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ENERGY, WORK, POWER

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Content
Page
Summary of Relevant Content
1–3
Worked Examples: Structured Questions
4–5
Worksheet 1: Multiple Choice Questions
6–8
Worksheet 2: Multiple Choice Questions
9 – 10
Worksheet 3: Multiple Choice Questions
11 – 13
Model Answers: Structured Questions
14 – 18
Answers: Worksheet 1, 2 and 3
19
© Science Education Centre 2002
Science Education Centre
!
SOWETO/DIEPKLOOF ◈ P.O.BOX 39067 ◈ BOOYSENS 2016 !!! " 011 9381666/7 # 011 9383603 email: sec@global.co.za
ENERGY, WORK, POWER
ENERGY
If something has ‘energy’ it can make things happen. To speak in more scientific terms:
Energy is the capacity of a system to perform work. The 'Law of Conservation of Energy'
states that energy can neither be destroyed nor created, but can be transformed from one form
to another. There are several types or forms of energy.
ENERGY FORMS
The energy forms most commonly used are:
• chemical energy
• potential energy (EP)
• kinetic energy (EK)
• heat energy
• magnetic energy
• electrical energy
• nuclear energy
• sound energy
• electromagnetic wave energy.
Energy is always measured in joules (J) no matter what form it is in.
Note: Sometimes EK and EP are considered together under the heading mechanical energy;
sound and electromagnetic waves are considered under the heading wave energy.
Chemical
Nuclear
Potential - EP
(EP = m·g·h)
Electrical
(E = I · V · t)
ENERGY FORMS
unit: Joule (J)
Magnetism
Kinetic - EK
(EK = ½ m · v2)
Energy waves (EMS)
Sound
Heat
Fig. 1: Energy forms: all energy resources result from the sun
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
1
Chemical energy
Chemicals such as food, oil, coal and gas would be included in this list. They can all be burnt
to provide heat. The chemicals inside a battery will react together to provide electricity.
Chemicals are stores of energy that can be released at a convenient time.
Potential energy (EP)
Potential means hidden or stored. A battery, a spring or a stretched elastic band are examples
of stored energy. Anything that is high up will have the ‘potential’ to be pulled down by
gravity, e.g., water behind a dam. This is sometimes called ‘gravitational energy’.
Kinetic energy (EK)
This is the energy of movement. The k.e. of a moving object increases with mass (kg) and/or
velocity (m/s). The equation used to calculate the k.e. of a moving object is:
Kinetic energy = ½ · Mass · Velocity2 = ½ m · v2
Electromagnetic waves (EMS)
Any wave form, such as light, X-rays, ultraviolet or gamma waves, belongs to a set known as
the electromagnetic spectrum (EMS). They are sometimes called waves, rays or even
radiations. These titles are sometimes even mixed together, e.g., ‘microwave radiation’. They
all transfer energy from place to place. Infrared radiation is a good example and is
responsible for the warming effect of a sunny day. Try to avoid phrases such as ‘heat wave’
when you really mean infrared radiation.
Heat energy
Ice melts when it is heated, an iron bar will expand when heated, etc.
Sound energy
This is an energy form because sound is the movement of air molecules.
Magnetic energy
Magnetism is an energy form and can make things happen (attraction/repulsion).
Electrical energy
The most easily converted energy form. A bulb will give out light (and heat) when provided
with electrical energy. The amount of electrical energy passing through a device depends on:
the current flow (in amperes), the potential difference (in volts) and the time (in seconds) for
which the circuit is switched on. Increase any of these and the total energy will increase. The
equation used to find electrical energy is:
Energy = Current x Potential Difference x Time = I x V x t
Nuclear energy
As atomic nuclei break up they have a heating effect on their surroundings. In nuclear power
plants nuclear energy is converted into heat, which is used to provide steam to drive a turbine
in order to generate electricity.
ENERGY TRANSFERS
Energy changes
Energy arrows (see figure 2 below) can be used to show the changes that take place when
energy is used. It is quite usual for energy to be wasted (in the form of heat) when one form
of energy is changed into another.
