Download Energy Unit Don 2013

Don
College
PHYSICAL SCIENCES 3
Work, Energy and
Power
1
Contents
1.
Forms and sources of energy
3
2.
Work
6
3.
Potential Energy (EP)
9
4.
Kinetic Energy (EK)
11
5.
Work done, energy transformations, etc
13
6.
Energy efficiency
20
7.
Power
22
Compiled by teachers of the Tasmanian Academy and Polytechnic Hellyer Campus,
with assistance from Jason Dicker (Launceston College).
2
1. FORMS & SOURCES OF ENERGY
„Energy‟ is a very common word in our everyday language. We talk about
the energy needed to run the machines in our factories, the energy to move a
cable car, energy to play a football game. Energy is supplied to us in the
form of fuels. Petrol is the fuel that makes a car move. Food is our source of
energy (our fuel!) enabling us to live.
Energy is something that can do useful jobs for us. There are many sources
of energy:
mechanical energy, electrical energy, chemical energy, heat energy, solar
energy, nuclear energy, etc.
Forms of energy
There are two major forms of energy, kinetic energy and potential energy.
Any object that is moving has kinetic energy.
A moving bowling ball can do work on the pins as it hits them. The
faster the ball is moving the more energy it has and the greater is the
amount of work it can do. Kinetic energy is the energy an object has
because of its movement.
Any object that is lifted above the ground gains (gravitational) potential energy. The higher it
is lifted the more potential energy it gains, and the more work it can do if it is released.
Potential energy is the energy an object possesses because of its change in height.
All energy forms can be classified as potential or kinetic energy, but sometimes people find
it easier to classify energy in terms of the source from which it comes.
Sources of energy
Electrical energy
Electrical energy is a form of energy that passes through wires due to the movement of
electrons. The flow of electrical current can be from a power station to an electrical appliance
in a home or from a car battery to the starter motor of the car. Electrical
energy can be used to do work in a variety of appliances. Almost every
conceivable kind of work can be done by electricity. Even the human
body uses electrical impulses to send messages back and forth between
various organs of the body.
Electrical currents can do work because they have energy.
3
Sources of energy (cont.)
Magnetic energy
Magnetic energy is the energy released when two magnetic fields interact.
Often electrical energy and magnetic energy are associated together
because magnetic fields are produced by electrical currents.
Chemical energy
Chemical energy is the energy stored when elements join together to form
compounds. When reactions occur the elements are rearranged and energy may
either be released or absorbed. A stick of dynamite releases chemical energy when
it explodes. Food contains stored chemical energy, and it is this energy
that keeps our bodies functioning.
Heat energy
Heat energy is a form of energy people are very aware of because it is used
in so many ways. It is the energy possessed by a substance in the form of kinetic energy of the
atoms or molecules which make up the substance. A change in the heat content
results in a change in temperature or a change of state. When any of these occur
work is done.
Light energy
Light energy, like heat energy, is a form of energy of which people are very aware
and that is critically important on Earth. Light energy from the Sun is the source of
life on Earth because it is converted to stored chemical energy in plants by the
process of photosynthesis.
For example, electrical energy is produced when light strikes a photovoltaic cell,
such as those on hand-held calculators and modern watches.
Sound energy
Sound energy is produced when air particles are forced to vibrate. The sound
energy in turn forces people‟s eardrums to vibrate, which in turn causes electrical
impulses to be sent to their brains.
Nuclear energy
Nuclear energy is released by the fission of nuclei in a nuclear reaction.
Energy from nuclear reactions can be released in a controlled way in nuclear reactors.
Questions
1.
What sources of energy are associated with each of the following? There may
be more than one type for each.
a
a bowl of breakfast cereal
b
a full tank of petrol
c
a bowling ball rolling down a bowling alley
d
a bow for firing arrows
e
an electric toaster
f
a dry cell (battery)
g
a nuclear warhead
4
2.
3.
All of the following devices convert energy from one form into another. What type of
energy does each receive, and what does it turn each into? For example, a solar cell
receives light energy and converts it into electrical energy.
a
a light bulb
b
a motor cycle engine
c
a loudspeaker
d
an electric kettle
e
a windmill
f.
a leaf
Name a device or appliance which converts the first-named form of energy into the
second.
a
electrical to light
d
electrical to kinetic
b
heat to electrical
e
chemical to electrical
c
sound to electrical
f
electrical to chemical
ANSWERS
1.
