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Work and Energy
Work
Power
Kinetic Energy
Potential Energy
Gravitational Potential Energy
Elastic Potential Energy
Energy Conservation
Non-Conservative Forces
More Examples
Summary
Homework
Circular Motion & Gravity
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Work
Work
Work = Force x distance
W = Fd
Great force exerted with no distance moved means no work is done!
Units?
Force in Newtons
distance in meters
1 N-m = 1 J = 1 joule
Actually, we can say "Work = Force x distance" only when the Force and
the distance moved are in the same direction.
If the force and the displacement do not lie along the same direction, then
we must take the component of the force parallel to the displacement and
multiply that by the displacement. It is only the component of the force
parallel to the displacement that contributes to the work done.
This means the general form of the definition of work is
W = F d cos
Another way of describing this is called the "dot product" or the "scalar
product",
Where we are using the mathematical notation of
A and B are vectors and S is a scalar. This is called the "scalar product of
two vectors" or the "dot product" since it is indicated by a dot.
where
is the angle between vectors
and
.
ToC
Power
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Power
However, power is power. Either type of units could be used. It would be -- and is -entirely proper and correct to state that a certain car has an engine which produces 98 kW
(that is 132 hp).
As you can see from these graphics from a 1983 Pontiac brochure, General Motors has
used SI (International System) units internally for many, many years. Of course, that is
true for all other large US companies.
Like the engineers in other multi-national companies, GM's engineers recorded data in SI
units. Power is measured in kilowatts (kW) and torque -- the rotational effect of a force -is measured in newton-meters (NM or N-m). This is not something new; remember these
graphics are from a 1983 Pontiac brochure!
Here are excerpts from Francophone Canadian automobile brochures that do, indeed,
show the engine power in kilowatts (kW) and in horsepower ("ch" for chavox in French).
James Watt patented the steam engine in 1769. To sell it, he needed to tell people how
many horses it would replace. Watt made the first measurements on how much power a
work horse could do -- he measured how quickly farm horses could do work.
Watt was somewhat conservative in his marketing. There are few horses that actually
produce one horsepower of power.
Work Kinetic Energy
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Kinetic Energy
Energy is the ability to do work.
If we apply a force to an object, it accelerates. We know this from
Newton's Second Law, F = m a.
We do work on an object when we exert a force F on it and move it
through a displacement s,
It has an acceleration a = F / m
and it has a final speed vf which we can find from
vf2 = vi2 + 2 a ( sf - si)
vf2 = vi2 + 2 a s
vf2 = vi2 + 2 ( F / m ) s
vf2 = vi2 + 2 ( F s ) / m
vf2 = vi2 + 2 W / m
W = (1/2) m vf2 - (1/2) m vi2
W = [ (1/2) m v2 ]
We define this quantity as the Kinetic Energy, KE,
KE = (1/2) m v2
W = KE
When we do work on an object, we change its Kinetic Energy. Because an
object moves it has energy or it can do work on something else. We call
this Kinetic Energy.
Power
Potential Energy
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Potential Energy
In addition to Kinetic Energy or "Energy of Motion", there are other forms
of energy. All these other forms may be classified as "Potential Energy",
PE. We will be interested mainly in
Gravitational Potential Energy, PEg
Elastic Potential Energy, PEel
Other forms of energy include nuclear energy and chemical energy.
Heat or thermal energy is a form of kinetic energy on a microscopic
level.
I will usually write potential energy as PE but it is also common to write
it a U (eg, The Mechanical Universe video tapes will always use U for
potential energy).
Kinetic Energy
Gravitational Potential Energy
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Gravitational Potential Energy
Consider lifting an object of mass m through a distance h,
To just lift the object, we must exert a force of F = m g.
In lifting it a distance h, we do work equal to
W=Fs
W = ( m g ) (h)
W=mgh
The object can now do that much work as it falls back to its original
position; this is Potential Energy,
PEg = m g h
or
Ug = m g h
Potential Energy
Elastic Potential Energy
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Elastic Potential Energy
How much work is done when we stretch a spring a distance x from its
equilibrium position?
First, we need to know about the general characteristics of a spring.
