Download c.o.e. tutorial

The Law of Conservation of Energy states simply that the energy of a system is a
constant, even if energy is transformed from one form to another within the system. In
other words, the energy put into a system will equal the extra energy that will be
available to do work by the system.
The energy in a system is always accounted for; it never disappears. What the Law of Conservation of
Energy does not state is that all that energy will be available to do work. In fact, in most cases, when an
energy transformation occurs, some of the energy transforms into heat because of friction, air resistance,
etc., a useless form of energy that cannot be harnessed effectively by humans (in most cases).
We will deal first with the idealized situation in which all the energy in an energy transformation
changes into a useful form such as kinetic energy or gravitational potential energy. Suppose a system is at
rest (meaning no work has been done on it - in other words, a force through a distance has not been applied
to it). If we do work on the system, we expend energy. This energy cannot disappear, and it doesn't. The
energy is transferred to the system (it might even change forms). In formula form, any of the following
apply:
Win = Wout
Ein = Eout
Einitial = Efinal
What does "work on the system" mean?
EXAMPLE: Consider the life of a rock. The rock has no
purpose in life if it is on the ground. If however, it is raised
to a higher plane, it is given (gravitational) potential energy
(GPE). How did it get up to that height? WORK (W)
was done on it. Someone or something must have applied a
force through a distance (W = F x d). Now the rock has the
"potential" of doing something in life (like falling)!
In formula form:
h
W in = GPEout
F x d = mgh
Suppose that you exert a force of 490 N upwards to lift the rock up to the ledge 3 meters
above. By exerting a force of 490 Newtons through a distance of 3 meters, the following
amount of work is done:
F = 490 J
Win =F x d = 490 N x 3 m = 1470 J
How much GPE will the rock have after being lifted?
Ein = Eout
Win = GPEout
1470 J = GPEout
Why, 1470 J, of course! The energy from Work has to go somewhere…
Now suppose you wanted to find out the mass of the rock. Well:
1470 J = GPEout
and…1470 J = mghout (because GPE = mgh)
Solving for "m" and plugging in what we know - gravity is 9.8 m/s2 and the height raised is 3 m
1470
m
gh
1470
m
(9.8 m 2 )(3 m)
s
50 kg = m
 PROBLEM SET 1:
1. 35 Joules of work is done to raise a 3 kg boulder to the top of a cliff. How high is the
cliff?
2. If a crate of Naranjilla of mass 14 kg is raised 10 m, how much work must be done to
accomplish this task?
CONSERVATION OF ENERGY can be expressed in other forms besides Win = GPEout..
The energy can change from Potential Energy (energy due to a change in state) to Kinetic
Energy (energy due to motion), for example. Following is an extremely effective strategy for
solving Conservation of Energy problems:
(a.) find from the problem statement what the initial state and final states of energy
are;
(b.) identify what quantity/variable the problem is asking you to find;
(c.) identify and label your given information/values;
(d.) set up the Conservation of Energy statement: Ein = Eout; decide what form(s) of
energy there are initially and finally
(e.) solve for the quantity in question (isolate it) without plugging in any values
(numbers) given;
(f.) plug in your given values once you have the quantity in question and solve;
(g.) don't forget your units (kgs, Joules, m/s);
Consider how energy is conserved in the following systems although it changes form:
a.
(a.)
(b.)
Ep
Ep
The system is at
rest. No work has
been done on the
system.
Kinetic
energy and potential
energy are zero.
(a.)
b. A hand has done
work on the sytem.
The work done
equals the amount of
elastic
potential
energy given to the
spring
(b.)
Ek
Ek
c.
(c.)
(d.)
Ep
Ep
Ek
Ek
(e.)
Ep
(f.)
(g.)
Ep
Ep
Ek
Ek
Ek
Ep
The spring is
released. Some of
the EPE of the
spring transforms
into TKE giving the
spring a velocity.
The total energy of
the system: EPE +
TKE
remains
constant.
f-g. The spring oscillates back and forth
in an endless cycle
in which energy,
although alternating
between EPE and
TKE,
remains
constant.
b. A hand has done work on
the system. The work
done equals the amount of
gravitational potential
energy given to the
pendulum bob.
Ep
Ek
Ep
Ek
c.
