Unit 4 Work, Energy and Power

Unit Four
Work, Energy and Power
Learning Outcome
After completing this unit, students are expected to:
• Understand the concepts of work, energy, and power.
• Express these quantities in mathematical formula.
• Solve problems based on these concepts.
• describe the necessary conditions for work to be done.
• calculate work done by variable force and work done against gravity and against friction.
• distinguish between positive and negative work
• explain the relationship between work and energy.
• derive the relationship between work and kinetic energy and use this to solve problems.
• show the relationship between work and potential energy W =
• describe the mechanical energy as the sum kinetic energy and potential energy.
• stale the law of conservation of mechanical energy.
• describe the use of energy resources.
• define the term power and calculate it using its formula.
• express the formula of mechanical power in terms of average velocity.
Introduction
This chapter introduces: meanings of work done, kinetic and potential energies, power and how
each of these are calculated using a given quantities. Problems are considered from different
corners of real life within the context of simplified models.
4.1 The Concept of Work
Work and energy are important concepts in physics as well as in our everyday lives. In physics,
a force does work if its point of application moves through a distance and there is a component
of the force in the direction of the velocity of the force's point of application. For a constant
force in one dimension, the work done equals the force component in the direction of the
displacement times the displacement. (This differs somewhat from the everyday use of the
word work. When you study hard for an exam, the only work you do according to the use of
the word in physics, is in pushing your pencil on the paper, or turning the pages of your book.)

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Unit 4 Work, Energy and Power
Energy is closely associated with work. They are like two faces of the coin. They cannot separate
to each other. The energy of an object is a mathematical representation of the amount of work
the object can do. It is ability (capacity) to do work. On the other hand, when work is done by
one system on another, energy is transferred between the two systems. For example, when you
do work pushing a swing, chemical energy of your body is transferred to the swing and appears
as kinetic energy of motion or as gravitational potential energy of the earth-swing system. There
are many forms of energy. Kinetic energy is associated with the motion of an object. Potential
energy is associated with the configuration of a system, such as the separation distance between
two objects that attract each other (gravitational potential energy) or due to stretching or
compressing of elastic materials (elastic potential energy). Thermal energy is associated with the
random motion of the molecules within a system and is closely connected with the temperature
of the system. Some other forms of energy are: Electrical, Nuclear, Chemical, sound, wind
energy, e.t.c.
4.1.1. Work done by a constant force
We consider an object that is displaced along the x axis by ∆𝒙 by applying a constant force 𝑭
that makes an angle 𝜽 with the direction of motion, as shown in Fig. 6.1.

Fig. 4.1.1 Work done by a constant force

Then the work done on the object by the constant force (constant in both magnitude and
direction) is defined as the product of the magnitude of the displacement times the component
of the force parallel to the displacement. That is,
𝑊 = 𝑥𝐹𝑐𝑜𝑠𝜃,
where 𝐹𝑐𝑜𝑠𝜃 is the component of the force in the direction of the displacement, W is work
done, and ∆𝑥 is the magnitude of the displacement. We notice that work done can be positive,
negative or zero depending on the angle 𝜃 between the force and the displacement. The work
done is positive when the force has a component in the same direction as the displacement (00
≤ 𝜃< 900). On the other hand, when the force has a component opposite to the displacement
(900 <𝜃 ≤ 1800), the work done is negative. When the force is perpendicular to the
displacement, 𝜃 = 900and the work done by the force is zero.
Work is said to be done on an object by a force if (a) the force is not perpendicular to the
displacement (𝜃 ≠ 900); and (b) the force displaces the object. For instant, a person pushing a
fixed wall is not doing work since there is no displacement.

