2021 Edition
Physical Sciences
Maths and Science Infinity
Work Energy and Power
Grade 12 Learner Booklet
Science Department, MSI
Second Floor, B1 Stewart Drive, Berea, East London, Eastern Cape,
RSA. Contact: MSI Head of Science (MR. P: +27655795185)
Contents
Page
1. WORK DONE BY A CONSTANT FORCE ................................................. 1
2. NET WORK DONE BY A FORCE .............................................................. 2
3. WORK – ENERGY THEOREM ................................................................... 5
4. CONSERVATIVE AND NON-CONSERVATIVE FORCES ........................ 9
4.1. Definitions ............................................................................................ 9
4.2. Work done by gravitational force ......................................................... 9
5. CONSERVATION OF ENERGY ............................................................... 11
5.1. A thinking routine: orange falling from a tree ..................................... 11
5.2 The principle of conservation of energy ............................................. 12
5.3. The principle of conservation of mechanical energy .......................... 12
5.4 Work done by non-conservative forces .............................................. 13
6. POWER ..................................................................................................... 17
6.1 Average power of a force ................................................................... 18
6.2 Power of an electric motor ................................................................. 19
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EXPECTED LEARNING OUTCOMES
By the end of this session, learners are expected to be able
to achieve the following:
1. Work
1.1.
1.2.
1.3.
1.4.
Define the work done on an object by a constant force
Draw a force diagram and free-body diagrams.
Calculate the net work done on an object
Distinguish between positive net work done and negative
net work done on the system
2. Work-Energy Theorem
2.1.
2.2.
State the Work-Energy Theorem
Apply the work-energy theorem to objects on frictionless
and rough surfaces:
 horizontal
 vertical and
 inclined planes
3. Conservative and non-conservative forces
3.1.
3.2.
3.3.
3.4.
Define a conservative force
Define a non-conservative force
State the principle of conservation of mechanical energy
Solve conservation of energy problems Wnc = ΔEk + ΔEp
4. Power
4.1.
4.2.
4.3.
4.4.
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Define power
Calculate the power involved when work is done
Perform calculations using Pave= Fvave
Calculate the power output for a pump lifting a mass moving
at constant speed
Work, Energy and Power by Mr. P (+27655795185)
ii | P a g e
1. Work done by a constant force
The term “work” carries different meanings depending on scenarios. The
meaning of work in everyday life is not the same as that in Physics. In
mechanics, work is done only when a force F moves an object through a
certain displacement. Hence, work is defined as the product of force,
displacement and the angle between the displacement and the force.
Scientifically, work is defined as:
W = F∆xcos𝜗
where:
F is the force acting on an object,
∆x is the displacement moved by the
object 𝜗 is the angle between F and ∆x
EXAMPLE 1
A box of mass 15kg is pulled by a force of 96N across a horizontal rough
surface through a displacement of 4,5m as shown below.
F = 96N
frictional force, 𝑓
pushing backwards
A
4,5m
B
The frictional force between the box and the surface is 20N.
Calculate the work done by each force.
SOLUTION
1.1
Work done by the applied force.
F
WF = F∆xcos𝜗
WF = 96 × 4,5𝜗
∆x
What is the value of 𝜗 ? Well, 𝜗 is the angle between displacement
and the applied force, F. Both ∆x and F are pointing in the same
direction (TO THE RIGHT). Thus, these are parallel lines whose
angle between them is 0. Thus, 𝜗 = 0. Hence, we have:
WF = 96 × 4,5 × cos 0 = 96 × 4,5 × 1 = 432 𝐽
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1.2 Work done by frictional force, f
Frictional force opposes motion. Hence the work done must be negative.
Therefore, from W f = F∆xcos𝜗, we have: Wf = 20 × 4,5 × 𝑐𝑜𝑠 𝜗. Let’s find
the value of 𝜗, the angle between displacement and frictional force.
𝜗= 180𝑜
f
Wf
∆x
= 20 × 4,5 × cos 𝜗
= 20 × 4,5 × 𝑐𝑜𝑠180
= 20 × 4,5 × (−1)
= −90 𝐽
2. Net Work done by a force
Net work done is the also called the total work done. It can be calculated
in two ways:
Method 1: Find net force first and then use it in the formula for work done.
The formula becomes: W net = Fnet∆xcos𝜗.