Experimentally, we find
This is known as Hooke's law
We might write this in equation form as F = k x. However, the force
exerted by the spring is always in the opposite direction to the stretch (or
compression) of the spring. Therefore, we write Hooke's law as
F=-kx
This is the force exerted by the spring. The external force we exert on
the spring is Fext = + k x.
The proportionality constant k is known as the spring constant and
describes how stiff the spring is.
Now we are again ready to ask
"How much work is done when we stretch a spring a distance x from its
equilibrium position?"
We know how to handle a constant force. For a constant force, we know
W=Fd
Our spring force varies, but we can think of it as being (nearly) constant as
we move through a (very) small distance,
The work done by a variable force is the area under the "curve" on a
Force - distance graph. For this Hooke's law force of Fext = k x, the work
done to the spring by the external force Fext is
W = (1/2) k x2
This work done on the spring as it is stretched (or compressed) can be
recovered. This is stored work that can be used to do work on something
else by this spring. That means the stretched (or compressed) spring has
energy -- potential energy. This is spring potential energy or elastic
potential energy.
PEel = (1/2) k x2
or
Uel = (1/2) k x2
Gravitational Potential Energy
Energy Conservation
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Nonconservative Forces
Friction is the main nonconservative force we will encounter. To the
extent that friction is present, energy will not be conserved. Well, that
means "large scale, mechanical energy" will not be conserved. Friction
produces heat and heat involves the microscopic motion of molecules.
Energy is always conserved when thermal energy or heat is included.
Usually, however, when we talk about Energy Conservation we mean
PEi + KE i = PEf + KE f
without including heat or thermal energy.
Energy Conservation
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More Examples
More Examples
A spring with spring constant k = 500 N/m is compressed a distance x
= 0.10 m. A block of mass m = 0.250 kg is placed next to the spring,
sitting on a frictionless, horizontal plane. When the spring and block
are released, how fast is the block moving as it leaves the spring?
When the spring is compressed, work is required and the
spring gains potential energy,
PEel = (1/2) k x2
PEel = (1/2) (500 N/m) (0.10 m)2
PEel = 2.5 J
As the spring and mass are released, this amount of work is
done to the mass to change its kinetic energy from zero to
a final value of
K = (1/2) m v2
K = (1/2) (0.25 kg) v2
K = (1/2) (0.25 kg) v2 = 2.5 J = PEel = Ws
v2 = 20 m2 / s2
v = 4.47 m/s
Explain the actions of a pole vaulter in terms of Energy Conservation.
As the pole vaulter runs toward the bar, he gains kinetic energy. As he
bends the pole, he does work on the pole and stores elastic potential
energy in the pole -- this is much like compressing a spring or pulling a
bow. That stored elastic potential energy is converted into gravitational
potential energy as he goes up and over the bar.
An interesting sidelight: A good pole vaulter -- or high jumper -- will
"curl" himself or herself "around" the bar so that his or her center of mass
may go under the bar so less energy is required.
A skier starts from rest at the top of a frictionless incline of heigh 20 m as shown
here. At the bottom of the incline, the skier encounters a horizontal surface where
the coefficient of kinetic friction between the skis and snow is 0.210. How far does
the skier travel on the horizontal surface before coming to rest?
While coming down the incline, energy is conserved. We will measure
gravitational potential energy with respect to the horizontal surface, so
PEg,f = 0. The skier starts from rest, so Ki = 0 . Then
Ei = Ef
Ki + PEel ,i = Kf + PEel ,f
0 + m g hi = (1/2) m vf2 + 0
m (9.8 m/s2) (20 m) = (1/2) m vf2
vf2 = 392 m2 / s2
vf = 19.8 m / s
As the skier moves across the "rough snow" on the horizontal plane,
friction does (negative) work and reduces the skier's Kinetic Energy to
zero when the skier stops. This work done by friction is
Wf = - Ff s
The friction force Ff is given by
Ff = Fn
where Fn is the normal force (perpendicular to the surface) On this
horizontal surface, it turns out that Fn = m g
Ff = Fn = = m g = (0.210) m (9.8 m/s2) = (2.06 m/s2) m
(Be careful, one of these m's stands for meters, as in m/s2, while the other
m is the mass of the skier! And we are also using s for seconds, as in m/s2,
and also for the distance moved!)