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Unit 4 Work, Energy and Power
Moreover, a person carrying a quintal of teff and moving horizontally is not doing work. In this
case the force applied on the
quintal is vertically upward and the F force (N)
displacement is in the horizontal
direction. If an applied force F is in the same F

direction as the displacement 𝑥, then Area under the graph = FS

𝜃= 0 and cos 0 =1. Area under the graph = Work done
In this case, the above Equation gives
𝑊=𝐹 𝑥
S

Distance moved against force (m)

Fig. 4.1.2 Plot of Force Vs distance

Work is a scalar quantity, and its units are force multiplied by length. Therefore, the
SI unit of work is the Newton. meter (N· m). This combination of units is used so frequently
that it has been given a name of its own: the joule ( J).
Plot of force against distance
We can plot a graph of the force applied against the distance travelled against the force. The
area under a force vs. distance moved graph is equal to the work done. The plot in the Figure
indicates graph of constant force acting over a distance.
Example 1
If you push a box 20 m forward by applying a force of 15 N in the forward direction, what is the
work you have done on the box?
Solution:
The force applied is F=15 N.
• The distance moved is s=20 m.
• The applied force and distance moved are in the same direction. Therefore, F||=15 N. Thus, the
work is
W = F||s
= (15 N)(20 m)
= 300 J
Example 2
What is the work done by you on a car, if you try to push the car up a
hill by applying a force of 40 N directed up the slope, but it slides downhill 30 cm?
Solution:
• The force applied is F=40 N
• The distance moved is s=30 cm=0.3m.

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Unit 4 Work, Energy and Power
• The applied force and distance moved are in opposite directions. Therefore, the angle between
them is 𝜃 = . Thus , the work done by the force is
W = Fs cos 1800
= (−40N)(0.3m)
= −12J
Example 3. F
Calculate the work done on a box, if it is pulled 5 m along the ground
by applying a force of F=10 N at an angle of 60◦ to the horizontal.
Solution:
• The force applied is F=10 N 600
• The distance moved is s=5 m along the ground
• The angle between the applied force and the motion is 600.
Thus the Work done is
W = Fs cos 600 Figure 4.1.3 Work done by constant force
= (5N)(5m)
= 25 J
Doing work against gravity
Work is often done against gravity. Whenever you lift up an object you are doing work against
the force of gravity. We can adapt our work done equation for working against gravity:
W=Fs
Work done against gravity = weight × vertical distance moved
Or Wgravity = w × h = mgh
Example: The work done in lifting a 60 kg mass vertically 3 m can be found using the work
done equation:
Wgravity = w × h
The weight is w = mg = 60 kg × 10 N/kg = 600 N
Wgravity= 600 N × 3 m
Wgravity= 1800 J
Work done against kinetic friction
We recall that whenever you push or pull an object along the ground you are doing work against
a force of kinetic friction. In Unit 3 we learnt that the kinetic friction is given as
=

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Unit 4 Work, Energy and Power
Kinetic friction always acts in the opposite direction
to motion‘ that is the kinetic friction and the
displacement makes 1800 .
Thus applying our work done equation, the
work done by friction is given as

=
= =

This result indicates that frictional force always does negative work. Its effect is always to
decrease the energy of the body on which it is doing work.
Note that the work done against friction is the negative of the work done by friction and is
expressible as
= =

Example: Determine the work done in pushing a 100 kg wooden block 30 m across a horizontal
concrete floor with = 0.48
Solution: The work done against friction is given as

W= Ns

In this case the normal contact force is equal to the weight (as the floor is horizontal) and so
N = w = mg
N = 100 kg × 10 N/kg = 1000 N
W= N×s
W = 0.48 × 1000 N × 30 m
W = 14 400 J or 14.4 kJ
This energy has been transformed into heat energy where the block and surface rub together.

Work done against Gravity and friction

If you were to push or pull on an object up a ramp
then you end up doing work against both friction
and gravity. In this case the total work done could
be found using the following equation:

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Unit 4 Work, Energy and Power
Total work done = work done against gravity
+ work done against friction
Wtotal= Wagainst gravity + Wagainst friction
Wtotal= (w × h) + (μkineticN × s)
Figure 4.1.4 Work done against gravity
We have to be very careful in considering the distances
we use in this equation; h has to be the vertical distance,
as this is the distance moved against gravity, whereas s must be the distance moved up the slope
as friction acts down the slope.
Example 1: A block of mass 4kg is pushed along an inclined plane that makes 450 from the
horizontal. If the coefficient of friction between the plane and the block is 0.2, find the total work
done on the in pushing it 4m along the plane.
Solution:
M=4kg, S=4m 𝜃=450, =0.2
What is required is Wtotal?
Wtotal= Wagainst gravity + Wagainst friction
Wtotal= (w × h) + (μkineticN × s)
= (mg s sin 𝜃) + (μkinetic mgcos 𝜃 s)
= (4kgx10m/s2x4m x 0.7)+(0.2x4kgx10m/s2x0.7x4m)
=112J+22.4 J=134.4J

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Unit 4 Work, Energy and Power

Review Questions

N
1. A 10 N force is applied to push a block across a friction
free surface for a displacement of 5.0 m to the right. Fig(a)
The block has a weight of 20 N. Determine the work done by
the following forces: normal force, weight, applied force.
Fapp

2. A 10 N frictional force slows a moving block to a stop after a)

a displacement of 5.0 m to the right. The block has a weight of 20 N. Fg
Determine the work done by the following forces: normal force, weight,
frictional force .Fig (b)
N

3. A 10 N force is applied to push a block across a frictional

surface at constant speed for a displacement of 5.0 m to the ffriction
right. The block has a weight of 20 N and the frictional
force is 10 N. Determine the work done by the following b)
forces: normal force, weight, frictional force. .Fig (c) Fg
N
4. A 20 N object is sliding at constant speed across a friction
free surface for a displacement of 5 m to the right. Determine Fapp
if there is any work done. .Fig (d)
ffriction
5. A 20 N object is pulled upward at constant speed
by a 20 N force for a vertical displacement of 5 m. c)
Fg
Determine if there is any work done. .Fig 4.12 (e)
T N

e) d)
Fg
Fg
Figure 4.1.5: Work done with
constant force

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4.1.2. Work done by a varying force
If a force is variable ( for example if it varies with position) , we cannot use W=F 𝑥 cos𝜃 to
calculate the work done by the force because this relationship applies only when F is constant in
magnitude and direction. However, for a force that varies uniformly with the displacement, the
work done can readily be calculated by multiply the average of the magnitude of the force over
the displacement by the magnitude of displacement if they are parallel
W= Fav 𝑥
If the block undergoes an arbitrary dis
Since the force varies uniformly Fav = , where 𝐹 𝑛 𝐹 are the initial and final force, thus
the work is given as
W= 𝑥
Example: A force acting a body varies uniformly from 20N to 40N in displacing the body by 5m
along the direction of the displacement. Calculate the work done by the force.
Solution: The average force is
Fav = = =
The work done by the force cann be calculated as
W= Fav 𝑥 = 𝑚=
Work done by a spring force
A block on a horizontal, frictionless surface is connected to a spring. If the spring is either
stretched or compressed a small distance from its unstretched (equilibrium) configuration, it
exerts a force on the block a force that can be expressed using Hook‘s law as
F= kx
where x is the position of the block relative to its equilibrium (x =0) position. The force varies
linearly with x.

Figure 4.1.6: Work done by a spring

If the block undergoes an arbitrary displacement from an initial position say xi to final position
say xf , then the work done by the spring Ws from xi to xf, is given as

W s= Fav 𝑥= W= 𝑥 , where Fi = kxi and Ff = kxf , 𝑥 = xf xi

Ws= 𝑘( ) (𝑥 𝑥 ) = 𝑘𝑥 𝑘𝑥 (work done by a spring force).