From example 1, net force is given by:
Fnet
=F–f
= 96 – 20
= 76N
Wnet = Fnet∆xcos𝜗
= 76 x 4,5 x cos 𝜗
The net force is in the same direction as the displacement. Hence the
angle between them is zero. Therefore, we have:
Wnet = Fnet∆xcos𝜗
= 76 x 4,5 x cos 𝜗
= 76 x 4,5 x cos 0
= 76 x 4,5 x 1
= 342 J
Method 2: Calculate work done by individual forces and sum up all of them
to find net work done.
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Net work done
= sum of work done by individual forces
Wnet = Wf + WF
= Wf + WF
= −90 + 432
= 342 J
EXAMPLE 2
A crate of mass 7kg is being pulled to the right across a rough horizontal
surface by a constant force F. The force F is applied at an angle of 26o to
the horizontal, as shown in the diagram below.
F
26
0o
friction force, 𝑓
2.1
Draw a labelled free-body diagram showing ALL the forces
acting on the crate.
(4)
A constant frictional force of 5N acts between the surface and the crate.
The coefficient of kinetic friction between the crate and the surface is 0,42
Calculate the magnitude of the:
2.2
Normal force acting on the crate
(3)
2.3
Force F
(4)
2.4
Acceleration of the crate
(3)
If the crate moves a distance of 19,2m. Calculate
2.5
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Net work done on the crate.
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SOLUTION
2.1
There are four forces acting on the crate: friction, f; Normal force
N, weight (Fg) and the applied force, F. The free body diagram is
shown below.
N
F
26o
f
mg
2.2
Calculating the normal force
The vertical forces are balanced and therefore cancel each other
out.
N
F
26o
f
Fy
Fx
mg
We have: N + Fy = mg. But Fy = F sin 26. But we cannot use
this equation because we have two unknowns Fy and F. However,
we know that 𝑓𝑘 = 𝜇𝑘 𝑁. Substituting values, we get: 5 = 0,42𝑁
Solving for N we get N = 11,9N.
2.3
We know that: Fy = F sin 26 and N + Fy = mg. Therefore, we have:
N + Fy = mg
N + F sin 26 = mg
F sin 26 = mg – N
F=
=
(mg – N)
sin 26
(7x9,8 – 11,9)
sin 26
= 127,74 N
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2.4
Calculating the acceleration of the crate
We can find the acceleration of the crate from Fnet = ma
where Fnet = Fx – f.
Fx = F cos 26
Fx = 127,74 cos 26
Fx = 114,81N
Thus, from Fnet = Fx – f. we have: Fnet = 114,81- 5 = 109,81 N
Substituting values in Fnet = ma, we get: 109,81= 7a.
Therefore, a = 15,69m/s2
2.5
Calculating the net work done
Wnet = Fnet∆xcos𝜗
= 109,81 x 19,2 x cos 𝜗
= 109,81 x 19,2 x cos 0
= 2108,35 J
3. Work – Energy Theorem
For us to understand the concept of work-energy theorem, do the following
activity.
Activity 1
Suppose that you push on 30 kg package with a constant force of 120 N
through a distance of 0.8 m, and that the opposing friction force averages 5
N as shown below.
𝑣𝑖 = 0
F = 120N
30kg
frictional force, 𝑓
pushing backwards
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𝑣𝑓 =?
A
0,8m
B
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1.1
Calculate:
1.1.1 the net work done on the package.
1.1.2 the final velocity of the package if it started from rest.
1.1.3 the change in the kinetic energy of the package.
1.2
What can you conclude from your answers to question 1.1.1 and
1.1.3?
Discussion
If you did your calculations correctly, you would notice that the answers to
1.1.1 and 1.1.3 are the same. We can verify this.
1.1.1 Net work done
Wnet = Fnet ∆x cos𝜗
= (F – f) ∆x cos 𝜗
= (120 – 5) ∆x cos 𝜗
= 115 x 0,8 x cos 0
= 92 J
This value is the net work done on the package. The person actually does more
work than this, because friction opposes the motion. Friction does negative work
and removes some of the energy the person expends and converts it to heat
energy. The net work equals the sum of the work done by each individual force.
1.1.2 the final velocity of the package if it started from rest
We can determine the final velocity from equations of
motion. 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑥.
We don’t know the acceleration of the package but we can
calculate it from net force using Fnet = ma. Thus, we have
Fnet = ma
115 = 30a
a = 3,83m/s2
We can now use 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑥 to find the final velocity.