2.
3.
(a)
Chemical energy
(b)
Chemical energy
(c)
Kinetic, heat and
sound energy
(d)
Potential energy
(e)
Electrical, heat and light
energy
(f)
Chemical and
electrical energy
(g)
Nuclear and heat, light,
kinetic energy
(a)
Electrical  light and heat energy
(b)
Chemical energy  heat, light, sound and
kinetic energy
(c)
Electrical  magnetic, kinetic and
sound energy
(d)
Electrical  heat and kinetic
(e)
Kinetic  heat and electrical energy
(?)
(f)
Light energy  chemical energy
(a)
Light bulb
(b)
Generator
(c)
Microphone
(d)
Fan
(e)
Battery
(f)
Battery recharger
5
2. WORK
A useful job involves energy transfer or energy change.
When a person carries a load up the stairs of a building, work is done. To carry a greater
load or to go up a few more floors, more work is done. Force and displacement are two
quantities involved with work. Work is done when a force produces a displacement.
W = Fs = mas
The unit for work is the joule (J). As 1 Joule is such a small measure of work we often use
kJ (kilo Joules) or MJ (Mega Joules) instead.
Work is a scalar quantity although force and displacement are vectors.
Please note that the use of the word „work‟ in science is different from that in ordinary
language.
CAUTION!!!
When a person holds a very heavy load and keeps it at rest, its displacement is
zero. If the person was to carry this load horizontally at a constant velocity the
force applied to it is zero.
In both cases the work done on the load is equal to zero.
The concept of work is quite abstract. It is very closely related to energy.
A useful relationship is:
Work Done = Change in Energy
We will look at how this relationship applies later.
Example 1
A force of 5.0 N moves a body 8.0 m in the direction of the force. What is the work done?
W = Fs
W = 5.0  8.0
W = 40 J
Example 2
Find the work done by a car of mass 1000 kg when it is accelerated from rest at 2.0 ms-2 in 20 s.
As we don‟t know displacement, this must be found first!
u=0
s = ut +
1 2
at
2
v = 2 ms-1
s=0+
1
 2  202
2
t = 20 s
s = 400 m
m = 1000 kg
Fs = mas = 1000  2.0  400
a = 2.0 ms-2
W = 800 000 J or 800 kJ
s t a v u!
6
Questions
6.
A force of 250 N is applied on a body of mass 71 kg, so that it moves through
0.900 m in the direction of the force. Find the work done on the body by the
force.
7.
How much work is done by a person who lifts a 7.5 kg object from the floor to
a shelf 2.0 m above the floor? (Remember ag!)
8.
A force is applied to a 5.0 kg object to bring it to rest from 6.0 ms-1 in 2.0 s.
Find
(a)
the acceleration and the force applied.
(b)
the distance that it travels.
(c)
the work done by the force.
9.
An oil drum is rolled along a smooth surface for a distance of 25.0 m with a force of 18.0
N. How much work has been done?
10.
A car is accelerated at 5.0 ms-2 for a distance of 20.0 m. If the mass of the car is 1500 kg,
how much work has been performed?
11.
If an object requires 200 J of work to be accelerated at 1.5 ms-2 for a distance of 40 m,
determine the mass of the object?
12.
How much work has been done if a car of mass 2 000 kg is decelerated at 1.5 ms-2 for a
distance of 80 m?
13.
A skater, initially at rest, does 250 J of work to accelerate at 1 ms-2. If
his mass is 70 kg, how far has he travelled during his acceleration?
14.
A box of mass 12 kg requires a force of 4 N to overcome friction so
that it can be moved at constant velocity over a distance of 1.5 m.
(a)
How much work is required?
(b)
What additional work would have been required if the box had
been accelerated at 0.2 ms-2 for the 1.5 m shift?
15.
A fork-lift raises crates of mass 600 kg a height of 3.0 m. How much
work is done?
16.
What is the mass of a lift if 400 000 J is required to raise the lift a height of 20.0 m?