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Unit 4 Work, Energy and Power
Ws = 𝑘𝑥 𝑘𝑥

This work done by the spring force Ws can have a positive or negative or zero value, depending
on whether the net transfer of energy is to or from the block as the block moves from xi to xf.
Caution: The final position xf appears in the second term on the right side.
Case-1: If xi= 0 and the final position x

Ws= = 𝑘𝑥

Case-2: When xi= xf, then W= 0

Case-3: When the spring is stretched, xi < xf, then Ws is negative or W< 0
Case-4: When the spring is compressed, xi > xf, then Ws is positive or W > 0
4.1.3 The work done by a force applied on a spring
Now let us consider the work done on the spring by an external agent that stretches the spring
very slowly and steadily (from rest) from xi to xf .The applied force increases uniformly up to the
elastic limit of the spring.

xi = 0

a) Unstretched

Fapp

b) Stretched

xf Figure 4.1.6 Work done by force applied on a spring

Then the applied force Fapp or Fp is equal in magnitude and opposite in direction to the spring
force FS = 𝑘𝑥 so that Fapp= ( kx) = kx. Therefore, the work done by this applied force (the
external agent) on the block–spring system is
Work done by the applied force = Work done by a spring
Wapp = Ws
= ( 𝑘𝑥 𝑘𝑥 )

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Unit 4 Work, Energy and Power
= 𝑘𝑥 𝑘𝑥
Graph of force that varies linearly position. For example the graph may represent the force
exerted on a block–spring system.
The work done by the force as the block moves from x = 0 to final x is the area of the shaded
triangle
Area under the graph = Work done
= 𝒙 .

Example 1: A spring has force constant 100N/m. How much work is done by a boy that
compress the spring by 20cm from its equilibrium position?
Solution:
K=100N/m , x= 0.2m
What is required is Wapp?
Wapp= = ⁄
Wapp=2J

4.2 Kinetic energy and work-energy theorem

Kinetic Energy
Kinetic energy KE is energy associated with the state of motion of an object. The faster the
object moves, the greater is its kinetic energy. When the object is stationary, its kinetic energy is
zero. For an object of mass m whose speed v is well below the speed of light
= 𝑚

Example 1.
A 3.0 kg duck flying past us at 2.0 m/s has a kinetic energy of 6.0 kg/m2/s2; that is, we associate
that number with the duck‘s motion. Thus, the flying duck has a kinetic energy of 6.0 J.
Example 2.
An1800kg car travelling at a steady speed is acquired a kinetic energy of 3240kJ. What was the
speed of the car?
KE= mv2
3240000J= (1800kg)v2
V2=3240000/900m2/s2
= 3600 m2/s2
V=√ m/s = 60m/s

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Unit 4 Work, Energy and Power
Work and kinetic energy
The total work done on a body by external forces is related to the body's displacement-that is,
to changes in its position. But the total work is also related to changes in the speed of the body.
Consider a particle with mass 𝑚 moving along the x-axis under the action of a constant net
force with magnitude 𝐹 directed along the positive x-axis (Fig. 4.2.1). Suppose the speed
changes from 1 to 2 while the particle undergoes a displacement 𝑥= 𝒙 −𝒙 from point 𝑥1
to 𝑥2.

Fig. 4.2.1 Block acted upon by net force

This constant net force on the object is given by Newton‘s second law as
𝑭 = 𝑚𝒂
Now since the force is constant, the acceleration will also remain constant and thus can be
written as

=
𝑠
Upon combining the above Equations, the net force is expressible as

𝐹=𝑚
or
𝐹𝑠 = 𝑚 𝑚

The product 𝐹𝑠 is the work done by the net force and the quantity

= 𝑚

is called kinetic energy of a mass 𝑚 moving with speed . It is energy due to the motion of an
object. Thus, 𝑚 is the final kinetic (KEf) energy of the mass and 𝑚 is its initial kinetic
energy (KEi). Thus one can write
𝑊𝑛𝑒𝑡= 𝑓− 𝑖

We therefore observe that this Equation relates the work done on an object by a net force to the
change in its kinetic energy and is known as work-energy theorem.