𝑣𝑓2 = 0 + 2(3,83)(0,8). Which gives 𝑣𝑓 = 2,48𝑚/𝑠
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1.1.3 The change in kinetic energy: ∆Ek = Ekfinal - Ekinitial
∆Ek
1
= 𝑚(𝑣𝑓2 − 𝑣𝑖2 ).
2
1
= (30)(2,482 − 0)
2
= 92,25 𝐽
1.2
Conclusion from 1.1.1 and 1.1.3. Net work done = Change in
kinetic energy.
NOTE
The conclusion made in 1.2 is a scientifically proven and in
particular it is called the work-energy theorem as officially
stated below.
Work Energy Theorem:
The work done on an object by a net force is equal to the
change in the object's kinetic energy.
EXAM TIP
During exams you will be required to apply the work-energy theorem in
various scenarios including: rough and frictionless surfaces on any of the
three
 horizontal
 vertical and
 inclined planes
A Challenge
How far does the package in the previous Activity 1 travel after the push,
assuming friction remains constant? Use work and energy
considerations.
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Strategy
We know that once the person stops pushing, friction will bring the package to
rest. In terms of energy, friction does negative work until it has removed all of
the package’s kinetic energy. The work done by friction is the force of friction
times the distance traveled times the cosine of the angle between the friction
force and displacement; hence, this gives us a way of finding the distance
traveled after the person stops pushing.
Solution
The normal force and force of gravity cancel in calculating the net force. The
horizontal friction force is then the net force, and it acts opposite to the
displacement, so θ = 180º. To reduce the kinetic energy of the package to zero,
the work Wf by friction must be minus the kinetic energy that the package started
with plus what the package accumulated due to the pushing.
Thus, work done by the frictional force is equal to −92,25 𝐽.
But we also know that:
Wf
= f∆xcos𝜗
Thus, we have:
−92,25 = 5∆xcos 180
−92,25 = 5∆x (-1)
Dividing both sides by minus 5 we get: ∆x = 18,45𝑚
Comment: This is a reasonable distance for a package to travel on a
relatively less rough surface (f = 5N is very small). Note that the work done by
friction is negative (the force is in the opposite direction of motion), so it
removes the kinetic energy.
IMPORTANT TO REMEMBER
Whenever work is done on an object by a force, energy is being transferred
from one object to another object. This is how energy transfer takes place
within a system. Therefore, amount of work done by a force on an object is
equal to the energy being transferred. Thus, WORK DONE = ENERGY
TRANSFERRED. When a force moves an object through a certain
displacement, kinetic energy is added to the system. In contrast, frictional force
removes kinetic energy from the system.
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4. Conservative and non-conservative forces
4.1.
Definitions
Conservative forces: a conservative force is a force for which the work
done in moving an object between two points is independent of the path
taken.
Examples of conservative forces
 gravitational force
 elastic force in a spring
 coulombic force.
Non-Conservative forces: a non-conservative force is a force for which
the work done in moving an object between two points depends on the
path taken.
Examples of nonconservative forces
 frictional force
 air resistance
 tension in a chord
4.2.
Work done by gravitational force
Let’s consider a block of mass 3 kg sliding down 1,5m an incline at 30
degrees to the horizontal as shown below. Friction force is 10N.
x-axis
N
F
∆x = 1,5m
60𝑜
𝑜
60
Fg = mg
θ = 30𝑜
30𝑜
Fgy
Fg
Fgx
Reminder: The angle between Fgy and the Fg is equal to the angle between
the horizontal and the incline. Hence, the gravitational force Fg
makes an angle of 60o with the x-axis (the surface in contact with
the block) and hence with the displacement.
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y-axis
Work done by gravitational force is given by:
WFg
= Fg∆xcos𝜗
where 𝜗 is the angle between displacement ∆x and the
gravitational force Fg.
and 𝜗 = 60𝑜 .
Thus,
WFg
= Fg∆xcos𝜗.
= mg∆xcos𝜗
= (3) (9,8) (1,5) cos 60
= 22,05 J
Thus, gravitational force does POSITIVE WORK on the block. The block gains
mechanical energy (kinetic energy) as it moves down the incline.
Work done by frictional force is calculated as follows:
Wf
= f∆xcos𝜗.
= (10) (1,5) cos𝜗
Frictional force and displacement are in the opposite direction. Hence the angle
between them is 180o. Thus, we have:
Wf
= (10) (1,5) cos 180
= (10) (1,5) (-1)
= -15 J
Net work done on the block is the sum of the two:
Wnet = WFg + Wf
= 22,05 + (-15)
= 7,05 J
Method 2: Calculation of net work done for an object on an incline.
