F - Uplift Education

download report

Transcript F - Uplift Education

Work and Energy
 Work: The word looks the same, it spells the same
but has different meaning in physics from the way it
is normally used in the everyday language
Definition of work W done by a constant force F
exerted on an object through distance d
F
θ
F
Fd
θ
Fd
d
 Only the component that acts in the same direction as the motion is
doing work on the box.
 Vertical component is just trying to lift the object up.
work = force along distance × the distance moved
W = Fdd = Fd cosθ
work = force × distance moved × cos of the angle between them
Units
 The SI unit for work is the newton–metre and is called
the joule named after the 19th Century physicist
James Prescott Joule.
 1 J (Joule) = 1N x 1 m
 Work is a scalar (add like ordinary numbers)
A force is applied. Question: Is the work done by that force?
Work - like studying very hard, trying to lift up the car and get
completely exausted, holding weights above head for half an hour is
no work worth mentioning in physics.
 According to the physics definition, you
are NOT doing work if you are just
holding the weight above your head
(no distance moved)
 you are doing work only while you are
lifting the weight above your head
(force in the direction of distance moved)
Who’s doing the work around here?
NO WORK
WORK
 If I carry a box across the room I do not do work on it because
the force is not in the direction of the motion (cos 900 = 0)
θ = 00
00< θ <900
θ = 900
900< θ <1800
cos θ = 1
cos θ = +
cos θ = 0
cos θ = –
θ = 1800
cos θ = –1
Work done by a force F is zero if:
 force is exerted but no motion is involved:
no distance moved, no work
 force is perpendicular to the direction of motion (cos 900 = 0)
d
d
F
F
F
motion
normal
force
tension in
the string
gravitational
force
Work done by force F is:
 positive
when the force and direction of motion are generally in the same directions
(the force helps the motion – can increase kinetic energy by increasing speed)
00< θ < 900 → cos θ = +
cos 00 = 1
W = Fd
 negative
when the force and direction of motion are generally in the opposite directions
(force opposes the motion – decreases kinetic energy by decreasing speed)
900< θ < 1800 → cos θ = –
cos 1800 = –1
W = - Fd
(the work done by friction
force is always negative)
Mike is cutting the grass using a human-powered lawn mower.
He pushes the mower with a force of 45 N directed at an angle
of 41° below the horizontal direction. Calculate the work that
Mike does on the mower in pushing it 9.1 m across the yard.
F = 45 N
d
410
F
W = Fd cos θ = 310 J
d = 9.1 m
θ = 410
Forward force is 200 N. Friction force is 200 N.
The distance moved is 200 km. Find
a. the work done by forward force F on the car.
b. the work done by friction force Ffr on the car.
c. the net work done on the car.
F = 200 N
a. WF = Fd cos 00
Ffr = 200 N
b. Wfr = Ffr d cos 1800 = - 4x107 J
d = 2x105 m
c. the net work done on the car means the
work done by net force on the car.
It can be found as:
= 4x107 J
W = WF + Wfr = 0
or
W = Fnet d cos θ = 0
(Fnet = 0)
Work done by a varying force - graphically

W = Fd cos θ applies only when the force is constant.

Force can vary in magnitude or direction during the action.

Examples: 1) rocket moving away from the Earth – force of gravity
decreases 2.) varying force of the golf club on a golf ball

Work done is determined graphically.