Light
(causes bulb to glow)
Battery
(Echem)
Electrical
Energy
Heat
(causes bulb to get hot: energy wasted)
Fig. 2: The major energy changes that take place in a torch light
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
2
WORK
In Physics, work requires a force to move. For instance: pushing a car to 'bump start' it
involves work. Work always involves an applied force moving a certain distance and is
measured by the product of the force and the distance it moves along its line of action. If the
force is perpendicular to the direction of motion, no work is done.
Work = F x s = force x distance (in the direction of the force)
"
"
"
NB
Work done equals energy gained.
It is a scalar.
It is measured in joules (J).
Total Work Done
Kinetic Energy
Potential Energy
Energy Dissipated
=
=
=
=
Applied Force x Distance*
Resultant Force x Distance*
Weight x Distance*
Frictional Force x Distance*
* Distance must always be in the direction of the force
POWER
Energy and power are not the same: do not confuse the terms 'energy' and 'power'. The word
energy has no connection with time but power does. To be powerful means to be able to use a
large amount of energy all the time. Power is the rate at which energy is used or transformed
power = energy ÷ time
"
"
or: power = work ÷ time
It is a scalar.
It is measured in watts (W).
FORMULAE
• F x s = work (provided F and S are in the same direction)
• Work = Power x Time
• Work done = energy gained.
• Kinetic energy:
EK = 1/2 m · v2
or: EK = acceleration · mass · distance* = a · m · s
(derived from:
v2 = u2 + ½ · a · t2
with u = 0 m/s)
• Potential energy:
EP = m · g · h
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
3
WORKED EXAMPLES
STRUCTURED QUESTIONS
Q1
A 2kg mass is accelerated horizontally from rest at 1 m/s2 for 20 m.
Find:
i)
the force required,
ii)
the work done,
iii)
the power required.
Q2
A mass of 10 kg is raised from rest to a height of 20 m using a cable with a tension of
140 N.
Find a) total work done
b) gain in EP
c) gain in EK
d) power required
Q3
A pump delivers 10000 kg of water into a dam 50 m vertically above a river in 10
minutes. If the voltage of the motor is 380 V find the current required to pump the
water (assume there are no frictional forces).
Q4
A bullet with a mass of 20 g is fired horizontally at 500 m/s into a stationary wooden
block with a mass of 4 kg. The block with the bullet embedded in it then slides for
1,55 m across a rough horizontal surface before it comes to a stop.
a) Show by calculation that the velocity of the bullet and the block immediately after
impact, is 2,5 m/s in the direction of motion.
b) Calculate the loss of kinetic energy of the system when the bullet strikes the block.
What happened to the energy?
c) Calculate the average magnitude of the frictional force between the block and the
surface when the block slides over it.
Q5
A force of 100 N acts on a 10 kg block at an angle of 600 from the horizontal, as
shown in the sketch.
100 N
The frictional force
experienced by the
block is 20 N.
20 N 10 kg 600
Determine the acceleration with which the block moves.
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
4
Q6
An object with mass 25 kg moves at a constant speed of 4 m/s along a horizontal
surface towards point A. It then moves up an incline towards point B which is 0,9 m
higher than point A (see sketch below). Provided all surfaces are frictionless, will the
object reach point B? Show all calculations to support your answer.
B
0,9 m
25kg
Q7
A
John, with a mass of 47kg, rides a skateboard with a mass of 3kg on a rough
horizontal road. At the bottom of an incline, his velocity is 4m/s. He rides up the
incline and reaches the top with a velocity of 1 m/s. the difference in height between
the top and the bottom of the incline is 0,6 m.
a) Calculate the work done against friction while John rode up the incline.
b) When John reaches the top of the incline with a velocity of 1 m/s, his cat with a
mass of 5 kg, drops from a tree into his arms. Calculate the velocity of John and the
cat as they move together on the skateboard.