17.
Convert the following to joules:
a
8 kJ
b
1.5 kJ
c
4.5 MJ
d
0.85 kJ
7
Relationship between Work and Energy
If a person does work they can change the energy of an object. This can either be kinetic,
potential or some other special form of these. As energy is conserved, it is perhaps useful to
think of the chemical energy provided by the food as being converted to other forms of energy.
On the other hand inanimate objects can be said to be doing work. An example of this is an
object with kinetic energy slowing down and coming to rest. In this case, its kinetic energy has
been converted to sound and heat and the frictional force that brought it to rest is said to have
done work.
Use the following:
∆E = Work Done = Fs
where F is the frictional force bringing the object to rest. If we know how far the object moves
before coming to rest we can use this concept to calculate the frictional force.
ANSWERS
6.
225 J
7.
147 J
8.(a)
–3 ms-2 & –15 N
(b)
6m
(c)
90 J
9.
450 J
10.
150 kJ (150 000 J)
11.
3.33 kg
12.
240 kJ (240 000 J)
13.
3.57 m
14.(a)
6J
(b)
3.6 J
15.
17 640 J (17.6 kJ)
16.
2040 kg
17.(a)
8 000 J
(b)
1 500 J
(c)
4 500 000 J
(d)
850 J
8
3. POTENTIAL ENERGY (EP)
Energy is the capacity to do work. Energy is measured by the quantity of work it can do, and
so it has the same unit as work: the joule.
Energy like work is also a scalar quantity.
To lift an object of mass m through a change in height (  h) on Earth, work has to be done. The
force is applied vertically upwards and is given by
F = mag
(where ag is the acceleration due to gravity = g)
The work done is therefore
W = Fs = mags = mg  h
This is called the gravitational or simply potential energy increase of the object.
EP = mg  h
Example 1
A person whose mass is 50 kg walks from the ground floor
to the second floor of a building 12 m above. What is the
person‟s gain in potential energy?
12 m
Solution
EP = mg  h
EP = 50  9.8  12
EP = 5880 J or 5.9 kJ
Example 2
It doesn‟t matter what path the boy takes to get from A to B. As the change in height (h) is the
same, the potential energy gain is the same though their paths are different.
B
B
A
A
Potential energy is independent of the path taken.
9
3. POTENTIAL ENERGY (EP) (cont.)
Example 3
How much work is required to lift a 2 kg mass from a height of 50 m to a height of 60 m above
the ground?
Solution
Work required = change in potential energy
EP = mg  h
(notice how this is identical to W = mag  h!)
EP = 2  9.8  10
EP = 196 J
QUESTIONS
20.
Find the potential energy of a 2.00 kg mass at a height of 40.0 m above the ground.
21.
How much work is required to raise a mass of 1.0 kg to a height of 20 m?
22.
How much work is done in lifting a 3.0 kg mass from 10 m to 15 m above the ground?
23.
An object falls to the ground from a height. Is potential energy gained or lost? Explain.
24.
An object is raised above the ground. Is potential energy gained or lost? Explain.
25.
An exercise machine is arranged so that a person lifts an iron block through a height of
1.50 m.
What is the mass of the block if the potential energy of it increases by 1200 J each lift?
26.
How high must a pile driver of mass one tonne be raised to gain 245 kJ of potential
energy?
27.
A loaded skip of total mass 2.0 tonne requires 2.94 MJ of work to be done to lift the load
from the working level of a mine to the surface. Use this information to calculate the
depth of the mine.
ANSWERS
20.
784 J
21.
196 J
22.
23.
EP is lost as height is
“lost”/converted to other
forms, Ek, heat….
24.
EP is gained as work must be done to gain height.
25.
81.6 kg
26.
25 m
27.
147 J
150 m
10
4. KINETIC ENERGY (EK)
Objects which are moving have kinetic energy.
The kinetic energy which the car had before the collision is
used to do work on the wall and the front panels of the car.
The result is obvious.
The car is brought to rest quickly as it strikes the wall as
the kinetic energy does work on the wall. The bricks, front
of the car and metal rails are rearranged.
If the car had been travelling much more slowly, not as much work would have been done on the
wall. Possibly the only damage might have been the buckling of the bumper of the car. If the car
had been travelling at a greater speed then the car and the wall may have been totally demolished.