1. ONLY consider the direction of motion along the x-plane (surface in
contact with the block).
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2. Calculate the net force in the x-plane and use that to find net work done.
Fnet
= Fgx – f
= Fg sin 30 – f
= mg sin 30 – f
= (3) (9,8) (0,5) – 10
= 4,7 N
Wnet = Fnet∆xcos𝜗
The angle between Fnet and displacement is zero degrees
because the two vectors are pointing in the SAME direction.
Therefore, we have:
Wnet = (4,7) (1,5) cos 0
= (4,7) (1,5) (1)
= 7,05 J
Therefore, net work done on the block is 7,05 J.
5. Conservation of Energy
5.1. A Thinking Routine: Orange falling from a tree
Think about this and write down your thoughts on a piece of paper.
Remember, you are only thinking, don’t worry about whether it’s
correct or not.
Let’s imagine a fruit, say orange, falling from a 50 m tall tree and hitting on
top on your head. Would it be painful? How would the pain compare if the
same orange fell down from 15 m high? What would make the impact more
severe in one case than the other? Does the height matter when an object
falls?
Discussion
When an object is at a certain height above the ground, it contains
mechanical energy of height. This energy is called gravitational potential
energy. This concept was introduced in Grade 10 Physics. The potential
exergy of an object can be calculated from the formula: Ep = mgh. The mass
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m and gravitational acceleration g are constant. Hence, the energy depends
on ONLY height h. Thus, an object at a greater height contains MORE
gravitational potential energy than an object at a lower height. Hence, the
one at a greater height causes more pain (impact) if it falls on our head or
shoulders.
In the following section, we will discuss more on energy changes and
conservation of mechanical energy.
5.2 The Principle of Conservation of Energy
The law of conservation of energy states that:
Energy cannot be created or destroyed, but is only
changed from one form into another.
When an object at a height, it contains potential energy, Ep. When it falls
don, it gains kinetic energy. No energy is lost. It is merely converted from
gravitational potential energy to kinetic energy.
5.3. The Principle of Conservation of Mechanical Energy
The law of conservation of mechanical energy (kinetic
and gravitational potential) states that:
The total amount of mechanical energy in an isolated
system remains constant.
A system is isolated when the net external force (excluding the gravitational
force) acting on the system is zero.
EXAMPLE 1
Let’s consider a 1,5kg small snook ball placed at a height of 2,5m above the
ground and allowed to roll down a frictionless curved surface as shown in
the diagram below.
P
2,5m
Q
R
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Energy changes from P to R
When the ball is at point P, it contains gravitational potential energy given by:
EP = mgh
= (1,5)(9,8)(2,5)
= 36,75 J.
The potential energy at R is ZERO. Why? The height in Ep = mgh is zero.
However, no energy is destroyed, it has only changed from potential to kinetic
energy. When the ball is at point R, all the potential energy is converted to
kinetic energy.
According to the Law of conservation of mechanical energy, we have:
EM(top) = EM(bottom)
(Ep + Ek)top = (Ep + Ek)bottom
1
1
(mgh + 𝑚𝑣 2 )top = (mgh + 𝑚𝑣 2 )bottom
2
2
1
1
(1,5)(9,8)(2,5) + (1,5)(0 )top = (1,5)(9,8)(0) + 𝑚𝑣 2 )bottom
36,75 =
1
2
(1,5)𝑣
2
2
2
2
bottom
49 = 𝑣 2 bottom
Hence, v = 7 m/s.
5.4
Work done by non-conservative forces
Remember, non-conservative force is the force whose work done
depends on the path taken by the moving object. An example is frictional
force.
Suppose that we have a trolly that is ROLLING DOWN a rough incline
from point A to B as shown in the diagram below.
A
𝝑
B
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When the trolley is pulled upwards from A to B, it experiences two things:
 Negative work done on the trolly (due to frictional force)
 Loss in potential energy (due to decrease in height)
 Gain in kinetic energy due to increased speed down the
slope
The presence of friction results in loss of mechanical energy. However, no
energy is destroyed or created. Therefore, this means that the total mechanical
energy that the trolly originally had (at A) should qual to the mechanical energy
at the bottom plus lost energy due to friction. Thus,
EM(top) > EM(bottom)
(Ep + Ek)top > (Ep + Ek)bottom
Let work done by non-conservative forces be W nc. Since energy cannot be
destroyed or created, then we have:
Wnc + (Ep + Ek)bottom = (Ep + Ek)top
Wnc = (Ep + Ek)top - (Ep + Ek)bottom
Wnc = (Ep(top) - Ep(bottom)) + (Ektop - Ekbottom)
Wnc = ∆Ep + ∆Ek
Thus,
Learners are expected to know ONLY how to use the last equation.