This means the work done by friction is
Wf = - Ff s = - [ (2.06 m/s2) m ] s
At the beginning of this horizontal section, the skier's original Kinetic
Energy is
Ko = (1/2) m v2 = (1/2) m (19.8 m / s)2 = 196 (m/s)2 m
When the skier comes to rest, the final Kinetic Energy will be zero,
Kf = 0
The only work done is the work due to friction so that is the net work.
And we know the net work equals the change in the Kinetic Energy,
Wf = K = Kf - Ko = 0 - 196 (m/s)2 m
- [ (2.06 m/s2) m ] s = - 196 (m/s)2 m
[ (2.06 m/s2)] s = 196 (m/s)2
s = 95.15 m/s
Nonconservative Forces
Summary
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Summary
Work is force multiplied by distance and has units of
J = joule = N-m
Work is the component of the force in the direction of the displacement,
multiplied by the displacement,
or
Energy is the ability to do work. You may think of energy as "stored up
work";
KE = (1/2) m v2
PEg = m g h
PEel = (1/2) k x2
We encounter many situations or many systems for which the total
energy remains constant. We describe this by saying Energy is
conserved.
E = PE + KE
E = constant
PE + KE = constant
Ei = Ef
PE i + KE i = PE f + KE f
More Examples
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Homework
PHY 1151
Chapter 6; Work, Power, and Energy
6.5, 14, 19, 22, 36, 38, 45, 55, 61, 72, 75, 88, 92
| ToC, Chapter 6 | Course Calendar |
6.5 How much work is done by gravity on a skier weighing 600 N who
slides 500 m down a hill (measured along the hill) which makes an angle of
30° with the horizontal as sketched in the figure?
W = F D cos = F D cos 60° = (600 N) (500 m) (0.50) = 150,000 J
6.14 In playing shuffleboard, a long wooden stick is used to give a push to a
0.3 kg puck and cause it to move across the floor. A player pushes with a
force of 25 N along the stick, inclined 40° to the ground. The coefficient of
friction between puck and floor is 0.10.
a) How much work is done on the puck by the stick as it
moves a distance of 0.7 m?
b) What will be the puck's speed after it has moved that 0.7 m,
assuming it started at rest?
c) From there the puck leaves the stick and slides across the
floor. How far will it slide?
a)
W = F D cos = (25 N) (0.7 m) (0.766) = 13.4 J
Remember, this is the work done by the stick. It is not the
total or net work done on the puck (for the friction force also
does work on the puck; however the weight and the normal
force do not do any work on the puck).
b) Now we need the total work done on the puck.
Wtot = Wnet = Wstick + Wf + WN + Ww
We just found the work done by the stick,
Wstick = 13.4 J
Wf = Ff D cos 180° = Ff (0.7 m) (- 1)
Now we must pause and determine the friction force Ff,
Ff = µ FN = (0.10) FN
And that means we must find the normal force, FN,
The sum of all the y-components of the forces must be zero
since there is no acceleration in the y-direction,
Fy = - (25 N)(0.643) + FN - 2.94 N = 0
FN = 2.94 N + 16.07 N = 19.01 N
Ff = µ FN = (0.10) FN = (0.10) (19.01 N) = 1.9 N
Wf = (1.9 N) (0.7 m) ( - 1) = - 1.33 J
The work done by the weight and by the normal force are
both zero since these forces are perpendicular to the distance
moved,
WN = 0
Ww = 0
Therefore,
Wtot = Wnet = Wstick + Wf + WN + Ww
Wtot = Wnet = 13.4 J - 1.33 J + 0 + 0
Wtot = Wnet = 12.1 J
What does this amount of work cause? From the work-energy
theorem (that is, from conservation of energy), we know that
work done on an object causes a change in its kinetic energy;
that is,
W = KEf - KEi
Since this puck starts at rest, we know
KEi = 0
so that
W = KEf = (1/2) m vf2 = (1/2) (0.3 kg) vf2 = 12.1 J
vf2 = 12.1 J/0.15 kg = 80.7 m2/s2
vf = 9.0 m/s
c) Now what happens? At the end of the 0.7 m the stick no
longer exerts a force and work is done only by the friction
force. The puck starts out with 12.1 J of KE and finally comes
to rest a distance D from where the stick released it. Again, we
apply the work-energy theorem,
W = - Ff D = - (1.9 N) D
W = KEf - KEi = 0 - 12.1 J
- ( 1.9 N ) D = - 12.1 J
D = ( 12.1 / 1.9 ) m
D = 6.4 m
6.19 An electric bill lists 1284 kW-h as the energy used for a month. If the
charge is 11.5 cents per kW-h, what is the total bill? How many joules of
energy were used that month?