If a smaller car had collided with the wall it is possible that less damage may have been done.
The amount of kinetic energy (EK) an object has depends both on its velocity and mass.
EK =
1 2
mv
2
where Ek is the kinetic energy in joules (J), m is the mass in kilograms (kg)
and v is the velocity of the object in metres per second (ms-1).
The work done to change the velocity of an object is equal to the change in kinetic
energy.
4. KINETIC ENERGY (EK)
Example 1
The kinetic energy of a 120 g baseball travelling at 35.0 ms-1 can be calculated
in the following way:
EK =?
EK =
1 2
mv
2
m = 0.120 kg
EK =
1
 0.120  (35.0)2
2
v = 35 ms-1
EK = 73.5 J
The ball has a kinetic energy of 73.5 J
Example 2
The kinetic energy of moving cars is of interest to people investigating road accidents. For
example, the kinetic energy of an average small car of mass 1.40 tonne when it is travelling at
60.0 kmh-1 can be calculated:
Ek =?
EK =
1 2
mv
2
m = 1.40 tonne = 1400 kg
EK =
1
 1400  16.72
2
v = 60.0 kmh-1  3.6 = 16.7 ms-1
EK = 195 000 J
The car has kinetic energy of 195 000 J or 195 kJ.
11
In some situations involving energy - calculations either the mass or the velocity are unknown.
Example 3
For example, the velocity of a shot put of mass 5.00 kg with a kinetic energy of 160 J can be
calculated as follows:
1 2
mv
2
v =?
EK =
EK= 160J
160 =
m = 5.0 kg
160 = 2.50  v2
1
 5.00  v2
2
160
= v2
2.50
v2 = 64.0
v = 8.00 ms-1forwards
The shot must be moving at 8.00 ms-1
QUESTIONS
30.
Calculate the kinetic energy of a 50.0 kg mass moving at a speed of 5.00 ms-1
31.
What is the kinetic energy of a bullet of mass 2.00 g travelling at a speed of 250 ms-1?
32.
What is the mass of an object possessing 200 J of kinetic energy and travelling at a
velocity of 5.00 ms-1?
33.
Determine the velocity of an airplane of mass 1 000 kg
which has a kinetic energy of 5.00 MJ.
34.
A metal ball of mass 4.00 kg moves along a smooth
horizontal surface at 3.00 ms-1. What loss in kinetic energy
that occurs as it comes to rest?
35.
Estimate as an order of magnitude the kinetic energy of:
(a) An adult athlete while running a 100 m sprint.
(b) A moving snail.
ANSWERS
30.
625 J
31.
62.5 J
33.
100 ms-1fwds
34.
18 J
35.
No answers given
32.
16 kg
12
5. WORK DONE, ENERGY TRANSFORMATIONS & THE
CONSERVATION OF ENERGY
The work done on an object is equal to the change in its energy.

If the increase in energy changes the velocity of an object, the work done is the same as
the increase in kinetic energy.

If the increase in energy results in a change in vertical position of an object, the work
done is the same as the increase in potential energy.
The change of work into kinetic energy, work into potential energy and kinetic energy into
potential energy, or the reverse of any of these changes are called energy transformations.
Energy is conserved in each case although in practice some energy is converted to non useful
heat energy.
Calculating work and energy change
A common application of work being done occurs when the speed of an object is reduced
because the brakes are applied.
Example 1
Determine the work done by the brakes of a bicycle to reduce its speed from 10.0 ms-1 to 4.0
ms-1 if the mass of the bicycle and cyclist is 85.0 kg.
v = 10.0 ms-1
W = Fs = decrease in kinetic energy
u = 4.0 ms
 EK = final EK – initial EK
m = 85.0 kg
 EK =
1 2 1 2
mv – mu
2
2
 EK =
1
1
 85.0  10.02 –
 85.0  4.02
2
2
-1
 EK = 4250 – 680
 EK = 3570 J
The braking system did 3570 J of work in slowing the bicycle
down.
13
5. WORK DONE, ENERGY
TRANSFORMATIONS & THE
CONSERVATION OF ENERGY (cont.)