EXAMPLE 1: Work done by non-conservative forces
Suppose a toy car of mass 2kg slides down a rough incline with an initial velocity
of 0,83m/s at point A (1,8m above the ground) and reaches the bottom at 3,5m/s
at point B.
A
1,8 m
𝝑
C
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Calculate:
1.1
the work done by nonconservative forces when the car moves from A to
B.
1.2
the gravitational potential energy when the car is at point A.
1.3
Work done by gravitational force.
The surface from B to C is frictionless.
1.4
What is the value of the net work done on the object as it slides from point
B to point C? Explain your answer.
Solution
1.1.
Friction is the only nonconservative force acting on the toy car. Using the
formula for nonconservative forces we get:
Wnc = ∆Ep + ∆Ek
1
Wnc = 𝑚𝑔(ℎ𝑓 − ℎ𝑖 ) + 𝑚(𝑣𝑓2 − 𝑣𝑖2 )
2
1
Wnc = (2)(9,8)(0 − 1,8) + (2)(3,52 − 0,832 )
2
Wnc = −35,25 + (12,25 − 0,6889)
Wnc = −35,25 + 11,5611
Wnc = −23,69 𝐽
Method 2
We can also use work-energy theorem to solve the problem.
Wnet = ∆Ek
It is easy to find the change in kinetic energy because we know the initial and
final velocities. However, the question is: how do we get net work done, Wnet?
Well, we find work done by all forces acting on the toy car and sum them up.
There are only two forces acting on the toy car, and these are:
 Gravitational force, Fg and
 Frictional force, f
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Therefore, from Wnet = ∆Ek, we have:
2
Wf + WFg = ½mvf - ½mvi
2
2
Wf + mgh = ½mvf - ½mvi
2
2
2
Wf + (2)(9,8)(1,8) = ½m(vf - vi )
2
2
Wf + (2)(9,8)(1,8) = (½)(2)(3,5 – 0,83 )
2
2
W + 35,28 = (½)(2)(3,5 – 0,83 )
f
Wf + 35,28 = 12,25 – 0,664
Wf = - 23,694 J
Method 3
We can also calculate the work done by friction (non-conservative force) by
considering mechanical energy at A and B and include frictional force in the
equation as follows:
E(mech)A = E(mech)B - Wf
(Ep + Ek)A = (Ep + Ek) B - Wf
2
2
(mgh + ½ mv ) A = (mgh + ½ mv ) B - Wf
2
2
(2)(9,8)(1,8) + ½(2)(0,83 ) = 0 + ½(2)(3,5 ) - Wf
35,28 + 0,6889 = 0 + 12,25 - Wf
Wf = - 23,71 J
1.2.
Gravitational potential energy is given by:
Ep = mgh
Ep = (2)(9,8)(1,8)
Ep = 35,28 J
1.3. Work done by gravitational force: W = FΔxcosθ, where F = Fg. Hence, we
have:
WFg = mgΔhcosθ
What’s the value of θ? Well, θ is the angle between Fg and
displacement.
Fg
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From the diagram of vectors above, Fg and ∆x are acting in the same
direction (downwards). Therefore, the angle between Fg and ∆x is zero.
Our equation becomes:
WFg
= mgΔhcos0
= (2)(9,8)(1,8)cos0o
= 35,28 J
Note:
1.4.
Work done by gravitational force will always equal to
gravitational potential energy for a similar problem as the
one above.
From B to C, there in is no friction. This means that velocity at B is equal
to the velocity at C. Eventually, there is going to be no change in kinetic
energy.
That is, Wnet = zero (because Wnet = ∆Ek = 0)
6. Power
In Physics, the term power refers to the rate of doing work. The SI units of
powers are watts (W).
Power =
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑡𝑖𝑚𝑒
Suppose a car is moving at a constant speed on a horizontal surface
covering a displacement of ∆x. The work done by the engine force is given
by:
WF = Fengine∆xcosθ
The average power from the engine is work done divided by time taken.
