Download Positive Work No Work Some Work Negative Work

NOTES ON WORK AND ENERGY
1.
Mechanical Work
Work = Fparallel *d = F * dparallel
Positive Work
Negative Work
Fparallel and d in the same direction
F=Fparallel (parallel to d)
d
Fparallel and d in opposite directions
F=Fparallel (anti-parallel to d)
d
Some Work
F=Fperpendicular (perpendicular to d)
Fperpendicular
No Work
F
Fparallel
d
No Work done because there is no Fparallel .
d
Only Fparallel does work
1a.
Mechanical Work = Fparallel d = F dparallel
Units: fp (foot pounds) and J (Joules = Nm Newton meter)
1E.
Examples:
1. Your weight, W= 150 p (pounds) and you climb or walk up 10 f (feet).
So, F= 150 p (up) is the force you apply against gravity and d=10 f (up).
The work you do is
Work = Fd = 150 p * 10 f = 1500 fp
2. Your mass m is m = 70 kg and you climb or walk up 3 m (meters).
So, F= your weight W = mg = 70 kg x 10.0 m/s2
(1)
You
d
F
You
(For ease of calculation, using g =10.0m/s2 instead of the standard 9.8
m/s2 - 2% error)
= 70x10 kg m/s2 = 700 N (Newtons). d = 3 m (up).
The work you do is
Work = Fd = 700 N x 3.0 m = 2100 J (Joules = Nm).
©Hulan E. Jack Jr. 2005 Revised 2006 ,2011,2012
Notes on Work and Energy
Page 1 of 8
W=mg
your weight
F
you push up
NOTES ON WORK AND ENERGY
1b.
Construcitons for Inclined Plane Problems
F
Movement along Inclined Plane with angle Fd between F and d
This is a collection
of constructs that
you will find
pieces useful when
doing various
problems of
motion on inclined
planes.
Fd is angle between F and d
l to
e
l
l
a
par
perpendicular to d
rt
a
l
u
end
p
r
pe
oF
d
Fd
F
d
F
F
F
d
F parallel to d
d
d
d
Power = Work/time = the rate of doing work
2.
(2)
Units: 1 hp (horsepower) = 550 fp/s = 33,000 fp/min ;
1 W (Watt) = 1 J/s .
2E.
Example: 1.
In 1., above Work = 1500 fp. Suppose you did that climb in 10 s (seconds).
The your power would be
Power = Work/time = 1500 fp /10s = 150 fp/s .
In terms of hp this becomes
Power = 150 (fp/s) ( 1/550 ) hp/(fp/s)
This is the fp/s -> hp conversion [fp/s]( )hp/[fp/s]
= 150/1500 hp = 1/10 hp =0.1 hp .
3.
Mechanical Energy is the ability to do mechanical work.
3.1
Mechanical Potential Energy , PE or EP , is the ability to do mechanical work because of
position.
3.1a Local Gravitational Potential Energy, Spring or elastic Potential Energy,
PEGrav or EP Grav = mgH (3)
PE spring or EP spring = ½ kx2 . (4)
m
H
“local” ground
x compressed
spring with
free
stiffness k
stretched
x
©Hulan E. Jack Jr. 2005 Revised 2006 ,2011
Page 2 of 8
Notes on Work and Energy
NOTES ON WORK AND ENERGY
3.2
Kinetic Energy , KE or EK , is the ability to do work because of motion.
KE = EK = ½ mv2
4.
.
v (velocity)
m
(5)
Work-Energy Theorem
Work = change in energy =E = Fd
(6)
4E.
Example:
A car travels a displacement d from point A to point B.
a:
When the force F and displacement d are in the same direction, positive work is done
on the car. This causes the Kinetic Energy to increase, increasing its speed.
Positive Work
F
v
v
d
b:
When the force F and displacement d are in opposite directions, negative work is
done on the car. This causes its Kinetic Energy to decrease, decreasing its speed v.
v
Negative Work
5.