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Unit 4 Work, Energy and Power
Then work-energy theorem can then be stated as the work done by a net force on an object is
equal to the change in its kinetic energy.
The work–kinetic energy theorem indicates that the speed of an object increases if the net work
done on it is positive because the final kinetic energy is greater than the initial kinetic energy.
The speed decreases if the network is negative because the final kinetic energy is less than the
initial kinetic energy. On the other hand, no work is done by the net force in moving an object
with a constant speed.
Example 1: A car of mass 2000kg is changing its speed uniformly from 10m/s to 30m/s through
a distance of 800m. a) What is the net work done by the car? b) What was the average net force
acting on the car?
Solution:
Given quantities are:
m= 2000kg u=10m/s2 V=30m/s2 S=800m
a) W=?
W=
= mv2 mu2
= 30)2 (2000)(10)2
=900000J-100000J
=800000J

b) F=?
W= FS
F=W/S= 800000J/800m
=1000N
Example 2: A 61 kg skier on level snow coasts 184 m to a stop from a speed of 12.0 m/s. Use
the work– energy principle to find the coefficient of kinetic friction between the skis and the
snow.
Solution:
𝑚 = 61 𝑘𝑔, 𝑠 = 184𝑚,= 12.0 𝑚/𝑠, 𝑓= 0 𝑚/𝑠, 𝑘= ?
According to the work-energy theorem, the change in the kinetic energy of the skis is equal to
the work done by the net force acting the skis. That is,
𝑊𝑛𝑒𝑡= 𝑓− 𝑖
The net horizontal force on the skis is the kinetic frictional force,
𝑓𝑘= 𝑘𝐹 = 𝑘𝑚𝑔
Consequently, the work done by the net force is equal to the work done by frictional force. We
thus have
𝑊𝑛𝑒𝑡= −𝑓𝑘𝑠 = − 𝑘𝑚gs
Using the given values, we obtain
2
−𝑚 𝑖 = − 𝑘𝑚 𝑠

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Unit 4 Work, Energy and Power

= =
𝑔𝑠

Review Questions
1. Describe the relationship between an object‘s kinetic energy and its:
(a) mass and (b) velocity
2. A stone with a mass of 100 g is thrown up into the air. It has an initial velocity of 3 m·s−1.
Calculate its kinetic energy
(a) as it leaves the thrower‘s hand. (b) when it reaches its turning point.
3. A car with a mass of 700 kg is travelling at a constant velocity of 100 km·hr−1. Calculate the
kinetic energy of the car.
4. A force of magnitude 5N acts on a 2kg body moving initially in the direction of the force with
a speed of 4m/s. Over what distance must the force act in order to change the body‘s speed to
6m/s?
5. A 50N force is applied horizontally to 10kg block which is initially at rest. After traveling
8m, the speed of the block is 5m/s. What is the coefficient of kinetic fraction?

4.3. Potential Energy and Conservation of Energy

In this section we will discuss another form of mechanical energy, called potential energy,
associated with the position or configuration (arrangement) of object. Thus, the potential energy
of a system of interacting objects represents the ability of the system to do work because of its
position or configuration.
Potential energy is present in the Universe in various forms, including gravitational, elastic,
electromagnetic, chemical, and nuclear. Furthermore, one form of energy in a system can be
converted to another. For example, when a system consists of an electric motor connected to a
battery, the chemical energy in the battery is converted to kinetic energy as the shaft of the motor
turns. The transformation of energy from one form to another is an essential part of the study of
physics, engineering, chemistry, biology, geology, and astronomy.

4.3.1. Gravitational potential energy

Gravitational potential energy U is the energy associated with the
state of separation between two objects that attract each other by the gravitational force. Here
we consider a physics book of mass 𝑚 lifted from an initial height
𝑦i to a final height above the surface of the earth, as indicated
in Fig. 4.3.1