Example 2
Sloping ramps are often used by people in supermarkets
with loaded trolleys as a smaller force is used over a
longer distance to do the same work as lifting the loaded
trolley the short vertical distance with a larger force.
h
In each case the work done and the potential energy
gained are the same.
The Potential energy gained = mg  h = Work done = Fs = mgh
Where s = displacement along the ramp and  h is the height gained.
Example 3
For an object to gain height work is done to gain EP. If the object is to fall it loses EP as it gains
EK. At all times energy is conserved and the total energy = EP + EK
ETotal = EP + EK
A swimmer of mass 50.0 kg steps off a
10.0 m diving platform, diving vertically
into the pool below.
10 m
At the top of the tower (10 m) she starts
with zero kinetic energy, and her
EP = mg  h = 50 x 9.8 x 10 = 4 900 J
5m
5 m into her dive
her total energy = 4 900 J of which
EP = mg  h = 50 x 9.8 x 5 = 2 450 J
and EK = 2 450 J. (ETotal = 4 900 J)
As the swimmer falls, her potential energy
decreases, but her kinetic energy
increases as she accelerates downwards.
0m
As she hits the water (0 m) all of her initial
EP has transformed to EK = 4900 J
To find her final velocity you can either use
equations of motion or EK =
EK = 4900 J =
4 900 = 25  v2
v2 = 196
v = 14.0 ms-1
1 2
mv
2
1 2
1
 50.0  v2
mv =
2
2
14
5. WORK DONE, ENERGY TRANSFORMATIONS & THE
CONSERVATION OF ENERGY (cont.)
As the swimmer enters the water, her velocity decreases, because the water slows her down.
What has happened to the energy? You hear a splash, so some of the energy has become sound
energy. The water splashes up, and waves spread out, so some has become kinetic energy of
the water. Eventually the waves die away and all the energy has been converted to random
kinetic energy of molecules. The internal energy of the water has increased, and its
temperature has risen very slightly.
If we were to add up all these forms of energy, we would find that the total is just the same as
when the swimmer started the jump.
The principle of conservation of energy states that:
energy can neither be created nor destroyed; it can only be
changed into other forms of energy.
Example 4
If a bike rider of mass 45.0 kg is coasting at 7.5 ms-1 along level ground and then starts up a
hill, it is possible to determine the vertical height the person travels up the hill before stopping.
To do this it is necessary to ignore the effects of friction and air resistance.
h =?
loss in EK = gain in EP = mg  h
m = 45.0kg
EK =
1 2
mv
2
v = 7.5ms
EK =
1
 45  7.52
2
g = 9.8 ms-2
EK = 1265 J
 EP = 1265 = mg  h
45  9.8   h = 1265
h =
1265
441
 h = 2.87 m
h= 2.87m
The cyclist rises 2.87 m vertically up the
hill.
Example 5
A swing oscillates through a height of 3m. How fast is the little girl going at the bottom
of the swing ?
In swinging, the energy changes from Grav. Pot . Energy to Kinetic Energy. So
Ep lost = Ek gained (conservation of energy )
mgh lost = 1/2 .mv2 gained
thus, gh = 1/2 v2 ( as the mass is common )
9.8  3 = 0.5 v2
v2 = 58.7
and so v = 7.65 ms-1
15
Questions
36.
How much work must be done to stop a car of mass 1 500 kg moving at 10.0 ms-1?
37.
A body of mass 25.0 kg is accelerated from rest to 40.0 ms-1. What is its gain in kinetic energy?
38.
How much work must be done on a motor cycle of mass 250 kg to accelerate it uniformly
from rest at 3.0 ms-2 for 4.0 s?
39.
What is the gain in kinetic energy of an object of mass 40 kg if it falls from rest for 4.0 s?
40.
What loss of kinetic energy occurs as the speed of a bicycle is reduced from 12 ms-1 to
4.0 ms-1. The mass of bicycle and cyclist is 100 kg.
41.
A body of mass 4.0 kg is moving at a constant velocity of 3.0 ms-1 west. It is then acted
upon by a force of 5.0 N west, while it moves 2.0 m west.
(a)
What is the work done by the force?