F
v
d
Conservation of Energy
As the body or system moves from initial to final position
If negative work is done on it, then
PE initial + KE initial = PE final + KE final + Work against system (Negative Work) . (7a)
Its final energy is less than its initial energy.
If positive work is done on it, then
PE initial + KE initial + Work on system (Positive Work) = PE final + KE final
Its final energy is greater than its initial energy.
(7b)
As the body or system under goes processes moving from point A to point B or states
PE A + KE A + Positive Work A->B
PEAA
PE
KEAA
Positive
WORKA->B
=
PE A + KE A
PEA
=
PEB
=
KEA
=
PE B + KE B .
PEAB
PE
KEB
(7c)
KE increased
PE B + KE B + Negative Work A->B. .
KE
KEAB
(7d)
Positive
Negative
KE decreased
WORKA->B
A->B
Note, since PE depends ONLY on position, then for a fixed position, work done can only
increase or decrease the KE.
©Hulan E. Jack Jr. 2005 Revised 2006 ,2011
Page 3 of 8
Notes on Work and Energy
NOTES ON WORK AND ENERGY
5a.
Energy Boxes - Accounting 101
The boxes shown are called Energy Boxes. They are a graphical representation of the
Conservation of Energy. There width represents the total energy at given point or position. This box
method convert Conservation of Energy problems to accounting. The roller coaster example, below,
illustrates this.
5E1.
W
=
J
000
B>C
=1
50
0J
>B
W A-
Example: The Roller Coaster
A roller coaster car of weight W =1000 N starts from rest at point A, a height HA = 30 m
above the local ground. It travels along
the roller coaster track to the valley point
m
A
B, then rises to point C, continues on the
C
point D. Friction does the negative work
shown as the car moves along track.
WC->D= 1200J
HA = 30 m B
D
WorkA->B = 1000J, WorkB->C = 500J, etc.
The pictures and the analysis below show
HC = 20m
the method.
HB = 10m
H = 10m
D
Local Ground
PEAB
PE
A
The scheme
Negative
Positive
WORKA->B
KE
KE
A B
PEA
=30000J
B
PEB
=10000J
C
PEC
=20000J
D
PED
=10000J
30000J =
PEA
=
WHA
= 30000J .
WorkA->B PEA = PEB
=1000J 30000J = WHB
+ KEB
+ WorkA->B
+ KEB
+ 1000J
30000J = 10000J + (19000J) + 1000J
KEB
=19000J
KEC= Work-A>C PEA = PEC
8500J =1500J 30000J = WHC
KED
=17300J
+ KE C + WorkA->C
+ KEC
+ 1500J
30000J = 20000J + (8500J) + 1500J
WorkA->D
=2700J
10000J + (17300J) + 2700J
PEA
= PED
+ KED
30000J = WHD + KED
+ WorkA->D
+ 2700J
Notice that value of the KE had to be the value needed to make the total equal to PEA =30000J.
This can be written, or solved, as the mathematical equation
KEX = PEX -
PEX - WorkA->X
(7e)
where X is the situation, or values, at any point X. (See 5b, below.)
Note that Work is cummulative.
At C Work A->C = Work A->B + Work B->C .
At D Work A->D = Work A->B + Work B->C + Work C->D .
©Hulan E. Jack Jr. 2005 Revised 2006 ,2011
Page 4 of 8
Notes on Work and Energy
(7f)
NOTES ON WORK AND ENERGY
5b. Let’s look at the Accounting features of these Energy Boxes.
The boxes have the following properties.
1. Each box has four sections, one each for KE, PE, Negative Work, Positive Work.
2. If a process has no Positive Work, then the sum of other three sections is constant equal to
the previous box.
3. If a process has Positive Work in a position, this value is added to the KE as shown in Eq.
(7c). Then the sum of other three sections is again constant , as in 2 above, for all following
boxes until the next instance of Positive Work.
Then if you know the value of two sections in a box, then the value of the third section must make the
sum of this box the same of that of the previous box. For visualization, the boxes are drawn to scale,
or sketched approximately scale.