1 m/s
4 m/s
0,6m
Q8
A skier of mass 70 kg starts from rest to move down a hill of slope 300 with a constant
acceleration of 3,2 m/s2 from a point P on the top to the foot Q of the hill.
The distance PQ is 100 m.
a) Calculate the velocity of the skier on
reaching Q.
b) Calculate the loss of mechanical
energy of the skier between the points P
and Q.
c) State the law of conservation of
energy.
___________________________
© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
5
Science Education Centre
!
SOWETO/DIEPKLOOF ◈ P.O.BOX 39067 ◈ BOOYSENS 2016 !!! " 011 9381666/7 # 011 9383603 email: sec@global.co.za
Topic:
Work, Energy, Power
Worksheet 1: Multiple Choice Questions
Time: 30 Minutes
Instructions: Make a cross over the letter A, B, C, D or E to show the correct answer.
1) The diagram below shows an object of mass 10 kg being pulled for a distance of 3 m by a
force of 30 N.
10 kg
30 N
3m
The work done is
A
B
C
D
E
9J
30 J
90 J
300 J
900 J
2) A box of weight 50 N is pulled 2 m along a horizontal floor by a force of 10 N and then
the box is lifted vertically through a height of 1 m (see sketch below).
1m
10 N
2m
50 N
What is the total work done on the box?
A
B
C
D
E
35 J
55 J
70 J
110 J
180 J
3) A girl weighing 400 N runs up a flight of stairs of vertical height 5 m in 4 seconds.
Her increase in gravitational potential energy is
A
1600 N
B
1600 J
C
1600 W
D
2000 J
E
2000 W
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
6
4) During the latter part of the motion of a rocket its mass is halved because the fuel is used
up, while its velocity increases by a factor of 10. By what factor does the kinetic energy
of the rocket increase during this stage?
A
B
C
D
5
10
50
100
5) A ball is projected vertically upwards and then returns to the ground. Ignore all friction.
Which one of the following statements about the kinetic energy and potential energy of
the ball is true?
A
B
C
D
The kinetic energy is always equal to the potential energy
The kinetic energy is always less than the potential energy
The kinetic energy is always more than the potential energy.
The sum of the kinetic energy and potential energy is always constant.
6) Two skiers S and T have identical masses. They begin from rest from the top of a hill at
point A and move to the ski resort. Skier S takes route 1 and skier T takes route 2, as
shown in the sketch. Which one of the following statements concerning the speed with
which S and T reach the ski resort is correct? Ignore all friction forces.
A
B
C
D
E
S and T both have a speed of zero.
The speed of S is smaller than the speed of T.
S and T both have the same speed.
The speed of S is greater than the speed of T.
The information given is insufficient to answer the question.
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
7
Questions 7 to 10:
Here are some possible ways in which energy can change:
A
B
C
D
E
chemical to heat
kinetic to sound
kinetic to heat
potential to heat
potential to kinetic
Which is the most important change in each of the following examples?
7) The release of an arrow from a bow.
8) The impact of the arrow in a target.
9) A cyclist riding along a level road at constant velocity.
10) A stone in mid-air, as it falls.
___________________________
© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
8
Science Education Centre
!
SOWETO/DIEPKLOOF ◈ P.O.BOX 39067 ◈ BOOYSENS 2016 !!! " 011 9381666/7 # 011 9383603 email: sec@global.co.za
Topic:
Work, Energy, Power
Worksheet 2: Multiple Choice Questions
Time: 30 Minutes
Instructions: Make a cross over the letter A, B, C, or D to show the correct answer.
1. An object moving in a straight line at constant velocity has kinetic energy E and
momentum p. If the speed of the object is doubled, the new value of kinetic energy and
momentum will be …
Kinetic Energy
2E
2E
4E
4E
A
B
C
D
Momentum
2p
4p
P
2p
2. A boy lifts a packet upwards by applying a constant force to it of magnitude greater than
the weight of the packet. The work done by this force equals the gain in …
A
B
C
D
potential energy of the package.
potential energy plus kinetic energy of the package.
kinetic energy of the package.
kinetic energy minus the gain in potential energy.
