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Honors Physics
Work and Energy
Objectives
You should be able to:
A. define and calculate work done by a constant force parallel to the displacement;
When you push on something and get it moving, it would be nice to be able to say,
quantitatively, how much of a push (“oomph”) you give to the object. Certainly, the magnitude
of the force exerted is important, but that's not enough...
If you take the force and multiply it by the length of time you exert that force, you get a
measure of the amount of push called the impulse. This is intimately related to momentum
and may be covered in the future.
In this unit we'll consider that measure of the amount of push that you get when you
multiply force by the distance the object moves while you're pushing it. This product is called
the work. Example: An object is moved 5.00 m by a force (pushing in the direction of motion;
that's important) of 7.00 N. The work done by this force is 35.0 N.m or 35.0 joules or 35.0 J.
(This energy unit is named after James P. Joule, an Englishman who made very precise energy
measurements and showed that heat is a form of energy.)
B. calculate work done by a constant force not parallel to the displacement;
For simplicity, let's say the force is constant, just not parallel to that path. Just multiply the
displacement by the vector by the parallel component of the force. In other words, multiply
the absolute value of the force times the absolute value of the distance times the cosine of the
smallest angle between them.
W = FDcos(Ө)
Work an example where you calculate the work done by each
of several forces parallel, antiparallel, oblique, and
perpendicular to the motion -- such as a block sliding down a
ramp (ooh how exciting!). Calculate each of the works
separately, then add them and use the "work-energy theorem" (see below) to find the change
in speed.
C. define kinetic energy and apply the work-energy theorem;
Kinetic energy (K) is energy due to the translational motion of an object and depends on
1
the mass and speed of the object. It is expressed mathematically as: 𝐾 = 2 𝑚𝑣 2
Suppose 60 J of work is done on an object. Does this allow us to say anything about
changes in the object's motion? Yes! There is a relationship between kinetic energy and
work, known as work-energy theorem. It defines the change in the object's kinetic energy as
the work done by each force acting on it.
Consider an object moving with speed v1 in the x-direction, acted on by a force F in the xdirection for a distance . Call the resulting new speed v2.
v
x
v v
W  Fx  max  m
x  m
v  mvavg v  m 1 2 v2  v1   12 mv22  v12  . Thus the work
t
t
2
2
1
is equal to the change in the quantity 2 mv (kinetic energy): 𝑊 = 𝐾𝑓 − 𝐾𝑖
Example: A 500 gram object moving at 12 m/s is acted on by a 5.0 N force in the same
direction for a distance of 38 m. What is the object's final speed? [Ans: 30 m/s]
D. define potential energy;
Potential energy is stored energy which a system has by virtue of its configuration or
position. It is called potential energy because the system has the potential for a force to do
work, thus converting the potential energy to kinetic energy. What force? Potential energy is
always defined with respect to some particular force; so there is gravitational potential
energy, electrical potential energy, potential energy of a spring, etc. Examples: lift a rock up in
the air, stretch a spring, separate two opposite electric charges. We’ll focus on gravitational
potential energy in this unit, but will work with other forms of potential energies later.
E. calculate work and potential energy changes for the gravitational force;
Consider gravitational force on a mass m near the Earth's surface, Fg = mg. The work done
by the force of gravity as the mass falls from a height h1 to h2 is:
𝑊𝑔 = 𝐹𝐷𝑐𝑜𝑠(0°) = 𝑚𝑔(ℎ1 − ℎ2 ) = −𝑚𝑔(ℎ2 − ℎ1 ) = −𝑚𝑔𝛥ℎ
The gravitational potential energy is 𝑃 = 𝑚𝑔ℎ where h is the height above a reference
line (usually the ground), and the work done by the force of gravity is equal to the negative
change in potential energy: 𝑊𝑔 = −(𝑃2 − 𝑃1 )
As an example, consider a 1 kg object sitting on a table. The mass is lifted one meter and
then dropped. What are the energy changes that occur? Note that the potential energy has
been defined to be zero on the tabletop. One can arbitrarily set the potential energy equal to
zero anywhere one likes. This is because it is only changes in PE that can be calculated or
measured. So the reference level can be anywhere convenient. To illustrate, let's re-label the
above drawing, with a different zero of gravitational potential energy:
F. Distinguish between conservative and non-conservative forces;
If the work done by a force doesn’t depend on the path taken by an object, the force is
considered to be conservative. An example is the gravitational force. In calculating the work
done, the initial and final heights are the only consideration; disregarding the path taken
between those two positions.
Further, for a conservative force, the net work done over a closed path is zero. See the
example in the objective F. The net work is zero and the change in total energy is zero for the
closed path. Also, note that when a conservative force acts on a moving object, the potential
energy associated with that force changes. As the object is moved upwards its gravitational
potential energy increases. And as the object moves downward its gravitational PE decreases.
Hence, the net work for a closed path is zero.
