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Lecture 5
Work
Energy
Work, Energy
Work and energy are fundamental physical
quantities in science.
Work
is done when a force moves an object through
a distance.
Energy
is the ability to do work
The unit of both work and energy is the Joule.
James Joule 1818—1889
English Physicist
4186 J = 1kcal
Work
F
distance (s)
The work done by a force F moving an object
by a distance s in the direction of the force is
defined as:
W = Fs
units of Newton.metre (Nm) = Joule
If the force points in a direction different to that of
the displacement,
the work done is given by the component of the
force parallel to the direction of displacement
multiplied by the displacement.
Work
If the force points in a direction
different than the displacement
F
q
FCosθ
displacement (s)
W = (FCosθ)s
Only the component of force parallel to the
direction of displacement does work on the
object.
Both work and energy are scalars.
Energy is always positive.
Work is positive if both force (or component)
and displacement point in the same direction
Work is negative if they point in opposite
directions.
Work
Work done only when energy is transferred
into or out of a system
Examples:
Horizontal motion:
Vertical motion:
object carried
F
mg
The work
done by the
d weight is:
W=Fd
=(mg)d
d
mg
F
mg
The work done by
the force F is:
W = (FCosθ)d
W=mg*0*d = 0 J
When force and motion are perpendicular to
each other the work done is zero
Work is in the direction of force
In general, only the
projection of the
force in the
direction of the
displacement does
work:
W = FHd
Cos30o = FH ÷ F
FH = FCos30o
FH =10*(0.866)N
FH = 8.66N
W = FCos30o *d = FH*d = 8.66*d joules
Work
Example
How much work is done on an object
of mass 10kg
(a) As it’s lifted straight up through a distance
of 1m
(b) as it’s lowered straight down through a
distance of 1m.
W = (FCosθ)s
W=(FCos0)s=Fs
(a)
F=mg
finish
W=mgs=10kg(9.8ms-2)1m
mg
W=98N.m = 98J
s
start
(lifting)
Work is positive since
the force is in same
direction as movement
Example contd
(b) as it’s lowered straight down through a
distance of 1m.
W = (FCosθ)s
start
s
W=FCos(180o)s = -Fs
F=mg
finish
mg
Force & s are in opposite
directions
W = -mgs
= -10kg(9.8ms-2)1m
W = -98N.m = -98J
(lowering)
Work is negative since the applied force
is in opposite direction to movement
Work is transfer of energy
V = 0km/h
V = 100km/h
W
The engine of the car does work:
It changes the ‘chemical’ energy of
the fuel into motion
 Work is transfer of energy
Energy
Sun is the ultimate source of energy on Earth
Heat and light
Almost all life on Earth is dependent
on the Sun
We get energy from food
Food grows by capturing sun’s energy
by a process called photosynthesis
Solar energy →chemical energy
Whatever we eat, energy within it comes
from the sun
Fossil fuels: turf, coal, oil, gas, all forms of energy
Energy needed to form
these came from the sun
Fossil fuels are made from plants and
animals that lived and died in swamps
millions of years ago.
Energy (J)
Energy is the ability to do work.
Work done by a system takes energy
out of the system
Work done on a system puts energy into it.
Energy can exist in many forms:
-Kinetic energy: the energy linked to
movement
-‘stored’ energies: elastic energy,
chemical energy, electrical energy,
gravitational potential energy,…
-‘Thermal energy: proportional to
temperature
-Radiation energy: electromagnetic waves
(light, etc…)
Work and Energy
Consider work done on a system,
producing motion.
If all the work done goes into causing motion then
the energy of motion is equal to the work done.
From Kinematics
v  v  2as
2
2
0
v 2  v02
a
2s
If we multiply a
by the mass m
of the object
mv 2  mv02
F  ma 
2s
mv 2 mv02
Fs 

2
2
mv 2
Fs 
2
If initial speed is zero
Work and Energy
mv 2
Fs 
2
Fs is just the work done on the object
by the applied force F.
The quantity
1 2
mv
2
is known as the Kinetic energy (KE)
of a object with mass m traveling at speed v.
1 2
KE  mv
2
Kinetic Energy
Kinetic energy is the energy associated with
an object in motion:
1 2
KE  mv
2
m: mass of the body
v: speed of the body
Example:
V = 0km/h
V = 100km/h
Mass of the car: 1ton =1000kg
Calculate final KE and work done by engine
Initial kinetic energy: KEinit = ½*1000*02 J = zeroJ
Final kinetic energy: KEfinal =
½*1000*(100/3.6)2 J = 386kJ
 The work done by the engine is:
W = 386kJ
Kinetic energy
Example
A father (mf=80kg) is
pushing a buggy
(mb=20kg). He exerts a
force of 10N directed 30°
below the horizontal.
Assume friction is
negligible. What is their
speed after 2m, starting
from rest?
W = FHd = F*cos(30°)*d
= 8.66N*2m = 17.32J
Einit = 0J
Efinal = ½(mf+mb)v2
Efinal = W = 17.32J
1
17.32Joules= (m f  mb )v 2
2
v
2 17.32
2 17.32