The lady from the first slide is pulling the car for 2 m with force of 160 N
at the angle of 60o , then she gets tired and lowers her arms behind her at
an angle of 45o pulling it now with 170 N for next 2 m. Finally seeing the
end of the journey she pulls it horizontally with the force of 40 N for 1 m.
Work done by her on the car is:
W = (160 N)(cos 60o)(2m) +(170 N)(cos 45o)(2m) + (40 N)(cos 0o)(1m)
W = 80x2 + 120x2 + 40x1 = pink area + green area + blue area = 440 J
http://www.kcvs.ca/map/java/applets/workE
nergy/applethelp/lesson/lesson.html#1
In general:
The area under a Force - distance graph equals
the work done by that force
Energy, E
In physics energy and work are very closed linked; in some senses
they are the same thing. If an object has energy it can do a work. On
the other hand the work done on an object is converted into energy.
work done = change in energy
W=∆E
Gravitational potential energy, PE = mg h
Imagine a mass m lifted a vertical distance h following two different paths.
F2
d
θ
F1
h
mg
1. straight up:
– minimum force needed is F1 = mg.
– the work done against gravitational force is:
W = F1 h cos 00
W = mg h
2. along a ramp a distance d (no friction)
– minimum force needed is F2 = mg sinθ.
– the work done against gravitational force is:
the work W = mg h is stored
as potential energy of the object.
PE = mg h
It depends on VERTICAL distance
from initial position not on the path taken.
W = F2 d cos 00 = mg sinθ d
W = mg h
A ramp can reduce the force – can make life easier
h
d
θ
h
1. W = F h = mg  h
2. W = F  d = mg sinθ  h/sinθ = mg  h
W = F h
or
F
d (along the ramp)
Kinetic energy, KE = ½ m v2
A moving object possesses the capacity to do work. A hammer by virtue
of its motion can be used to do work in driving a nail into a piece of wood.
How much work a moving object is capable of doing? Let’s find out.
Imagine an object of mass m accelerated from rest by a resultant
force, F. After traveling a distance d the mass has velocity v.
The work done is:
W = Fd cos 00 = Fd = mad
From v2 = u2 + 2ad → ad = v2/2
and W = ½ mv2
The work done (Fd) by the force F in moving the object a
distance d is now expressed in terms of the properties of the body
and its motion. The quantity ½ mv2 is called the kinetic energy
(KE) of the body and is the energy that a body possesses by
virtue of its motion. The kinetic energy of a body essentially tells
us how much work the body is capable of doing.
Summary:
 Work done on an object by applied force against gravitational force
(when the net force is zero, so there is no acceleration) is stored
as gravitational potential energy.
 PE depends on the reference frame in which it is measured, so we
keep in mind some reference level, like desk, lower reservoir of
water, ground,…. PE = mg ∆h
 If the net force acting on a body is not zero, then:
The work done by net force on a body is equal (results)
to the change in the kinetic energy of the body:
W = ∆KE = KEf – KEi
Units
(W)
PE = mgh
(PE) = (1kg)(1 m s-2 )(1m) = 1 kg m2 s-2 = 1 J
KE = ½ mv2
= (1N)(1m)
= 1 kg m2 s-2 = 1 J
W = Fd
(KE) = (1kg)(1m/s)2
= 1 kg m2 s-2 = 1 J
Even in units we see that the work and energy are equivalent.
(1)
A father pushes his child on a sled on level ice, a distance 5 m
from rest, giving a final speed of 2 m/s. If the mass of the child
and sled is 30 kg, how much work did he do?
W = ∆ KE = ½ m v2 – 0 = ½ (30 kg)(2)2 = 60 J
other way: W = Fd cosθ = Fd
F = ma
v 2 -u2
a=
= 0.4 m/s2
2d
(2)
F = 12 N
W = 60 J
What is the average force he exerted on the child?
W = Fd = 60 J, and d = 5 m, so F = 60/5 = 12 N
(3) A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the
road does work on the car through a backward-directed friction force.
How much work must this friction force do in order to stop the car?
W = ∆ KE = 0 – ½ m u2 = – ½ (1000 kg) (45 x1000 m/3600 s)2
W = – 78125 J = – 78 kJ
(the – sign just means the work leads to a decrease in KE)
u
v=0
Ffr
d
(work done
by friction
force)
(4) Instead of slamming on the brakes the work required to stop the car is
provided by a tree!!!
What average force is required to stop a 1000-kg car going at 45 km/h if the
car collapses one foot (0.3 m) upon impact?
W = – 78125 J
W=–Fd
F=
-W
78125
=
d
0.3
F = 260 x 103 N
The net force acting on the car is F
corresponds to 29 tons hitting you OUCH
Do you see why the cars should not be rigid. Smaller collapse distance,
greater force, greater acceleration. For half the distance force would
double!!!!
OUCH, OUCH
a = (v2 - u2 )/2d = - 520 m/s2 HUGE!!!!
The principle of energy conservation !!!!!!
 we shall see in nuclear physics that this should
actually be the principle of mass-energy conservation
 To demonstrate the so-called principle of energy conservation
we will solve a dynamics problem in two different ways, one
using Newton’s laws and the kinematics equations and
the other using the principle of energy conservation.
An object of mass 4.0 kg slides from rest without friction down an
inclined plane. The plane makes an angle of 30° with the horizontal
and the object starts from a vertical height of 0.5 m. Determine the
speed of the object when it reaches the bottom of the plane.
h
d=
sinθ
v=?
h=0 B
N
u=0
A
point A: h = 0.5 m, KE = 0 (u = 0)
point B: h = 0 , object has gained KE
d
h=0.5m
300
Method 1: Newton’s law
a=
F
mgsinθ
=
= gsinθ
m
m
v2
u2
=
 h 
+ 2ad = 2gsinθ 