3. An object is dropped from the top of a high building and falls freely to the ground. Which
one of the graphs below best represents its potential energy (Ep) as a function of the
distance (s) fallen by the object?
Ep
Ep
A
B
distance
Ep
distance
Ep
C
distance
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
D
distance
9
4. A large and a small sphere are released at the same time from the same height above the
ground. Which one of the following quantities associated with the spheres will be the
same for both after 1 second, if frictional effects are ignored?
A speed
B momentum
C potential energy
D kinetic energy
5. Which one of the following expressions has the same units as power?
A force x distance
B work x time
C force x acceleration
D force x velocity
6. A person lifts a heavy load to a vertical height of 2,0 m in 3 seconds. If he/she had done
this more slowly in 6 seconds, the work on the load would have been:
A twice as great
B four times as great
C the same
D half as great.
7. At what height above the ground must a mass of 10 kg be to have a potential energy equal
in value to the kinetic energy possessed by a mass of 10 kg moving with a velocity of 20
m/s?
A 10 m
B 20 m
C 50 m
D 100 m
8. A girl runs up one flight of steps. Which one of the following factors does not affect the
work done by the girl against gravity?
A mass of the girl
B height of the steps
C speed of the girl
D acceleration due to gravity
9. A girl weighing 400 N runs up the a flight of stairs (height 5 m) in a time of 4 seconds.
Her rate of working against gravity is
A 320 W
B 400 W
C 500 W
D 2000 W
10. A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after
falling for 4 seconds. The stone's potential energy with respect to the ground is equal to
its kinetic energy …
A at the moment of impact.
B 2 seconds after the stone is released.
C after the stone has fallen 40 m.
D when the stone is moving at 20 m/s.
___________________________
© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
10
Science Education Centre
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SOWETO/DIEPKLOOF ◈ P.O.BOX 39067 ◈ BOOYSENS 2016 !!! " 011 9381666/7 # 011 9383603 email: sec@global.co.za
Topic:
Work, Energy, Power
Worksheet 3: Multiple Choice Questions
Time: 30 Minutes
Instructions: Make a cross over the letter A, B, C, or D to show the correct answer.
T
1. An object of weight W is lifted vertically through a distance h
by a cable hanging from a helicopter. The helicopter accelerates
upwards and the tension in the cable is T.
The work done on the object (in J) and the type of energy that
this work is converted into, is …
A
B
C
D
Work done on the object
Th
(T – W) h
Th
(T – W) h
W
Work done converted into
potential energy only
potential energy only
potential and kinetic energy
potential and kinetic energy
2. Thembi does exercises in the gym. The amount of work she does is measured in various
time frames. In which of the following cases will her power output be greatest?
When she does
A
B
C
D
10 J work in 10 s
60 J work in 20 s
80 J work in 30 s
100 J work in 40 s
3. A car moving at a speed of v has kinetic energy Ek. If the speed of the car increases to 2v,
the kinetic energy will then be …
A
B
C
D
2 Ek
4 Ek
8 Ek
16 Ek
4. Two toy cars X and Y of mass m and 2m
respectively are at rest at point A. They are allowed
to run down a smooth frictionless track. As the cars
pass point B, how will their velocities compare
A
B
C
D
vx = vy
vx = 2 vy
2 vx = vy
4 vx = vy
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
Y
X
A
B
11
5. In the diagram to the right a simple pendulum with
mass m swings to and fro. At position B the value
of the potential energy and kinetic energy are as
follows:
EP
maximum
maximum
minimum
minimum
A
B
C
D
EK
maximum
minimum
minimum
maximum
A
B
C
6. When a racing car brakes heavily, coming to a halt, into what form of energy is its kinetic
energy transformed?