(b)
What is the resulting increase in the kinetic energy of the body?
(c)
What was the initial kinetic energy of the body?
(d)
What is the final kinetic energy of the body?
(e)
Determine its final speed.
42.
How much work is done in changing the speed of a vehicle of mass 5 000 kg from 20.0
ms-1 to 30.0 ms-1?
43
A body of mass 3.0 kg is released from rest from the top of a building 25 m high.
(a) What is its kinetic energy when it hits the ground?
(b) With what velocity does it hit the ground?
44.
An arrow is shot vertically upwards with an initial velocity of 20 ms-1. How high will the
arrow fly? (HINT: equations of motion).
45.
Assuming the skier starts from rest at A describe their changes in potential energy and
kinetic energy at positions A, B, C, D, E and F below.
A
46.
a
Describe the energy changes as the pole vaulter moves from rest at A to B, from B
to C, from C to D and from D to E.
b
Why is it necessary for the vaulter to have quite a long run-up?
c
Describe the role the pole plays in the energy conversion processes.
d
In terms of energy transfer explain why is it necessary to have a thick sponge safety
mat for the jumper to land in?
16
47.
A car is driven accidentally over the edge of a 12.0 m high cliff, as shown.
a
What is the potential energy of the car at the top of the cliff if it has a mass of 1.5
tonne?
b
Describe the relationship between kinetic and potential energy of the car as it falls.
c
Calculate the velocity of the car the moment before it strikes the ground.
d
What is the source of the energy that re-models the front
of the car and scatters stones in all directions.
48.
A 300 kg roller-coaster hits the bottom of a dip and runs along
a level section of track at 18.0 ms-1. At the top of the next rise
its speed has reduced to 8.0 ms-1. Determine the height of the rise. (Assume there is no
work done against friction).
49.
A roller-coaster of total mass 800 kg moves so it is travelling at 15 ms-1 over the crest of
hill B, as shown. Use the information provided on the diagram to calculate the velocity it
will be travelling at points A, C and D. Assume there are no energy losses due to friction.
6m
8m
C
50.
Find the mass of a stone which when dropped down a shaft of
height 50 m strikes the bottom with 200 J of kinetic energy.
17
5. WORK DONE, ENERGY TRANSFORMATIONS & THE
CONSERVATION OF ENERGY (cont.)
ANSWERS
36.
75 000 J
37.
20 000 J
39.
30 730 J
40.
6 400 J
41(a)
10 J
(b)
10 J
(d)
28 J
(e)
3.74 ms-1
42.
1 250 000 J
43(a)
735 J
44.
20.4 m
45
At A all Ep. At B Ep decreases as EK increases.
38.
18 000 J
(c)
18 J
(b)
22.1 ms-1
At C EK is increasing as Ep decreases.
At D EK decreases as Ep increases.
At E EK increases as Ep decreases – same as C.
At F EK decreases as Ep increases – same as E..
AT ALL TIMES TOTAL ENERGY IS CONSTANT
46(a)
At A zero EK and zero Ep. At B all EK  elastic Ep.
At C EK is decreasing as Ep increases. At D all Ep.
At E all EK until work is done on the mat.
AT ALL TIMES TOTAL ENERGY IS CONSTANT
(b)
To gain EK to convert to elastic Ep for the pole.
(c)
EK to convert to elastic Ep for the pole so that upwards EK can be achieved.
(d)
So that the athlete‟s EK can be converted to work done on the mat and not the
ground (ouch!)..
47(a)
176 400 J
(d)
EK  work done on stones.
48.
13.3 m
49A
14.8 ms-1
50.
0.41 kg
(b)
C
ET = EP + EK
(c)
15.3 ms-1
19.4 ms-1
D
14.8 ms-1
18
Extra Problems involving work done against friction.
1. A skier of mass 65 kg stands at the top of a 75 m slope (see right). A small frictional force
of 8.5 N acts between the skier and the snow.
Determine:
a. The GPE of the skier at the top of the slope
b. The work done in overcoming friction as the skier
skies down the 150 m slope
c. The KE of the skier when she reaches point X
d. The velocity of the skier when she reaches point X
2. A loaded roller coaster of mass 1750 kg has a velocity of 12.3 ms-1 at point A (see below).
Determine the velocity of the roller coaster at points B and C, assuming that the track is
frictionless.