Fig. 4.3.1 A physics book lifted up

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The work done by gravitational force as the book is raised from the initial height 𝑦𝑖 to a final
height 𝑦𝑓 from the ground is given by
𝑊𝑔=|𝐹𝑔||Δ𝑟|cos𝜃,
where 𝐹𝑔=−𝑚𝑔𝑗 is the gravitational force on the book and Δ𝑟=(𝑦𝑓−𝑦𝑖)𝑗 is the displacement and
𝜃 is the angle between the gravitational force and the displacement. In Figure , 𝜃=180°. It then
follows that
𝑊𝑔=−[𝑚𝑔𝑦𝑓−𝑚𝑔𝑦𝑖].
The quantity 𝑔=𝑚𝑔𝑦 is called gravitational potential energy. Thus 𝑔𝑓=𝑚𝑔𝑦𝑓 is the final
gravitational potential energy and 𝑔𝑖=𝑚𝑔𝑦𝑖 is the initial gravitational potential energy of the
earth-book system. Accordingly, the work done by gravitational force can be expressed as
𝑊𝑔=−[ 𝑔𝑓− 𝑔𝑖]=−Δ 𝑔.
We observe that work done by gravitational force is equal to the negative of the change in
gravitational potential energy. When the object moves down y decreases, the gravitational force
does positive work, and the potential energy decreases. When the object moves up, the work
done by the gravitational force is negative, and the potential energy increases.
It is often convenient to choose as the reference configuration for zero potential energy the
configuration in which an object is at the surface of the Earth, but this is not essential.
Example 1: A 2kg book is placed on a shelf 2.5m height above the ground. What is the
Gravitational potential energy of the book?
Ug= mgh
Ug= (2kg)(10m/s2)(2.5m)
=50J
Example 2: A 1kg apple is falling down freely through a distance of 1.5m from the top of the
tree.
a) What is loss in Gravitational Potential Energy of the apple?
b) Where does it go the loss in GPE?
Solution
a) The loss in GPE= Δ 𝑔 =mgy
=1kgx10m/s2x1.5m
=30J
b) It changes into kinetic energy

Review Questions
1. A boy, of mass 30 kg, climbs onto the roof of their garage. The roof is 2.5 m from the ground.
He now jumps off the roof and lands on the ground.
(a) How much potential energy has the boy gained by climbing on the roof?
(b) He now jumps down. What is his potential energy when he is 1 m from the ground?
(c) What is the potential energy of the boy when he lands on the ground?

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2. A hiker walks up a mountain, 800 m above sea level, to spend the night at the top in the first
overnight hut. The second day he walks to the second overnight hut, 500 m above sea level. The
third day he returns to his starting point, 200 m above sea level.
(a) What is the potential energy of the hiker at the first hut (relative to sea level)?
(b) How much potential energy has the hiker lost during the second day?
(c) How much potential energy did the hiker have when he started his journey (relative to sea
level)?
(d) How much potential energy did the hiker have at the end of his journey?
4.3. 2. Elastic potential energy
We are familiar now with gravitational potential energy; let us explore a second type of potential
energy. When a spring is stretched (or compressed) from its equilibrium position, it has ability to
do work as it returns to the equilibrium position. Thus, the spring may have the potential for
doing work because of its stretch (or compression). Energy that is stored in an elastic object
when you stretch, compress, twist, or otherwise deform it is called elastic potential energy.
Consider an object of mass 𝒎 attached to a spring of spring constant , as shown in Fig.4.3.2

Fig. 4.3.2 A block attached to a spring and acted upon by a force

For a force that is a linear function of position, such as spring force (𝐹𝑠𝑝=−𝑘𝑥), the work done by
the force is the average force multiplied by the displacement. Accordingly, the work done by

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spring force when the spring is stretched from 𝑥𝑖 to 𝑥𝑓 is the product of the average spring force
and the displacement. That is;

The quantity = 𝑘𝑥 is elastic potential energy of spring-mass system when the spring is
stretched or compressed by 𝑥. We thus notice that = 𝑘𝑥 is the final elastic potential
energy and = 𝑘𝑥 is the initial elastic potential energy. Accordingly, we observe from
the above equation that the work done by spring force when the spring is stretched from 𝑥𝑖 to 𝑥𝑓
is equal to the negative of the change in elastic potential energy. That is,