 sinθ 
v = 2gh
v = 3.2 m/s
Method 2: Energy conservation
As the object slides down the plane its
PE becomes transformed into KE.
If we assume that no energy is ‘lost’ we
can write
PEA = KEB
mgh = ½ mv2
v = 2gh
v = 3.2 m/s
Note that the mass of the object does not come into the question, nor does the
distance travelled down the plane, only vertical height.
 The second solution involves making the assumption that potential
energy is transformed into kinetic energy and that no energy is converted
into other forms of energy (heat, sound,…).
This is ‘Law of conservation of energy’, this means the energy is
conserved. Energy cannot be created or destroyed; it may be
transformed from one form into another, but the total amount of energy
never changes.
 When using the energy principle we are only concerned with the initial
and final conditions and not with what happens in between.
 No time included in energy principle.
 Very powerful tool of enormous importance.
 Clearly in this example it is much quicker to use the energy principle.
This is often the case with many problems and in fact with some
problems the solution can only be achieved using energy considerations.
 If it is transformed into thermal energy, we say it is
dissipated. The object can’t keep that energy. It is
shared with environment.
 and now problems and examples. Some of them
would be very, very hard to solve using kinematic
equations and Newton’s laws.
Dropping down from a pole. As he dives,
PE becomes KE. Total energy is always constant,
equal to initial energy.
If accounted for air resistance, then how
would the numbers change?
In presence of air, some energy gets transformed to
heat (which is random motion of the air molecules).
Total energy at any height would be PE + KE + heat,
so at a given height, the KE would be less than in
vacuum.
What happens when he hits the ground?
Just before he hits ground, he has large KE (large
speed). This gets transformed into heat energy of
his hands and the ground on impact, sound energy,
and energy associated with deformation .
Amusement park physics
 the roller coaster is an excellent
example of the conversion of
energy from one form into another
 work must first be done in lifting the
cars to the top of the first hill.
 the work is stored as gravitational
potential energy
 you are then on your way!
PE is being transformed into
KE and vice versa
Up and down the track
PE
PE
PE + KE
If friction is negligible the ball will get
up to the same height on the right side.
Three balls are thrown from the top of the cliff along paths A, B, and C with
the same initial speed (air resistance is negligible). Which ball strikes the
ground below with the greatest speed?
a. A
b. B
c. C
d. All strike with the same speed
h
The initial PE + KE of each ball is: mgh + ½ mu2.
This amount of energy becomes KE before impact ½ mv2.
mgh + ½ mu2 = ½ mv2
m cancels
The speed of impact for each ball is the same: v =
2gh + u2
It depends ONLY on HEIGHT and initial SPEED, not mass, not path !!!!!
A child of mass m is released from rest at the top of a water slide, at
height h = 8.5 m above the bottom of the slide. Assuming that the slide
is frictionless because of the water on it, find the child’s speed at the
bottom of the slide.
Do you think you could use kinematics
equations and Newton’s laws? Try it. Good luck !
mgh = ½ mv2
v = 2gh = 13 m/s
m cancels
on both
sides
a baby, an elephant and you would reach
the bottom at the same speed !!!!!
And, by the way, it is the same speed as you were to fall straight down
from the same height.
v2 = u2 + 2gh = 2gh
Work is a way of transferring energy from one form to another, but
itself is not a form of energy. All types of energies and work have the
same units. Everything can be transformed into each other.
When work is done on an object, energy is
transferred to that object.
Example: A spring at the bottom of a slide is
compressed by an external force. A parent
releases the spring when a child sits on it.
EPE is transferred to the child by spring
force doing the work which is transformed
into KE of the child. On the way up, KE is
being transformed into PE.
This energy is what enables that child to
then do work. How? First that PE has to be
transformed back into KE – child slides
down a slide – KE of the child can do work
on the waiting parent – it can knock him
over.
Parent does the work on the ground.
A car (toy car – no engine) is at the top of a hill on a
frictionless track. What must the car’s speed be at the top of
the first hill if it can just make it to the top of the second hill?
v2 = 0
PE1 + KE1 = PE2 + KE2
mgh1 + ½ mv12 = mgh2 + ½ mv22
v1
m cancels out ; v2 = 0
gh1 + ½ v12 = gh2
40m
20m
v12 = 2g( h2 – h1)
v1 = 22 m/s
A simple pendulum consists of a 2.0 kg mass attached