The pendulum bob is
released. Some of the
GPE of the pendulum bob
transforms into TKE
giving the pendulum a
velocity (the pendulum
has gained velocity but
lost height). The total
energy of the system:
GPE + TKE remains
constant.
d.
The pendulum, at the
bottom of the swing, has
only TKE (it is at
maximum velocity) and
no GPE (because it has no
height).
Consequently,
the pendulum has reached
its maximum velocity.
e.
The pendulum swings
passed the middle and
gains some GPE (because
it has height). It begins to
slow
down
(loses
velocity) and so its TKE
begins to lessen. TKE +
GPE, however, is still a
constant.
(c.)
d. The spring reaches
equilibrium position.
All of its energy is
TKE. The spring
experiences no forces
(either to stretch it or
compress it) for the
moment.
e. The spring is
completely
compressed and all of its
energy
becomes
EPE again.
The
spring
has
no
velocity therefore,
the TKE is zero.
Ek
a. The system is at rest. No
work has been done on
the system.
Kinetic
energy and potential
energy are zero.
(d.)
Ep
Ek
Ep
Ek
Ep
Ek
(e.)
(f.)
(g.)
Ep
Ek
Ep
Ek
(h.)
f.-h. The pendulum oscillates
back and forth in an
endless cycle in which
energy, although alternating between GPE and
TKE, remains constant.
 PROBLEM SET 2:
1. How does a system "at rest" gain energy?
2. If you do 25 Joules of WORK stretching a spring,
(a.) how much elastic potential energy will it have
available?
(b.) what is the spring constant k of that spring if
it was stretched 1.2 m?
(c.) what is the fastest speed the spring will have
when it is released?
(d.) what would the elastic potential energy be when the spring is moving at its fastest
speed?
3. If a pendulum bob has a mass of 2.5 kg, and its maximum velocity is 8 m/s,
(a.) from what height was it released?
(b.) what is the height at maximum velocity?
(c.) do you need the mass to solve the problems above? Explain.
(d.) how high is the pendulum bob when its velocity is 2 m/s?
(e.) how fast is the pendulum bob moving when the height is 0.75 m?
 PROBLEM SET 3:
1. A 100 kg man jumps out of a burning
building into a fire rescue net 15 m below.
The net stretches 2.5 m before bringing
the man momentarily to rest (and then
tossing him back up).
What is the
potential energy of the stretched net if
mechanical energy is conserved?
icecube
2.
An ice cube is released from the edge of a
hemispherical frictionless bowl whose radius is
0.22 m. How fast is the ice cube moving at the
bottom of the bowl?
r
3. A frictionless roller-coaster car tops the first hill in the Figure below with
speed vo m/s. What is the speed at
(a.)
(b.)
(c.)
(d.)
Point A?
Point B?
Point C?
How high will the car go on the last hill, which is too high for it to
cross?
vo
first
hill
A
B
h
h
h/2
C
4. The figure at right shows an 8 kg stone resting on a spring. The spring is
compressed 0.10 m by the stone.
(a.) What is the spring constant k of the spring?
(b.) If the stone is pushed down an additional
30 cm. and released, what is the potential
energy of the compressed spring just
before that release?
(c.) How high above the release position will
the stone rise?
5. A ball with a mass of m kg is attached to the end of a very light rod with
B
l
A
length "l" meters and a negligible mass. The
other end of the rod is pivoted so that the ball
can move in a vertical circle (see Figure). The
rod is held in the horizontal position and then
given enough of a downward push so that the
ball swings down and around and just reaches
the vertically upright position, having zero speed
there.
(a.) What is the change in the potential energy of the ball from Point A to
Point B?
(b.) What initial speed must have been given to the ball?
 THE REAL WORLD
Up to now we've been considering situations in IDEAL CONDITIONS where no friction or
air resistance applies. We will now explore conservation of energy and its formulas in a more
realistic scenario: when part of the energy of the system transforms into heat. When two
objects are in contact or when the particles (or molecules) of a substance pass each other by or
move, heat is generated. In the real world, heat is a byproduct of work and, when produced,
is normally lost or diffused into the surrounding environment. Since it is lost, this heat
generated is sometimes called "useless" energy. Nevertheless, it IS energy and must
somehow figure into our formula for conservation of energy: Ein = Eout.
