Transcript energy - Learning

Work
Work is the product of the ………………applied in the direction
of motion, and the ………………of the object.
W = … x ….
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The unit of work is ……… called a Joule (J).
One joule is the amount of work done when a force of …N acts over
…. m.
1J =
…………….
F
=
1…………-2.m
=
1kg.m2.s-2
Work is a ………… quantity. (…………….. Not necessary)
From the formula we see that if there is no ………………….. then there
is no work done.
If the displacement is …………………………… to the force then there is
no work done either.
Work
Work is the product of the resultant force applied in the direction
of motion, and the displacement of the object.
W = F • xcos
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The unit of work is N.m called a Joule(J).
The joule is the amount of work done when a force of 1N acts over
1m.
1J =
1N.1m
=
1kg.m.s-2.m

2
-2
=
1kg.m .s
x
Work is a scalar quantity. ( Direction not necessary)
From the formula we see that if there is no movement then there is
no work done.
If the displacement is perpendicular to the force then there is no
work done either.
Energy – Kinetic Energy
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Energy is the
capacity to do
…………..
Energy is a scalar
quantity and has the
same unit as work
viz. ………...
When ENERGY is
transferred ………….
is done.
If an object is
accelerated from rest.
FORCE
m
-2
a m.sm
m m
Energy of movement
W = f….
F = m….
=>W = ………
vf2 = vi2 +2a x
but
vi = 0
=> x = vf2 ..…..
W = ………………
=>
EK = ………………...
Energy – Kinetic Energy
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Energy is the
capacity to do
work.
Energy is a scalar
quantity and has
the same unit as
work viz. Joule.
When ENERGY is
transferred
WORK is done.
FORCE
m
-2
a m.sm
m m
Energy of movement
W = f.x
F = m.a
=>W = max
vf2 = vi2 +2ax
but
vi = 0
=> x = vf2 /2a
W = m.a.(vf2/2a)
=>
EK = 1/2mvf2
Potential energy
Potential energy is the energy that an object has
as a result of its ……………..or ………………...
Object lifted to a height “h”.
Energy gained (or transferred) = ……………….
Ep of an object = ………….. to lift the object to that height
W = ………
F = ………
W = ………….
Ep = ………….
Where h is the displacement of lifted object.
h
Potential energy
Potential energy is the energy that an object has
as a result of its position or state.
Object lifted to a height “h”.
Energy gained (or transferred) = work done
Ep of an object = work done to lift the object to that height
W = f.x
Fg = mg
W = mgx
Ep = mgh
Where h is the displacement of lifted object.
h
Potential energy Bouncing Efficiency
Observation: A ball dropped from a height bounces back to a lower
height than it was released from.
Task:
Investigate the efficiency of the bounce of a ball.
Theory:
The efficiency of the bounce can be expressed as the % of energy
of the bounce over the original potential energy of the ball.
h
initial
Efficiency = Initial Potential energy (top) x100
Potential energy of the bounce
Outcomes:
Write up all the steps of the scientific method and include:
Calculations of energy and efficiency from at least five
different heights. Also show at least one calculation of the
velocity of the ball on impact with the surface. Compare the
efficiency of at least two different balls.
h
bounce
Mechanical Energy
An object dropped from
REST
EK = ____
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Mechanical energy is the SUM of the
and
energy of an
object at any given time.
E=
.
The energy, which is converted or
transferred, is equal to the
to do
this.
If an object is moving and is brought to rest
then the
to do this is equal
to the
.
At any point during the fall of an object the
is equal to the
potential energy of the object
it
began to fall.
Ep =
.
Ep = ____
Ek =
.
Mechanical Energy
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Mechanical energy is the sum of the kinetic
and potential energy of an object at any
given time.
E = Ep + E`k
The energy, which is converted or
transferred, is equal to the work done to do
this.
If an object is moving and is brought to rest
then the work done to do this is equal to the
change in kinetic energy. Ek = work done
At any point during the fall of an object the
mechanical energy is equal to the potential
energy of the object before it began to fall.