A
B
C
D
potential energy
kinetic energy
internal energy
elastic energy
7. Which one of the following is a measure of power?
A
B
C
D
N/s
kg · m/s
J/C
J/s
8. Observe the pendulum in the drawing to the right. At the highest point A of its swing the
ball has 500 J potential energy in respect to its lowest point.
At the lowest point B of its swing the ball has 500 J kinetic
energy. The total mechanical energy of this system is …
A
B
C
D
0J
250 J
500 J
1000 J
A
9. Two children A and B of equal
masses are at a swimming bath. Child
A drops vertically from a diving
board 5 m high. Child B slides from
the same height down a slide into the
water.
A
5m
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Physics/FET/Revision: Work, Energy, Power
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B
5m
B
The children start moving at the same instant. Which one of the following statements is
true? (Ignore air resistance and friction on the slide.)
The children hit the water …
A
B
C
D
at the same time with the same speed
at the same time with different speeds
at different times with the same speed
at different times with different speeds
10. An object is dropped from the top of a high building and falls freely to the ground. Which
one of the graphs below best represents both its potential energy and kinetic energy as
functions of the time fallen by the object?
energy
energy
A
EP
energy
EP
time
energy
C
EP
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
B
D
EP
time
13
time
time
Model Answers Structured Questions:
Q1:
i)
Since there is only one force acting on the mass, this force is the resultant force (FRes).
2 kg
2 kg
a = 1m/s2
FRes
FRes
ii)
Force x distance = work
2N x 20m
= 40 J = work done
iii)
Power = work/time = 40J/time
=m·a
= 2kg · 1m/s2
=2N
Since we don't know the time, we have to apply the equations of motion:
given:
u = 0 m/s
s = u t + 1/2 · a · t2
v = ____
20 = 0 + 1/2 · 1 · t2
a = 1 m/s2
t2 = 40
t = ???
∴ t = √40 = 6,3 seconds
s = 20 m
therefore:
Power = 40J/6,3s = 6,3 W
Answer to Q2
140 N
140 N
Start with a force diagram:
10kg
Weight = mg = 100 N
40 N↑
10 kg
= a = 4 m/s2↑ (keep this for later)
a) Work done = applied force x distance = 140N x 20m = 2800 J
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
14
FRes = 40 N↑ = m · a
b) EP = m · g · h = 10kg · 10m/s2 · 20m = 2000 J
c) EK = a · m · s = 4m/s2 · 10kg · 20m = 800 J
or: EK = FRes · s = 40N · 20m = 800 J
d) Power = work done / time taken
1. Find time using equations of motion:
TABLE:
s
u = 0 m/s
v =
a = 4 m/s2
s = 20 m
t = ???
= u t + 1/2 a t2
20m
= 0 + 1/2 · 4 · t2
40/4
= 10 = t2
t
= 3,16 seconds
Power = work done / time taken = 2800/3,16 = 886 W
Answer to Q3
This is a case of pump lifting a certain mass of water (10000 kg) up a certain vertical distance
(50 m) in a certain time (600 s). Using Power = Voltage · Current we can find I.
Work = Force · distance = 100000 N · 50m = 5 ·106 J
Power = work/time = 5 ·106 / 600 = 5/6 ·104 = 8333 W
8333 W = 380 V · I
∴ I = 8333/380 = 21.9 A
block of wood
comes to rest
1,55 m from initial position
Answer to Q4
+ ve
500m/s
Start with a diagram:
given:
mass of bullet mB = 20 g = 0,02 kg
velocity of bullet vB = 500 m/s
mass of wood mW = 4kg
velocity of wood before impact vW = 0 m/s
a) This is a momentum problem. Take bullet direction as positive.