Cost = (1284 kW-h)($0.115/kW-h) = $147.66
E = 1284 kW-h = 1284 x 103 W-h [ (J/s)/W] ( 3600 s/h )
E = 4.62 x 106 J
6.22 The output of a common student laboratory laser is about 250
milliwatts. How much energy is delivered in one hour?
E = P t = (0.25 mW) (1 h) (
)(
)(
) = 0.9 J
6.36 The power of a boat’s motor at a speed of 40.0 km/h is 22.0 kW. If the
power is 15% greater while pulling a skier at the same speed, what is the
tension in the tow rope?
P = F v (assuming F and v are parallel)
P=Fv
As usual, we must change the speed from units of km/h to units of m/s
v = 40
(
)(
) = 11.11
F= =
F = 1980 N
This is Fwater, the force the water exerts on the boat.
When a skier is towed the power is 15% greater,
P' = (1.15)(22 kW) = 25.3 kW
Now the net force is
Fnet = Ftow + Fwater
and that is the force that goes into
P=Fv
F= =
= 2277 N
Ftow = Fnet - Fwater = 2277 N - 1980 N
Ftow = 297 N
This is the tension in the ski tow rope
6.38 What is the kinetic energy of a 1750 kg car traveling at 80 km/h?
KE = (1/2) m v2
As usual, we must change the speed from units of km/h to units of m/s
v = 80 (km/h) (h/3600 s) (1000 m/km) = 22.22 m/s
KE = (1/2) m v2
KE = (1/2) (1750 kg) (22.22 m/s)2 = 4.32 x 105 J
6.45 Below are data for eight air track gliders with different masses and
speeds. Calculate the values of their kinetic energies.
mass (kg) speed (m/s)
A 0.225 kg 0.85 m/s
B 0.225
0.90
C 0.250
0.35
D 0.250
0.50
E 0.267
0.65
F 0.315
0.82
G 0.450
0.45
H 0.450
0.52
KE = (1/2) m v2
A) KE = (1/2) (0.225) (0.85 m/s)2 = 0.0813 J
B) KE = (1/2) (0.225) (0.90 m/s)2 = 0.0911 J
C) KE = (1/2) (0.250) (0.35 m/s)2 = 0.0153 J
D) KE = (1/2) (0.250) (0.50 m/s)2 = 0.0313 J
E) KE = (1/2) (0.267) (0.65 m/s)2 = 0.0564 J
F) KE = (1/2) (0.315) (0.82 m/s)2 = 0.1059 J
G) KE = (1/2) (0.450) (0.45 m/s)2 = 0.0456 J
H) KE = (1/2) (0.450) (0.52 m/s)2 = 0.0608 J
6.55 A ski slope drops at an angle of 20° and is 600 m long. Find the change
in potential energy of a 70-kg skier who goes down this slope.
PE = m g h = (70 kg)(9.8 m/s2)(205 m) = 141,000 J = 141 kJ
6.61 In loading a certain dart gun, the dart is pushed in 12 cm before the
gun is cocked. A force of 20 N is required at the end. Find the spring
constant of the spring in the gun. Find the total amount of energy stored in
the spring.
F=kx
k= =
= 167
PEel = (1/2) k x2
PEel = ( /2) (167 N/m)(0.12 m)2
PEel = 1.2 J
1
6.72 An arrow of mass 0.1 kg is fired directly upward by an archer who
exerts an average force of 75 N on a bowstring that is pulled back 0.6 m.