Example 1: Calculate the energy stored in a spring when it is stretched 8cm by a 100N force?
Solution:
F= kx and Fav=1/2(kx)=1/2x100N =50N
W= (Fav)(x)= 50Nx0.08m= 4J
Example 2: A spring with a spring constant of 1500N/m is compressed by 5cm. How much
elastic potential energy will be stored in the spring?
Solution:
k=1500N/m, x=5cm =5x10-2 m
Us=?
Us = kx2
= (1500N/m)(5x10-2 m)2
=1.875J
4.3.3 The Isolated System–Conservation of Mechanical Energy
The introduction of potential energy allows
us to generate a powerful and universally
applicable principle for solving problems
that are difficult to solve with Newton‘s
laws. Let us develop this new principle by
thinking about the book–Earth system
in the Figure again

Figure 4.3.3 Conservation of Mechanical

Energy

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After we have lifted the book, there is gravitational potential energy stored in the system, which
we can calculate from the work done by the external agent on the system, using W= Δ 𝑔.
Let us now shift our focus to the work done on the book alone by the gravitational force (Figure)
as the book falls back to its original height. As the book falls from yf to yi, the work done by the
gravitational force on the book is
𝑊on book =|𝐹 𝑔||Δ𝑟|cos𝜃 =−[𝑚𝑔𝑦i−𝑚𝑔𝑦f]= 𝑚𝑔𝑦f−𝑚𝑔𝑦i
From the work–kinetic energy theorem, the work done on the book is equal to the change in the
kinetic energy of the book:
Won book =ΔKbook
Therefore, equating these two expressions for the work done on the book,
ΔKbook = 𝑚𝑔𝑦f−𝑚𝑔𝑦i
Now, let us relate each side of this equation to the system of the book and the Earth. For
the right-hand side,
𝑚𝑔𝑦f−𝑚𝑔𝑦i= −[𝑚𝑔𝑦i−𝑚𝑔𝑦f]=-(Uf-Ui)= - ΔUg
where Ug is the gravitational potential energy of the system. For the left-hand side of the
Equation, because the book is the only part of the system that is moving, we see that ΔKbook =ΔK,
where K is the kinetic energy of the system. Thus, with each side of the Equation replaced with
its system equivalent, the equation becomes
ΔK= -ΔUg
This equation can be manipulated to provide a very important general result for solving
problems. First, we bring the change in potential energy to the left side of the equation:
ΔK+ΔUg=0
On the left, we have a sum of changes of the energy stored in the system. The right hand is zero
because there are no transfers of energy across the boundary of the system—the book–Earth
system is isolated from the environment.
We define the sum of kinetic and potential energies as mechanical energy:
Emech+K+Ug
We will encounter other types of potential energy besides gravitational later in the text, so we
can write the general form of the definition for mechanical energy without a subscript on U:
Emech=K+U
where U represents the total of all types of potential energy.
Let us now write the changes in energy explicitly:
(Kf -Ki)+(Uf -Ui)=0
Kf +Uf =Ki +Ui
For the gravitational situation that we have described, one can written as
𝑚 𝑚𝑔𝑦 = 𝑚 𝑚𝑔𝑦
As the book falls to the Earth, the book–Earth system loses potential energy and gains kinetic
energy, such that the total of the two types of energy always remains constant.

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Unit 4 Work, Energy and Power
The above Equation is a statement of conservation of mechanical energy for an isolated system.
An isolated system is one for which there are no energy transfers across the boundary. The
energy in such a system is conserved—the sum of the kinetic and potential energies remains
constant. (This statement assumes that no nonconservative forces act within the system; such as
frictional force)

Example 1: A bird of 5000g is flying at a speed of 20m/s at an altitude of 600m above the
ground. Determine the mechanical energy of this bird?
Mechanical energy = ∑ 𝑜𝑡𝑒𝑛𝑡𝑖 𝑛𝑒𝑟𝑔𝑦 ∑ 𝑖𝑛𝑒𝑡𝑖𝑐 𝑛𝑒𝑟𝑔𝑦
ME= PE + KE
K= mv = (5kg)(20m/s)2=1000J and Ug= mgh =5kgx10m/s2 x600m =30000J
2

