Download Introductory Lectures on Work and Energy (Note: these lectures will

Introductory Lectures on Work and Energy
(Note: these lectures will be accompanied by illustrations and demonstrations wherever possible).
Day 1:
We begin by introducing work: W = FdcosƟ, where F = force, d = displacement (how far object travels
under influence of force), and Ɵ is the angle between the directions of F and d. Note that if Ɵ is zero
(meaning the force is parallel to the motion), then cosƟ = 1, and the equation reduces to the more
familiar W = Fd (work = force x distance). Also note that if Ɵ = 90° (force perpendicular to motion, as in
the case of circular motion), then cosƟ = 0 and hence work is zero (no work is done by a force
perpendicular to motion). The units of work are Joules: following the equation, Newtons x meters =
Joules. (7 minutes)
Consider an object, initially traveling at velocity v0, being pushed horizontally by a constant force F on a
level frictionless surface. Here Ɵ= 0 (thus W = Fd), and since there is no friction, the force F is the net
force (Fnet) on the object. We can then use two previously-learned equations: Fnet = ma where m =
mass and a = acceleration (Newton’s Second Law); and v 2= v02 + 2ad (one of the kinematic equations
which describes constant acceleration motion). Combining the equations together, we find:
W = Fd = mad = m(v2 – v02)/2
(equation 1)
We now define kinetic energy: KE = mv2/2
For this example the quantity mv02/2 would then refer to the initial kinetic energy, KE0.
Equation 1 then becomes W = KE – KE0 = Δ(KE) (work equals change in kinetic energy). This is
known as the “work-energy theorem”, and is more formally written as Wnet = Δ(KE) where Wnet
is the net work done on an object (work done by all forces acting on the object).
This theorem indicates that work and energy are equivalent quantities. Thus energy has the
same units as work (Joules). Sometimes energy is referred to as “the ability to do work”.
(Note: although simplifications were made in order to arrive at the conclusion
Wnet = Δ(KE), it can be shown with more mathematics that the same conclusion is true for any number
of forces on mass m, applied at any angles: the net work done on an object always equals its change in
kinetic energy). (14 minutes)
Kinetic energy (KE = mv2/2) is often referred to as “energy of motion”, since clearly KE only exists if v is
not zero. Going forward we will see that it is only one of several different types of energy. Consider
now lifting an object from the ground to height h above the ground. In order to do so we would need to
apply a force equals to the object’s weight (mg) where g is the Earth’s gravitational field strength (9.8
meters per second squared), over a distance h. Since the force is parallel to the motion, the work done
by the lifter in this case is simply W =force x distance = mgh .
We now define this quantity as gravitational potential energy (PEg): PEg = mgh.
In examples which do not involve any other kinds of potential energy, we can drop the subscript g and
simply write PE = mgh.
Here h = height measured above some arbitrary reference level. (Note: it may seem strange that this
reference level is arbitrary, implying that PEg can take on virtually any value; however we shall see that,
in any practical application, the important quantity is the change in PEg (ΔPEg = mgΔh), which doesn’t
depend on the reference level, since the change in height Δh doesn’t depend on the reference level).
(5 minutes)
Teacher introduce practice exercise): (1 minute)
Suppose a tabletop is 0.9 meters above the ground. A 1.0-kg book is lifted to a height 0.5 meters above
the table, then dropped. Calculate the change of potential energy of the book (a) using the tabletop as
the reference level, and (b) using the floor as the reference level.
Students work on calculation, consulting classmates if they wish (6 minutes).
Teacher present solution (3 minutes):
(a) At 0.5 meters above the tabletop, the potential energy is
mgh = (1.0 kg)(9.8 m/s2)(0.5 m) = 4.9 Joules
On the tabletop, the potential energy is zero since that was defined as the reference level.
Thus the change of potential energy is 0 – 4.9 Joules = - 4.9 Joules
(b) Using the floor as the reference level, the initial height is 0.9 m + 0.5 m = 1.4 m, so the potential
energy is mgh = (1.0 kg)(9.8 m/s2)(1.4 m) = 13.72 Joules
After the book has fallen to the tabletop, it is 0.9 meters above the floor. Thus the potential
energy is mgh = (1.0 kg)(9.8 m/s2)(0.9 m) = 8.82 Joules
Thus the change in potential energy is 8.82 Joules – 13.72 Joules = - 4.9 Joules,
Which is the same result as obtained in part (a).
So far we have encountered KE and gravitational PE. We shall see that there are numerous other types
of energy as well; however many problems involve only KE and PE. As an example, consider an object of
mass m falling freely from rest from height h above the floor (which we shall use as the reference level).
Before it falls, it is at rest so its KE is zero. At height h its PE is equal to mgh. When it lands, its PE is zero
since it is now at height zero; what about its KE? We can solve for its velocity using the kinematic
equation previously referenced: v2 = v02 + 2ad. Since it starts from rest (v0 = 0), it follows that v2 = 2gh,
since the acceleration is –g and the displacement is –h (note that the negatives cancel out). Then the
kinetic energy is
KE = mv2/2 = m(2gh)/2 = mgh, which is the same as the initial potential energy.
What have we just concluded? At the beginning of the fall, all of the energy is potential. At the bottom,
all of the energy is kinetic. The KE at the bottom has been proven to be equal to the PE at the top. Thus
it is correct to say that the sum of the KE plus PE is the same at the top and bottom (equal to mgh).
Using the same set of equations, it can be shown that the sum of KE and PE is equal to the same value
(mgh) at any point during the fall. (10 minutes)
Finally we define the sum of KE plus PE as Mechanical Energy:
Mechanical Energy = KE + PE
The example just completed is a simple example of the Law of Conservation of Mechanical Energy: In
the absence of friction and air resistance, the sum of kinetic plus potential energy is conserved. (Note:
if there are springs or similar elastic objects in a problem, then there exists another type of potential
energy - elastic PE – which must be included in the conservation law. We shall encounter elastic PE a bit
later).