3. A 65 kg backstroke swimmer uses up 3.00 x 10 3 J of energy during a 100.0 m race in
which his final speed is 2.00 ms-1. How much energy is expended against fluid friction with
the water?
4. A skier of mass 70 kg is poised at the top lf a 50.0 m high slope which as a frictional force
of 45.9 N acting between the skis and the snow (see below). Determine:
a. The total energy of the skier at the top of the slope
b. The energy lost in opposing friction in traveling down the length of the slope
c. The velocity of the skier at the bottom of the slope
Answers
Q1
a) 4.8 x 104 J
b) 1.3 x 103 J
Q2
B 9.62 ms-1 C 18.6 ms-1
Q3 2.87 x 103 J
Q4
a) 3.43 x 104 J
c) 28.7 ms-1
b) 5430 J
c) 4.7 x 104 J
d) 38 ms-1
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6. ENERGY EFFICENCY
In any situation where energy is being used to do work, some of the energy is always
transformed into thermal energy, and we can never turn all of the original energy into useful
work. Although no energy has been destroyed, it has been degraded, in the sense that it is no
longer available for doing useful work. The ratio of useful work output to the total energy input
is called the percentage efficiency.
Energy Efficiency =
" useful" energy output
 100%
energy input
Because of the degradation of energy, no machine can ever be made 100% efficient.
In our body our muscles cannot turn all the chemical energy of food into useful work. About
75% of this energy is just turned into internal energy in the muscles, raising their temperature.
Device
Desired energy transfer
Efficiency%
Large electric motor
Electric to kinetic
90
High-efficiency solar cell
Solar to electric
35
Human body
Chemical to kinetic
25
Compact low-energy fluorescent light
Electric to light
25
Open fireplace
Chemical to heat
15
Light bulb
Electric to light
5
20
6. DEGRADATION OF ENERGY (cont.)
Example
A dog team of huskies pull a sled with a total force of 2 800 N. After they have pulled on the
traces for 25 m, the 200 kg sled is moving at a speed of 6.0 ms-1.
Calculate the efficiency with which the huskies are using their energy to pull the sled.
Energy input
Energy input = work = Fs =2800  25
Energy input = 70 000 J
Energy output = kinetic energy of the sled =
1
mv 2
2
Energy output
Energy output =
1
 200  62
2
Energy output =3600 J
Energy efficiency (%)
(Energy) Efficiency =
" useful" energy output
 100%
energy input
=
3600
 100%
70000
Efficiency = 5.1%
This first part of the motion is when the dogs have the most friction to overcome between sled
and snow (freezing to the surface, etc.)
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7. POWER
You can walk slowly up a flight of stairs and not feel very tired, but if you run up the stairs, you
may feel quite exhausted. In both cases you have raised the same mass (your own) through the
same height, against the pull of gravity; so you have done the same amount of work. The
difference is in the time it took you to do that work.
Doing the work in a shorter time requires more power. The rate at which the work was done
gives us a measure of the power.
Power is the rate at which work is done (or the rate of change in energy).
The unit of power is the watt, W which is the Js-1.
P=
W
mgh
E
=
=
= Fvav
t
t
t
P = power (Watts or W)
W = work or energy (J)
F = force (N)
vav = average velocity (ms-1)
Example 1
A student of mass 75 kg climbs stairs to a height of 3.0 m in 5.0 seconds. What is their power?
m = 75 kg
P=
ma g h
W
=
t
t
 h = 3.0 m
P=
75  9.8  3.0
5.0
t = 5.0 s
P = 441 W
P=?
Example 2
At what speed will a vehicle of power output 3 000 W be moving horizontally against a
frictional force of 200 N?
P = 3000 W
P = Fvav
F = 200 N
3000 = 200  vav
v or vav = ?
vav =
3000
= 15 ms-1
200
vav = 15 ms-1
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7. POWER (cont.)
QUESTIONS
51.
What power is required to raise a bucket of water (total mass 25 kg)
from the bottom of a12 m well in l5 seconds?
52.
A pump is rated 1.5 kW. Find:
(a)
the work the pump is capable of doing per second.