If any energy is lost through work then
E = Ep + Ek + Work done
EK = 0
Ep = mgh
Ep = 0
Ek = (mgh)
Mechanical Energy
A 4kg ball is dropped 8m.
Ep = ……
m
4kg
Ek = ….J
Ep Potential
Energy LOST
8m
4m
EK kinetic energy
………..= Ep ……!
Ep = …
m
Ek = ……
E = Ep + Ek
E = Ep + Ek
= …+ …
= ……J
E = Ep + Ek
= … + ….
= 320J
E = Ep + Ek
=…+…
= …..J
Any object that is above the
ground has the POTENTIAL
to fall Down.
The potential energy EP
=mgh (Height) lost would be
converted into velocity
(kinetic energy) EK.
Ep(……) = Ek(……….)
If the object is lifted up then
it ……….. potential energy.
At any time the …….of these
two is called the
MECHANICAL ENERGY and it stays ……………….
Mechanical Energy
Ep = mgh
m
4kg
Ek = 0J
Ep Potential
Energy LOST
E = Ep + Ek
= 320 + 0
= 320J
Any object that is above the
ground has the POTENTIAL to
fall Down.
The potential energy EP =mgh
(Height) lost would be converted
into speed (kinetic energy) EK.
Ep(Top) = Ek(Bottom)
8m
4m
E = Ep + Ek
= 160 + 160
If the object is lifted up then it
= 320J
gains potential energy.
EK kinetic energy
gained = Ep lost!
Ep = 0
m
Ek = 320J
E = Ep + Ek
= 0 + 320
= 320J
At any time the sum of these two
is called the MECHANICAL
ENERGY - and it stays constant.
E = Ep + Ek
Conservation of energy - Pendulum
Energy cannot be created or
destroyed but merely
transferred from one form
to another.
Whenever there is energy
conversion there is …………
done.
A
C
h
B
The …………of the
pendulum is not required to
calculate the velocity.
Ep (Top) = Ek (Bottom)
………… = ………….2
=>
V = √………..
Conservation of energy - Pendulum
Energy cannot be created or
destroyed but merely
transferred from one form
to another.
Whenever there is energy
conversion there is work
done.
A
C
h
B
The mass of the pendulum
is not required to calculate
the velocity.
Ep (Top) = Ek (Bottom)
Mgh = 1/2mv2
=>
V = √2gh
A pendulum question
• A person of 60kg is lifted to a height of 30m on a slingshot
pendulum and then released what is its maximum speed?
Total Mechanical Energy
(Top)
ETop = Ep + EK
m
= mgh + 1/2mv2
= (60)(10)(30) + 0
h
= 18000J
EBottom = Ep + Ek
= mgh + 1/2mv2
18 000 = 0 + 1/2 (60)v2
v2 = 18000/30 = 600
v = 600 = 24.5 m.s-1
A pendulum
• A person of 60kg is lifted to a height of 30m on a slingshot
pendulum and then released what is its maximum speed?
Total Mechanical Energy
(Top)
ETop = Ep + EK
= mgh + 1/2mv2
m
= (60)(10)(30) + 0
= 18000J
h
EBottom = Ep + Ek
= mgh + 1/2mv2
18 000 = 0 + 1/2 (60)v2
v2 = 18000/30 = 600
v = 600 = 24.5 m.s-1
Power
• Power is the ………………….. at which
WORK is done.
Power =
Units: Watts (W) =
Since v = ……….., Power can be found by
P = ………….
If a force of 20N is exerted over a distance of 5m for a time of 30s the
power used would be.
Power
• Power is the RATE at which WORK is done.
Power =
Units: Watts (W) =
Work
time
Joules (J)
Time (s)
Since v = s/t, Power can be found by
P = F.v
If a force of 20N is exerted over a distance of 5m for a time of 30s the power
used would be. W = F.x = (20).5 = 100J P = W/t = 100/30 = 3.3 W