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
15
Momentum before = Momentum after
mB · vB + mW · vW = (mB + mW) v
0,02 · (+ 500) + 4 · 0 = (0,02 + 4) v
10 + 0 = 4,02 · v
∴ v = 10/4,02 = 2,5 m/s (rounded up to 2,5)
b)
bullet + block
bullet only
EK
before
impact:
∴
c)
= 1/2 m · v2
EK
after
impact:
= 1/2 · 0,02 · (5·102)2
= 1/2 (mB + mW) v2
= 1/2 (00,2 + 4,0) · 2,52
= 0,01 · 25 · 104
= 2,01 · 6,25
= 2500 J
= 12, 6 J
EK lost (in the form of heat and sound)
= EK (before) - EK (after)
= 2500 J - 12,6 J
= 2487,4 J
table:
u = 2,5 m/s
v = 0 m/s
a = ?
t = ?
s = 1,55 m
frictional force
frictional force
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
equation of motion:
=
u2 + 2 · a · s
v2
0
=
2,52 + 2 · a · 1,55
-6,25 =
3,1 · a
∴a
=
-2 m/s2
=m·a
= 4,02 · 2
= 8,04 N in the opposite direction to movement
16
Answer to Q5
The block can only move horizontally, therefore all forces must be in the horizontal direction,
which the 100 N is not. So, first one has to find the component of the 100 N in the horizontal
direction.
100 N
cos 600 = adj/hyp = F/100N
hence: F = 100N · cos 600 = 100N · 0,5 = 50 N
600
F
Now we have:
10 kg
50 N
20 N
FRes = 50N - 20N = 30N = a · m = a · 10 kg
∴
a = 3 m/s2 (in the opposite direction to the friction)
Answer to Q6
To reach B the object must have an energy greater or at least equal to Potential energy at
point B:
EP = m g h = 25 · 10 · 0,9 = 225 J
EP > EK
But kinetic energy at point A:
EK = 1/2 m v2 = 1/2 · 25 · 42 = 8 · 25 = 200 J
As a result the object will not reach point B since it does not have sufficient energy at point A
to do so.
Answer to Q7
a) At the bottom John has only kinetic energy:
EK = 1/2 m v2 = 1/2 · (47 + 3) · 42 = 25 · 16 = 400 J
At the top he has potential energy (he has risen 0,6 m higher) and kinetic energy (he is still
moving at 1 m/s).
therefore:
EP = m g h = (47 + 3) · 10 · 0,6 = 50 · 6 = 300 J
EK = 1/2 m v2 = 1/2 · (47 + 3) · 12 = 25 J
∴ He still has (300 + 25) J = 325 J of energy at the top.
∴ Energy lost to friction = 400 J - 325 J = 75 J
∴ Work done against friction = 75 J
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
17
b)
John's direction of movement taken as positive)
Momentum before = momentum after
mjohn · vjohn + mcat · vcat = (mjohn + mcat ) · v
50 · 1 + 5 · 0 = (50 + 5) · v
v = 50/55 = 0,9 m/s in the direction John is moving.
Answer to Q8
a) let's use equation of motion:
u = 0 (starting from rest)
v=?
a = 3,2 m/s2
t=?
s = 100 m
velocity at Q:
v2 = u2 + 2 · a · s
= 0 + 2 · 3,2 · 100
= 640
∴ v = 25,3 m/s
v = 25,3 m/s
b) Loss of energy = EP (top) - EK (bottom)
= m · g · h - 1/2 m · v2
vertical distance between P and Q:
P
= 70 · 10 · 50* - 1/2 · 70 · 640
= 35000 J - 22400 J
100m
= 12600 J
300
Q
sin 300 = opp/hyp = opp/100
∴ opp (height) = 0,5 · 100 = 50 m
c) In an isolated system the energy is always conserved. It may be converted from one form
to another, but it is never lost.
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
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ANSWERS TO MULTIPLE CHOICE QUESTIONS
Worksheet 1:
Worksheet 2:
Worksheet 3:
1
C
1
D
1
C
2
C
2
B
2
B
3
D
3
A
3
B
4
C
4
A
4
A
5
D
5
D
5
D
6
C
6
C
6
C
7
E
7
B
7
D
8
C
8
C
8
C
9
A
9
C
9
C
10
E
10
C
10
A
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© Science Education Centre 2002
Physics/FET/Revision: Work, Energy, Power
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