Thus,
P
=
=
The term
∆x
∆t
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑡𝑖𝑚𝑒
𝐹∆𝑥𝑐𝑜𝑠 𝜃
∆𝑡
represents average velocity (speed). The angle between F and
displacement is zero, hence cos 0 = 1. The equation for power then reduces
to:
Pave = F𝒗𝒂𝒗𝒆
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6.1
Average Power of a force
Problem 1: Consider a car of mass 750kg going up an incline plane at 30o to
the horizontal at a constant velocity of 15m/s. Calculate the average power of
the engine if the frictional force experienced by the car is 1200N.
B
30o
A
Finding the Solution
The average power of the engine of the car can be calculated from:
Pave = F𝑣𝑎𝑣𝑒
We know average velocity (=15m/s). But we don’t know F. how do we get this?
Well, lets consider all forces acting on the car as it moves up the inclined plane.
Let’s start by drawing force diagram of the situation.
x-axis
N
F
f
30
Fg
30
Fgy
Fg
Fgx
y-axis
The car experiences two forces:
 The x-component of gravitational force
 frictional force
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Work, Energy and Power 18 | P a g e
x-component of Fg is
Fgx
= Fgsin30
= mgsin30
= (750)(9,8)(0,5)
= 3675 J
Both Fgx and f are acting downwards (in the opposite direction to the engine
force). Since the car is moving at constant velocity, it means that there is no
acceleration. Hence, the net force is zero. Fnet = Fengine - (Fgx + f) = 0. Thus,
Fengine = Fgx + f.
Therefore, F engine = 3675 + 1200 = 4875 J.
The average power from the engine is therefore:
Pave
= F𝑣𝑎𝑣𝑒
= 4875 x 15
= 73125 W
6.2
Power of an Electric Motor
Electric motors are useful in daily life. We collect rain water into Jojo tanks at
home. What if there is no rain to collect the water? Well, we need another way
of collecting water from the ground to the tanks. We can use electric motors to
pump water from the ground to Jojo Tanks in our homes.
If you want to pump water using electric motors, the water will be pumped in
litres per minute or kilograms per minute. For example, a motor can pump
150kg per minute. Meaning that after two minutes, the tank will have 150 x 2 =
300 kg, after 3 minutes the tank will have 150 x 3 = 450kg of water.
In real life, we need powerful pumps that fill in our water tanks in the shortest
time. The pumps therefore need more power.
Let us look at the following example.
Problem 1
A water pump is used to fill in a tank 5m from underground as shown below.
Calculate the minimum power required of an electric motor to pump water at a
rate of 60kg per minute from a depth of 5m.
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by Mr. P (+27655795185)
Work, Energy and Power 19 | P a g e
Maximum
height
Water out for
home use
JoJo.
5m
Water pump
Underground water
Zero Point
Solving the problem
The upward force exerted by the pump is a non-conservative force. The work
done by the nonconservative force is equal to the change in mechanical energy
of the water
Wnc = ∆Ep + ∆Ek
The electric motor pumps water at a constant rate, therefore the water will move
through the pipe at a constant speed. Since there is no change in the speed
of the water then the change in kinetic energy is zero (∆Ek = 0). This means
that the work done by the electric motor is equal to the change in gravitational
potential energy of the water only:
Wnc = ∆Ep
(since ∆Ek = 0)
𝑊𝑛𝑐 = 𝐸𝑝𝑓 − 𝐸𝑝𝑖
𝑊𝑛𝑐 = 𝑚𝑔ℎ𝑓 − 𝑚𝑔ℎ𝑖
The initial height is zero (taking the position of the pump as a reference point).
Hence, we have:
Wnc
= mghf − 0
= 60×9,8×5
= 2940 J
The minimum power of the electric motor is:
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by Mr. P (+27655795185)
Work, Energy and Power 20 | P a g e
P
Work done
=
=
time
2940 J
60 sec
= 49 W
P
Method 2
The water pump is pumping the water a constant velocity 5m divided by 60
seconds = 0,083m/s. Therefore, the upward force F of the pump is equal in
magnitude to the weight of the water and thus:
F
= mg
= (60)(9,8)
= 588 N
The average power required to keep an object moving at constant velocity is:
Pave
= F𝑣𝑎𝑣𝑒
= (588)(0,083)
= 48,804 W
MSI
by Mr. P (+27655795185)
Work, Energy and Power 21 | P a g e