One may be tempted to ask: what is the good of this law, since we were able to calculate the speed of
the object at the bottom of its fall by other means? If the only problem that this law could be used to
solve were that of freely falling objects, indeed it would not be terribly useful. However, consider also
an object of mass m sliding down a curved frictionless track which begins from rest at height h above the
ground. At the top its KE is zero (since it is motionless), and its PE is mgh. At the bottom its PE is zero. A
quick comparison reveals that, from the point of view of mechanical energy, this problem is exactly the
same as that of a freely falling object, which we just solved! Therefore its kinetic energy at the bottom
is, once again, equal to mgh. This is a problem that could not be solved using kinematic equations, since
for the curved track the acceleration is not constant.
In fact, throughout history the Law of Conservation of Mechanical Energy has been demonstrated to be
true for a wide variety of systems. In the coming days you will apply this Law to a device that you will
build in class. (7 minutes)
Day 2:
The story so far: KE = mv2/2
PE = mgh
In the absence of friction and air resistance, the sum of KE plus PE is constant.
Let’s return for a moment to the case of the freely falling object. Note that the PE does not change
instantly into KE at any point during the fall; rather, during the fall the PE gradually decreases, since the
height gradually decreases. Thus the KE gradually increases, which is the result you would intuitively
expect (it doesn’t speed up all at once, but gradually). So another way to interpret the Law of
Conservation of Mechanical Energy is to say that between any two times in the problem, the amount of
PE lost equals the amount of KE gained (or vice versa). As the object falls, PE is converting into KE. (5
minutes)
Practice calculation: if a 1-kilogram object freely falls from a height of 2.0 meters (starting from rest),
how fast will it be going when it is at height 0.8 meters?
Students attempt to solve, consulting with classmates if they wish (10 minutes).
Present solution (4 minutes):
At height 2 meters, the PE = mgh = (1 kg) (9.8 m/s2)(2 meters) = 19.6 Joules
At height 0.8 meters, PE = (1 kg)(9.8 m/s2)(0.8 meters) = 7.84 Joules
KE gained equals PE lost, so at height 0.8 meters, KE = 19.6 Joules – 7.84 Joules = 11.76 Joules
KE = mv2/2, so v = sqrt(2*KE/m) = sqrt (2*11.76/1) = 4.85 m/s (meters per second)
Another example of this Law is the simple pendulum. Without air resistance, the pendulum would
swing to its lowest point, rise again to its initial height on the opposite side, then reverse path back to its
lowest point, then back to its starting point, and keep repeating the process forever. (Realistically, the
maximum height will gradually reduce due to air resistance. But a pendulum in outer space, or in a
vacuum chamber, would behave as just described). Using the bottom of the swing as the reference
level, we can say that at the top of the swing (either side) all the energy is potential. At the bottom, all
the energy is kinetic. On the way down, PE is converting to KE; and on the way up, KE is converting into
PE. (5 minutes)
Day 3 (this lecture may spill over into a later period):
The Law of Conservation of Mechanical Energy says that, in the absence of friction and air resistance,
the sum of KE plus PE is constant. The last couple of days we have seen, among others, the example of a
freely falling object. You will eventually apply this same Law to your gravity car. Be aware that this is a
more complicated example, in that while the potential energy is provided by a falling mass of one
kilogram or less, the kinetic energy involves the entire mass of the vehicle moving. That is, the value of
mass in the PE equation is not the same as the value of mass in the KE equation. (2 minutes)
As a separate example: what if friction or air resistance is present? Such examples can be extremely
complicated; we shall look at one simple case here. The key is to note that friction or air resistance
produces heat, which is yet another form of energy (sometimes referred to as thermal energy).
Consider giving your physics book a shove across your desk, then letting go. Initially you give it some
kinetic energy; but the book quickly stops, so then its KE equals zero. What happened to it? Well, in the
process of stopping, the friction produced a little bit of heat. If we could accurately measure the
amount of heat produced, we would find that it is equal to the amount of KE the book initially had. In
this example, KE has been converted to thermal energy. (3 minutes)
The process just mentioned is a simple illustration of the Law of Conservation of Energy, which is
arguably the most important Law in all of science. In its full glory, this Law states:
Energy can neither be created nor destroyed, but merely converted from one or more forms into one
or more other forms.
Note that this no longer refers only to mechanical energy, but rather to any form of energy. (These
forms include KE, PEg, PE (elastic), and thermal energy, all of which have been referred to in recent days;
and also chemical energy, sound, light, electromagnetic energy, and nuclear energy, all of which we shall
discuss later in the course).
As mentioned above, one simple example of this is the conversion of KE into thermal energy caused by
friction. Let’s get more quantitative about this: consider an object of mass m moving with initial
velocity v0, on a surface which has coefficient of kinetic friction µ. How far will this object travel before
coming to rest? You may recall wrestling with this problem, or a similar one, in the unit on Newton’s
Laws. Here we shall see how to solve it using the work-energy theorem previously introduced.
You may recall that the frictional force is given by Ff =µFN where FN is the normal force. For this case FN
is simply the weight of the object (mg), so the frictional force becomes Ff = µmg. We can then calculate
the work done by friction in moving a distance d: using the formula W = FdcosƟ, we find
W = - µmgd , where the minus sign arises because the frictional force is in the opposite direction to the
motion, hence Ɵ is 180° and cosƟ= -1. Now applying the work-energy theorem,
Δ(KE) = 0 – mv02/2 = -µmgd, which can be solved to yield d = v02/2µg.
Type equation here.