Let’s use the box to analyze point C.
C
PEC
=20000J
C
PEC
=20000J
C
PEC
=20000J
10000J
PEA = PEC + KEC + Work A->C = 30000J
The potential energy at C is determined by the height at
C. So PEC = WHC = 1000J*20m = 20000J . So, there is
10000J left to account for.
Work A>C
=1500J
In PEA = PEC + KEC + Work A->C , the PE’s are
accounted for. So, the remaining 10000J is split
between KEC and Work A->C . Work A->C =
Work A->B + Work B->C = 1000J+500J = 1500J
KEC= Work A>C
In PEA = PEC + KEC + Work A->C only KEC is left.
8500J =1500J
Hence KEC = 30000J-20000J-1500J = 8500J
Finally, finding the KE is just an accounting mandate of make the books (boxes) balance.
This is really Eq(7e), above.
6.
The Force that changes energy over a displacement
From the Work-Energy Theorem, Eq. (6) ,
Work = change in energy =E = Fd = Fx,
has the view that a a force F applied to a body over a displacement d causes a change in
energy, E, of that body . We can rewrite this as
F
F = E/d
where
= E/x ,
(8)
x = d
©Hulan E. Jack Jr. 2005 Revised 2006 ,2011
F
x
Work = E = Fx
Page 5 of 8
Notes on Work and Energy
NOTES ON WORK AND ENERGY
the displacement d has been replaced by the change in displacement x. This just puts the
expression in more customary form.
Let’s return to rewrite Eq.(6) as change the view This change of view gives the force F
required to change energy of a body by E over a change in displacement x.
Let’s return to rewrite Eq.(6) as
E = F *x = F * x ,
(8a)
This demonstrates that a small (short) stopping change in displacement, x , requires a large
force F , while a large (long) change in displacement , x, requires a small force F.
Let’s apply this to a car coming to a stop by running into a big tree or wall. Before the 1950's
cars were very rigid. So, in such a crash the car stopped over a very short displacement.
There were no seat belts. So, by Newton’s 1st Law of motion, the passengers kept moving
until they crashed into wind shields and steering wheels, and so on. When they hit these items
the passengers would come to a sudden stop over a short displacement. The large stopping
force smashed up the passengers. What a mess!!! In the 1950's auto makers started putting
“crumple” points in cars so that they would somewhat “fold up” in a collision. This caused the
car to travel a long distance as it “folded up” to a stop . The large x requiring a small F to
stop the passengers inside. The car was “totalled”, but the passengers survived quite well.
Dynamic Loading
When you jump to the ground the force between you and the floor as you stop is more than
your weight. Let’s say you jump from a height H, (a). Here
you have a potential energy PE = mgH = WH. When your
feet hit the ground (b), as you come to a stop you bend your
knees by a height x, ( c). At (b) your PE is now kinetic
energy KE = PE = WH. At ( c) you have no energy, the
work of stopping , E =F x, has eaten it up. That is
E =F x = - mgH = WH.
(8b)
x
So, the force to stop you is
F = mgH/x = W (H/x),
(9)
where W = mg .
If you jump from height H = 3 m, at (a), and after your feet
touch the ground, at (b), you bend your knees by x = 1/3
m , at ( c), the stopping force F is
F = W(
H
(a)
3 x3
3 x3
H
3m
3
) = W(
) = W(
) = W(
) = W(
) = 9W
1
1
1x 3
Δx
(1)
( )m
( )
(
)
3
3
3
(a)
(b)
c
In (a) above, the m (mass) is cancelled from top and bottom.
In (b) above, multiply top and bottom by 3 to get rid of the 1/3 in the denominator.
©Hulan E. Jack Jr. 2005 Revised 2006 ,2011
Page 6 of 8
Notes on Work and Energy
(b) (c)
NOTES ON WORK AND ENERGY
In ( c) above, cancel the 3's in the denominator (1 x 3/3) = (1 ) .