After it comes down, the arrow penetrates 20 cm vertically into the
ground. Use the ideas of work and energy to answer the following:
a) With what speed does the arrow leave the bow?
b) How high does the arrow go?
c) What average upward force does the ground exert on the
arrow while stopping it?
a)
Ef = (1/2) m v2 = Ei = W = (75 N)(0.6 m) =
Yikes! My apologies for that arithmetic error.
75 x 0.6 = 45
Ef = (1/2) m v2 = Ei = W = (75 N)(0.6 m) =45.00
(1/2) (0.1 kg) v2 = 45.0 J
v2 = 900 m2/s2
v = 30 m/s
b)
Ef = m g h = 45 J
(0.1 kg) (9.8 m/s2) h = 45 J
h = 45.9 m
c)
W = F D = 45.0 J
F (0.2 m) = 45.0 J
F = 225 N
6.75 A ball is attached to a cord of length L which is initially held
horizontal from a fixed support at P as sketched in the figure. The ball is
released from rest at point A.
a) Find its speed at point B, the bottom of the swing.
A peg is fixed at point O, a distance y directly below point P.
The cord hits the peg and the ball swings upward in an arc as
shown.
b) Find the ball's speed at point C when the cord between it
and the peg is horizontal.
c) Find the ball's height (relative to its starting position) at
point D, the top of its swing after hitting the peg (Assume y >
L/2).
d) Find the ball's speed at point D, the top of its swing after
hitting the peg.
a) At position A, vA = 0 or KEA = 0
If we measure all the vertical distances from yB = 0 at position
B, then at A, the mass has a height yA = L, so that
PEA = m g yA = m g L
( This choice for yB = 0 means PEB = 0 ).
Therefore, the total energy at A, EA, is
EA = KEA + PEA = 0 + m g L = m g L
Energy is conserved so
EB = EA
KEB = (1/2) m vB2
PEB = m g yB = 0
Remember, we have measured vertical distances here so we
have set yB = 0. That means our total energy at position B is
EB = KEB + PEB = (1/2) m vB2 + 0 = (1/2) m vB2
EB = (1/2) m vB2 = m g L = EA
vB2 = 2 g L
vB =
A peg is fixed at point O, a distance y directly below point P.
The cord hits the peg and the ball swings upward in an arc as
shown.
c) At position C, the (vertical) height above position B is
yC = (L - y)
Energy is conserved so
EC = EA
KEC = m vC2
PEC = m g yC = m g (L - y)
That means our total energy at position C is
EC = KEC + PEC = (1/2) m vC2 + m g (L - y)
EB = (1/2) m vC2 + m g (L - y) = m g L = EA
(1/2) m vC2 + m g L - m g y = m g L
(1/2) m vC2 = m g y
v C2 = 2 g y
vC =
d) At position D, the (vertical) height above position B is
yD = 2 (L - y)
as can be seen by taking a more careful look at the diagram.
Energy is conserved so
ED = EA
KED = (1/2) m vD2
PED = m g yD = m g [2 (L - y)]
That means our total energy at position D is
ED = KEC + PEC = (1/2) m vC2 + m g [2 (L - y)]
ED = (1/2) m vD2 + m g [ 2 (L - y) ] = m g L = EA
(1/2) m vD2 + 2 m g L - 2 m g y = m g L
(1/2) m vD2 = 2 m g y - m g L
v D2 = 2 g ( 2 y - L )
vD =
6.88 Calculate escape velocity from the surface of Mars.
From Equation 6.23, we know the escape speed in general,
vesc =
vesc =
where we have used values of M and R from Table 5.1 on page
162 of the text and been careful to put numbers in with
consistent units. Thus,
vesc = 5010
vesc = 5010
(
)(
) = 18,000
vesc = 11,000
Remember, escape velocity from Earth's surface is about
25,000 mi/hr, about twice escape velocity from Mars.
6.92 The pilot of the first successful human-powered, heavier-than-air
flight could generate an average of 0.32 kW (0.43 hp) during the flight. The
aircraft covered 2.4 km at an average speed of 4.1 m/s. How much energy
was expended during the flight?
Energy = power x time
v=
t= =
=
= 585 s
Energy = (0.32 kW)(585 s)
Energy = (320 W)(585 s)(
Energy = 187,000 J
)
| ToC, Chapter 6 | Course Calendar |
(c) Doug Davis, 2000; all rights reserved.