(b)
the time taken to perform 75 000 J of work if operating at the maximum 1.5 kW.
53
Determine the power required if a wooden crate is steadily pushed 4.0 m in 8.0 s across a
horizontal floor by a force of 8.0 N.
54.
An engine has a possible power output of 8.0 kW. How much work can this engine do in
10 minutes, assuming 100% efficiency?
55.
A car of mass 1 000 kg is accelerated from rest at 2.0 ms-2 over a distance of 25 m.
Calculate:
(a)
the time to travel 25 m.
(b)
the work done in travelling 25 m.
(c)
the power at which the car is operating.
56.
A bicycle travels at a constant speed covering 1 000 m in 40 s against a frictional
resistance of 160 N. What power is the rider developing?
57.
A crane lifts a mass of 1 500 kg through a height of 50 m in 15 s. At what power is the
crane operating?
58.
What is the power of an airplane of mass 3 000 kg if when on the runway it is capable of
reaching a speed of 80 ms-1 from rest in 4.0 seconds?
59.
A car with a power output of 2.5 kW is moving against a constant speed against a
frictional resistance of 180 N. What distance does it travel in 1.0 minute?
60.
To what height can a rocket of mass 20 000 kg reach in 10.0 seconds if it
is capable of a power output of 5.0 kW?
ANSWERS
51.
196 W
52.
50 s
53.
4.0 W
54.
4.8 MJ
55(a)
5.0 s
(b)
5 000 J
(c)
10 000 W
56.
533 W
57.
49 kW
58.
2.4 MW
59.
833 m
60
0.255 m
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Problems (W, E, P)
1) How much work is done when
a) A bricklayer lifts a brick weighing 40N a vertical height of 1.5m?
b) 1000kg of coal is raised from the bottom of a mine 300m deep?
2) An upward force of 6.0kN raises a mine cage a vertical distance of 100m in 40s at a constant rate.
a) how much work is done by the force in the 40s?
b) what work is done by the force in one second?
c) at what power is the force working?
3) Determine the kinetic energy of
a) a 0.2 kg billiard ball moving at 3ms-1;
b)a 50g golf ball moving at 80ms-1.
4) Taking the gravitational potential energy at ground level as zero, determine the gravitational
potential energy of
a) a 1.4kg book while carried 0.80m above the ground;
b) a 50g golf ball at the top of its flight 9.0m above the fairway.
5) A particle of mass 2kg moving at 50ms-1 is acted upon by an unbalanced force giving it an
acceleration of 10ms-2 for 12 minutes. What is the power developed by the force and how much
work is done?
6) A ball of mass 2kg is held 10m above the ground and then allowed to fall freely. It accelerates
downwards at 9.8ms-2. Assuming no air resistance. What is:
a) the weight of the ball?
b) the gravitational potential energy on release?
c) the kinetic energy after 5m of fall?
d) the total of potential and kinetic energy at ground level?
7) A car of mass 4000kg travelling at 20 metres per second is brought to rest by its brakes. Calculate
the initial kinetic energy of the car. What happens to this energy as the car is brought to rest?
8) A ball of mass 100g strikes a wall and rebounds in the opposite direction without loss of speed. Its
speed throughout is 20ms-1 Calculate:
a) the initial momentum of the ball
b) the change in momentum
c) the initial kinetic energy of the ball
A
E
C D
B
F
Figure 1
9) An amusement park roller coaster moves from A to F. The coaster is not powered in any way after it
leaves A and there is some friction acting. Over which section is:
a) kinetic energy increasing
b) kinetic energy being transformed into potential energy
c) the coaster accelerating
d) heat being produced.
e) At which position marked A to F would the force exerted by the coaster down on the track be
greatest? Explain your answer.
10) Discuss the energy changes occurring as a person executes a series of jumps on a trampoline.
Answers
1.a)60J b)2940kJ
2.a)600kJ b)15kJ c) 15kW 3.a) 0.9J b) 160J
4.a) 10.976J b) 4.41J
5. 73kW
6.a) 19.6N b) 196J
6.c) 98J d) 196J
7. 800kJ
8.a) 2kgm/s b) -4kgm/s c) 20J
24