So, here F = 9W , nine times your weight! This stopping force is often called dynamic
loading. The total force Ftotal you exert on the floor is your weight plus this stopping force F,
Ftotal = W(to support you in gravity) + F (to stop you) =
W + 9W = 10 W = 10 times your weight.
Example - the Curb of Pain
This is an example explained by the Dynamic Loading . As you walk along crossing streets,
your “system” gets to expect a “fixed range” of drop
sidewalk 6 in
sidewalk 8 in
distance from sidewalk to street. Let’s suppose that you
had been encountering curbs that had about 6 inch drops. usual curb
bad curb
Your system gets to expect the street to be about 6 inches
below the sidewalk. Suddenly you encounter a bad curb with street 8 inches below the
sidewalk. Let’s follow the picture “STEPPING OFF THE CURB”, below. In (a) you start
stepping off the curb by moving your front leg forward off the side walk while keeping the
back leg in contact with the sidewalk. In (b)
your front foot reaches 6 inches below the
sidewalk . Your system expects the front foot to
be just contacting the street, so your system
releases your back foot from contact with
sidewalk. OOPS! There is nothing there. So, as
shown in (c) you go into free fall for the
remaining 2 inches. During that time of free
sidewalk
sidewalk 8 in 6 in
H=2 in
fall your body accelerates downward at the rate sidewalk 8 in
Free
2
a = g =9.8m/s . When you hit the street, you
Fall
bad curb
bad curb
bad curb
(a)
come to a stop. You must loose the KE at street
(b)
(c)
stepping
release
equal to the PE at the 2 inch top. So,
Free Fall
off curb
back leg
F = WH/x
WH is your PE at the top. x short displacement for stopping after you make contact with
the street. The x comes from shoe sole and leg bone and street material compressing, and
some knee bending. For this example let’s assume that x = 1/4 inch. So,
F = WH/x = W[2in] / [(1/4in)] = W[4*2in] / [(4*1)/4in] =W(4*2) = 8W.
This is a momentary stopping force acting on you - 8 times your WEIGHT. That’s why your
back screams out, it has to support 8 times its normal load!
©Hulan E. Jack Jr. 2005 Revised 2006 ,2011
Page 7 of 8
Notes on Work and Energy
NOTES ON WORK AND ENERGY
7.
Linear Momentum
Definition
m
Linear Momentum p = mv , a vector.
v
Impulse I = Ft.
FROM HERE ON t is specifically the time the bodies actually interact.
Impulse Momentum Theorem.
I = p ,
Hence Ft = p.
So if p is constant, the
t
Ft
t
p =
F
Ft ,
=
long interaction time causes small force
F
small (short) interaction time causes a large force.
(This is similar to the Dynamic Loading in Section 6, Eq.(9) above)
p = p1 + p = 0
2
p=0
All is at rest
m1
m2
INITIAL
a1
m1
F
BOOM!!
moving in opposite directions
-F
m2
BOOM!!
a2
v1
p1=
m1
m1
p2=m
v1
1
v2
v2
2
m1a t =m1v t)
(p1=
m2
1
(p2=m
2
a2t
= m v2 )
2
FINAL
Init
ially the masses m1 and m2 are just touching each other, in no way stuck together. Then an explosion
, “BOOM” , occurs between them causing them to fly apart.
During “BOOM!!” , by Newtom’s 3rd Law of Motion equal and opposite , and action-reaction pair,
act on the masses as shown. By Newton’s 2nd Law of Motion they have the accelerations shown.
Finally, after a short interaction time t, the two masses have the velocities, v1 = a1 t and v2 = a2 t
, shown. Since p1 and p2 are in opposite directions and equal magnitudes, their sum is zero, p1 + p2 =
0 . Hence the final momentum equals 0 as the initial momentum.
So
LINEAR MOMENTUM IS CONSERVED.
©Hulan E. Jack Jr. 2005 Revised 2006 ,2011
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Notes on Work and Energy