The kinetic friction force, however, is non-conservative. The length of the path over which
the frictional force acts affects how much work is done by friction. The longer the path, the
more interaction between the two surfaces, results in greater work done.
Further, over a closed path the net work done by friction would NOT be zero. The kinetic
frictional force always acts opposite the object’s displacement resulting in a total negative net
work. Moreover, there is no such thing as potential energy associated with a nonconservative
force.
To illustrate: 1) Toss an object straight up. It spontaneously comes back down! That's
because at the higher position it had gravitational potential energy, which let the force of
gravity do work on the object, increasing its KE and decreasing its PE. However, 2) slide an
object horizontally across a table. It doesn't spontaneously come back! That's because the
friction between table and object always does negative work (why?) and so can only decrease
the object's KE.
G. Conservation of mechanical energy
“The total mechanical energy (E = K + P) of an object remains constant as the object moves,
provided that the net work done by the external nonconservative forces is zero.”
Mathematically, this statement can be expressed as: 𝑃𝑖 + 𝐾𝑖 = 𝑃𝑓 + 𝐾𝑓
If non-conservative forces are acting, then the total mechanical energy (E = K + P) does not
remain constant. The work done by the non-conservative force would be equal to the change
in the mechanical energy, most likely a reduction. Mathematically this would be expressed as:
𝑃𝑖 + 𝐾𝑖 + 𝑊𝑛𝑐 = 𝑃𝑓 + 𝐾𝑓
Example: A 950kg car slows from 30m/s to 10m/s over a distance of 100m. The magnitude
of the frictional force can be determined from conservation of energy.
𝑃𝑖 + 𝐾𝑖 + 𝑊𝑛𝑐 = 𝑃𝑓 + 𝐾𝑓
1
1
𝑚𝑔ℎ𝑖 + 𝑚𝑣𝑖 2 + 𝑊𝑛𝑐 = 𝑚𝑔ℎ𝑓 + 𝑚𝑣𝑓 2
2
2
(950𝑘𝑔) (9.8
𝑚
1
𝑚 2
𝑚
1
𝑚 2
(0𝑚)
(950𝑘𝑔)
(0𝑚)
(950𝑘𝑔)
)
+
(30
)
+
𝑊
=
(950𝑘𝑔)
(9.8
)
+
(10
)
𝑛𝑐
𝑠2
2
𝑠
𝑠2
2
𝑠
Wnc = 3.8x105
Wnc = -FD = -F(100m), so F = 3.8x103
The car had a change in mechanical energy of -3.8x105 Joules as it was slowed down. This
loss of energy is equal to the work done by friction.
H. define and calculate power
Power is the rate at which work is done or the rate at which energy is transferred or
𝑊
transformed. 𝑃 = 𝛥𝑡 (note, this is the average power). An interesting and useful alternative
𝑊
𝐹𝐷
for calculating power is 𝑃 = 𝛥𝑡 = 𝛥𝑡 = 𝐹𝑣. Power is equal to the force times the average
speed. You may wish to distinguish between Power and Potential Energy by notating Potential
Energy as “PE” instead of “P”.
Often, the amount of work done is not as interesting as the rate at which the work is done.
Imagine running up a flight of stairs. You can run up the stairs at steady pace or leisurely
climb the stairs. In both cases, the change in your gravitational potential energy is the same
meaning the work you did was the same. However, your power would be different. In the fast
ascent you did the work in a shorter amount of time making your power greater than the time
consuming leisurely climb.
The S.I. unit of power is the watt (named after the steam engine developer James Watt). A
watt is equal to a joule per second and can be further reduced to fundamental S.I. units:
kg  m
m
2
J N m
kg  m 2
W (watt)  
 s

s
s
s
s3
The BE system unit of power is the horsepower (hp). A horsepower is equal to the foot pound
ft  pound
per second. hp(horsepower)  1
s
 the horsepower.
power compared to

 745.7watts As you can see, the watt is a small unit of
I. (Enrichment) define and calculate work done by a force that varies with position or is
not parallel to the displacement;
First, how do you calculate work done by a force parallel to
displacement, but which varies with position? An example of this
might be the work done in stretching ("cocking"?) a bow (as in
"bow and arrow"). A compound bow could exert a force, F, which
depends on the displacement, x, as in the first graph at right. If the
force were the same everywhere, it would be easy to calculate the
work by just multiplying the one force value by the displacement,
say, W = (136N)(0.45m) = 61.20J. Note that this product can be
represented on the second graph by the shaded area. In the case of
a position-varying force, it is still true that the work is given by the area between the force-vs.position curve and the position axis, as shown in the third graph.
If you knew the force as a mathematical function of position, you could determine the work
by integrating the force over position (if you knew calculus). Failing that, you would estimate
the area under the curve, possibly as follows: 1 square = (0.05 m)(20 N) = 1.0 J. Counting the
number of squares in the third graph yields an estimate for the work of roughly 60 J. (This
can be done with varying degrees of precision.)