to a string. It is released from rest at A
as shown. Its speed at the lowest point B is:
PEA + KEA = PEB + KEB
A
1.85 m
B
vB =
2ghA =
1
1
2
mv A + mghA = mv B 2 + mghB
2
2
1
0 + ghA = mv B 2 + 0
2
2×9.80m/s2×1.85m = 6m/s
● Conservation of energy law AGAIN – WITH FRICTION INCLUDED
All previous examples were neglecting friction force, air resistance …
What happens if the friction force acts?
Can we still use conservation of energy principle?
YES, WE CAN
● Example: An object of mass 4.0 kg slides 1.0 m down
an inclined plane starting from rest. Determine the
speed of the object when it reaches the bottom of
the plane if
a.
friction is neglected
all PE was transformed into KE
initial energy = final energy
b. constant friction force of 16 N acts
on the object as it slides down.
Friction converts part of KE of the object
into heat energy. This energy equals to
the work done by the friction. We say that
the frictional force has dissipated energy.
initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object
PEA = KEB
mgh = ½ mv2
v = 2gh = 3.2 m/s
PEA – Ffr d = KEB
mgh – Ffr d = ½ mv2
20 – 16 = 2.5 v2
v = 1.4 m/s
Types of energy and energy transformations
We usually classify energy as the following forms:
1 Kinetic energy
2 Gravitational potential energy
3 Elastic potential energy
4 Thermal or heat energy
5 Light energy
6 Sound energy
7 Chemical energy
8 Electrical energy
9 Magnetic energy
10 Nuclear energy
There are energy changes happening all around us all of
the time. Here are a few examples to think about.
In any given transformation, some of the energy
is almost inevitably changed to heat.
• A bouncing ball
The gravitational potential energy is changed to kinetic
energy and back again with each bounce losing energy
due to work done by air friction and nonelastic contact
with the ground . Eventually the ball comes to rest and
all the energy has become low grade heat.
• An aeroplane taking off
As fuel is burned in the engines, chemical energy is converted
to heat, light and sound. The plane accelerates down the
runway and its kinetic energy increases. There will be heat
energy due to friction between the tyres and the runway. As
the plane takes off and climbs into the sky, its gravitational
potential energy increases.
• A nuclear power station
Nuclear energy from the uranium fuel changes to thermal
energy that is used to boil water. The kinetic energy of the
steam molecules drives turbines, and as the kinetic energy of
the turbines increases, it interacts with magnetic energy to
give electrical energy.
• A laptop computer
The electrical energy form the mains or chemtcal energy from
the battery is changed to light on the screen and sound through
the speakers. There is kinetic energy in the fan, magnetic energy
in the motors, and plenty of heat is generated.
• A human body
Most of the chemical energy from our food is changed to heat
to keep us alive. Some is changed to kinetic energy as we
move, or gravitational potential energy if we climb stairs. There
is also elastic potential energy in our muscles, sound energy
when we talk and electrical energy in our nerves and brain.
Power
◘
Power is the work done in unit time or energy converted in unit time
P=
W
t
or P =
E
t
measures how fast work is done or how quickly energy is converted.
Power is a scalar quantity.
◘
Units:
1 W(Watt) =
1J(joule)
1s
A 100 W light bulb converts electrical energy to heat and light at
the rate of 100 J every second.
Sometimes you’ll see power given in kW or even MW.
Calculate the power of a worker in a supermarket who
stacks shelves 1.5 m high with cartons of orange juice,
each of mass 6.0 kg, at the rate of 30 cartons per minute.
W
Fdcos00 (30×60N)×1.5m
P=
=
=
t
t
60s
P = 45 W
There is another way to calculate power
W
Fdcos00
d
P=
=
=F
t
t
t
P = Fv
Power is equal to force times velocity, providing that both
force and velocity are constant and in the same direction.
Constant velocity with a force applied ??????
That’s because we are interested in one force only. Net force
is obviously zero. Like power of the engine of the car.
Efficiency
◘
Efficiency is the ratio of how much work, energy or power we get
out of a system compared to how much is put in.
useful output
efficiency =
total input
eff =
Wout
E
P
= out = out
Win
Ein
Pin
◘
No units
◘
Efficiency can be expressed as percentage by multiplying by 100%.
◘
No real machine or sysem can ever be 100% efficient,
because there will always be some energy changed to
heat due to friction, air resistance or other causes.
A car engine has an efficiency of 20 % and produces an
average of 25 kJ of useful work per second.
How much energy is converted into heat per second.
Eout
eff =
Ein
0.2 =
25000J
Ein
Ein = 125000 J
heat = 125 kJ – 25 kJ = 100 kJ