Let us see how…
EXAMPLE: If a block-spring system is set into
motion on a surface on which friction exists, the back
and forth oscillations will diminish until the blockspring system is at rest again. The mechanical energy
has decreased and if we had a sensitive thermometer
nearby, we would see that the temperature in the
environment has risen, and consequently that heat
energy has been transferred. The increase in the amount of heat energy will be precisely
the decrease in the amount of mechanical energy.
A byproduct of the work done by friction against the motion is heat! We will call this
change in the thermal energy of the system Q…and we will stick it into our conservation
of energy formula:
EPEin
input useful
energy
=
EPEout +
TKEout
output useful
energy
+
Q
by-product
useless energy
EXAMPLE: A steel ball whose mass is 0.52 kg is fired vertically
downward from a height h of 18 m with an initial speed vo of 14 m/s.
If the ball hits the ground with a force of 2500 N,
(a.) what is the total mechanical energy available to the ball right
before it hits the ground?
(b.) in the absence of friction/resistance, how far should the steel ball
be able to penetrate into the surface?
(c.) if the ball penetrates a distance of 0.025 m into the ground, how
much of the ball's energy became heat?
h = 18 m
?
Ans. (a.) When shot, the ball has both TKE and GPE. Right before it
hits the ground, all the ball's energy is kinetic. To find the total kinetic energy
right before the ball hits the ground use the COE formula:
Ein = Eout
TKEin + GPEin = TKEout
mghin + ½mvin2 = TKEout
(0.52 kg)(9.8 m/s2)(18m) + ½(0.52 kg)(14 m/s)2 = TKEout
142.688 J = TKEout
Ans. (b.) In the absence of friction/resistance, in other words, in the ideal situation, the
steel ball does work penetrating the ground and that's it. So…
By formula:
Ein = Eout
TKEin = Wout
TKEin = F x d
TKE in
d
F
142.688 J
d
2500 N
0.0570752 m = d
Ans. (c.) Since the ball penetrated somewhat less distance than it did originally, some of
the ball's energy must have dissipated as heat. We need to find out with how
much:
Ein = Eout
TKEin = Wout + Qout
TKEin = F x d + Qout
TKE in
 Q out
F x d
142.688 J
 Q out
2500 x (0.025 m)
142.688 J
 Q out
62.5 m)
2.283008 J = Qout
 PROBLEM SET 4:
1. A dog of mass 6.0 kg runs onto a
curved ramp as shown in the figure
at right with a speed of 7.8 m/s.
The initial perro height is 8.5 m.
The dog slides down the ramp and
comes to a momentary stop when
he reaches a height of 11.1 m h1
from the floor at the other end of
the ramp. What is the increase in
heat energy of the dog/ramp due to
the sliding?
vo
h2
2. A 0.075 kg FRISBEE is thrown
from a height of 1.1 m above the
ground with a speed of 12 m/s. When it has reached a height of 2.1 m,
its speed is 10.5 m/s.
(a.)
How much gravitational potential
energy did the FRISBEE gain?
(b.) How much of the FRISBEE'S
mechanical energy was dissipated by
air drag?
v = 12 m/s
h = 2.1 m
h = 1.1 m
3. A baseball player throws a baseball with
an initial speed of 20 m/s.
Just before the catcher catches the ball at the same
height, the ball's speed has been reduced to 11.0 m/s. By
how much did air drag reduce the mechanical energy of
the ball? (the baseball has a mass of 0.09 kg).
4. A river descends 15 m through rapids. The speed of the
river is 3.2 m/s upon entering the rapids and 13 m/s as it
leaves the rapids. Consider 10 kg of water going down
that river…
(a.) What percentage of gravitational potential energy
lost by the 10 kg of water becomes kinetic energy
of water downstream?
(b.) What happens to the rest of the energy?
5. Fozzy, a 25 kg bear slides, from rest, 12 m
down a pine tree. Right before the bear hits the
ground, it is moving with a speed of 5. 6 m/s.
(a.) How much gravitational potential energy
does Fozzy lose before hitting the ground?
(b.) How much mechanical energy dissipates
due to friction?