 0.59m / s
m f  mb
80  20
v  0.59ms
1
Potential Energy
We have seen that Kinetic energy is the
energy related to an object’s motion.
There is another type of energy which is
related to a object’s shape or position, this is
known as Potential energy (PE).
Examples:
Shape:
Coiled Watch spring,
Position
Water at the top of a hydroelectric dam
We will consider potential energy mainly
related to position.
Potential Energy
Example:
A man lifts a box of mass m
from the ground to a height h.
What is the work done on the
box?
h Work done = energy given to box
W = force x distance = Fh
W = mgh
Initial and final kinetic energy is zero (v=0)
Energy of box is that associated with its position
Gravitational potential energy
PE = mgh
Potential Energy
Example
A man lifts a box of mass
20kg from the ground to a
height of 1.5m. What is the
work done on the box?
Initial potential energy: PEinit = 20*9.8*0 J = 0 J
Final potential energy: PEfinal 20*9.8*1.5J = 294J
 The work done by the man is: W = 294J
PE = m g h
J kg m.s-2
m: mass of the body
h: altitude of the
object
m
Linking KE and PE
Total energy, kinetic, potential, chemical
etc in a closed system is constant or
conserved.
This is an example of the
principle of conservation of energy.
System is closed if no energy enters or leaves
it by any method
Energy can be transferred only between
components within the system
principle of conservation of energy
KE +PE+OE =constant
OE is the sum of all other forms of energy
Energy can be transferred within the system
Initial total energy equals the final total energy
KEi +PEi+OEi = KEf +PEf+OEf
Linking KE and PE
Initial energy
=
Final energy
(1/2)mvi2 + mghi +OEi = (1/2)mvf2 + mghf +OEf
initial
kinetic
energy
initial
initial all final
kinetic
potential other
energy energy energy
final
potential
energy
Final all
other
energy
Just consider KE and PE then
(1/2)mvi2 + mghi = (1/2)mvf2 + mghf
In other words the sum of Kinetic and Potential
energy is the same before and after the force
acted on the object.
Linking KE and PE
An object drops vertically, starting
with zero velocity. If air resistance
(friction) is negligible, what is its
speed after 10m?
mg
h
Initial kinetic energy = ½mvi2 = 0
Initial potential energy =mgh
Potential energy decreases:
Kinetic energy increases:
energy transfered from potential to kinetic:
Using conservation of energy KE = PE
mgh = ½mv2
v  2 gh
v  v  2as
2
2
0
v  2as
v = 14m.s-1
Linking KE and PE
A sledge with zero initial
velocity slides down a
bumpy slope. What is
its final speed, after
loosing 5m of altitude?
Assume
friction
is
negligible
h
KEi +PEi+OEi = KEf +PEf+OEf
Since there is no friction
OEi = OEf
0 + PEi = KEf + 0
KE= ½mv2
PE= mgh
v  2gh
mgh = ½mv2
v =√2*9.8*5 ms-1
v = 9.9m.s-1
This approach works whatever the complexity
of the trajectory.
Conservation of energy
If the final velocity of the sledge (10kg) is 8 ms-1,
instead of 9.9 ms-1? How much energy is
converted to thermal energy by friction on the
way down?
KEi +PEi+OEi = KEf +PEf+OEf
0 + mghi +0 =1/2mvf2 + 0 + thermal energy
Thermal energy = mgh -1/2mvf2
= 10*9.8*5 - ½*10*82 = 170J
Friction transfers mechanical energy into
thermal energy
Thermal energy is another form of energy:
associated with the random kinetic energy
of atoms and molecules in an object.
Roller coaster?
Exercise:
A roller coaster, starting with
1m/s, 20m above ground.
What is its final speed if
friction is neglected?
Start
20m
End
KEi +PEi+OEi = KEf +PEf+OEf
(1/2)mvi2 + mghi = (1/2)mvf2 + mghf
(1/2)mvi2 + mghi = (1/2)mvf2 + 0

2 2gh
v vinit
 v = 19.8ms-1