^2)} = 2200\,\mbox{m} = 1.37\,\mbox{miles}
\]
Twelve Watts will lift a one-kilogram mass a vertical height of 1.37
miles in thirty minutes. The speed is 2.74 MPH. Twelve Watts is
needed to lift a one kilogram mass with a constant speed of
2.74 MPH.
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jones.3 (R Jones) Last Modified 12:37:33 26-Feb-2022