Emech= 30000J + 1000J =31000J

Example 2: During a flood a tree truck of mass 100 kg falls down a waterfall. The waterfall is
5 m high. If air resistance is ignored, calculate
a. the potential energy of the tree trunk at the top of the waterfall.
b. the kinetic energy of the tree trunk at the bottom of the waterfall.
c. the magnitude of the velocity of the tree trunk at the bottom of the waterfall.
m=100kg h=5m
a) Ug=?
Ug=mgh
=100KgX10m/sX5m=5000J
b) Using conservation of mechanical energy
Ebottom= Etop
At the top the energy is purely potential and at the bottom the
energy is purely kinetic. Thus
K=Ug=mgh=5000J
c) Using the expression for kinetic energy
mv2 =mgh
= √ 𝑔𝑕=10m/s
Conservative and Non-conservative (Dissipative) Forces
Conservative Forces
Conservative forces have these two equivalent properties:
1. The work done by a conservative force on a particle moving between any two points is
independent of the path taken by the particle.
2. The work done by a conservative force on a particle moving through any closed path is zero.
(A closed path is one in which the beginning and end points are identical.) The gravitational
force is one example of a conservative force, and the force that a spring exerts on any object
attached to the spring is another.
In general, the work Wc done by a conservative force on an object that is a member of a system
as the object moves from one position to another is equal to the initial value of the potential
energy of the system minus the final value:

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Unit 4 Work, Energy and Power
𝑊 = =
Nonconservative Forces
A force is non-conservative if it does not satisfy properties 1 and 2 for conservative forces. Non-
conservative forces acting within a system cause a change in the mechanical energy E of the
system.
Energy in oscillating systems
We have seen that when an object falls its GPE is converted into kinetic energy. The same is true
if you throw an object into the air. Here the kinetic energy is transformed into GPE as it rises.

In oscillating systems kinetic energy is continuously being transformed into potential energy and
vice-versa. If there are no energy losses, then the total mechanical energy will remain constant
forever.
Consider the following examples
A) The oscillation of Simple pendulum
In oscillation of simple pendulum, in the absence of friction, the
mechanical energy is conserved when the bob swings between
A and B. The two points are called amplitude position.
Point C refers to the equilibrium position of the bob.
At amplitudepositions( at A and B) A B
KEA=KEB=0 ME A = KE A + GPE A =mgh D h
GPEA=GPEB=mgh ME B=KE B + GPE B =mgh
MEA= MEB=mgh
C
At equilibrium positions(at C) MEA= KE A+ GPEA=mgh 4.3.4 A simple pendulum
KEC= m MEB=KEB + GPEB=mgh transforms GPE into K.E
GPEc= 0 ME c = KE B+ GPE B= m and then back again.
At positions D

Position D is any position on the curve between C and B or C and A. At this position both
Kinetic energy and potential energy are existed.
MEA= KE A+ GPEA=mgh
KED = m MEB=KEB + GPEB=mgh
GPED= mghD MEc= KEB+ GPEB= m
The mechanical energy at any position is conserved.
MED= KED + GPED
Therefore, the mechanical energy at each position is
the same. = m + mghD
MEA=MEB=MEC=MED
mgh= m = m + mghD

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Unit 4 Work, Energy and Power
The total mechanical energy = kinetic energy + potential energy
Another example of an oscillating system is a mass–spring system. In simple terms this is just a
mass on the end of a spring. However, the suspension in a car is a more complex example of a
mass–spring system. In this case the potential energy may not be GPE, instead it may be EPE.

4.3.5 Graph showing how the potential energy and kinetic

energy of oscillating systems are related

As the spring is compressed the EPE increases

and the mass slows down (its Ek decreases).
Eventually the mass will stop; at this point
the EPE is at its maximum and the
Ek is zero. The mass then accelerates as EPE
is converted into Ek. This process continues. Fig. 4.3.6 An example of a mass–spring
system

A more complex example might be a mass–spring

system oscillating vertically like the one shown in
Figure 4.3.5 In this case the kinetic energy is changed into
GPE and EPE. In any case the total mechanical energy
of the system remains then same

Fig. 4.3.7 A vertically oscillating

mass–spring system

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Unit 4 Work, Energy and Power
Energy resources
Every country demands a huge amount of energy, from fuel to run cars and other vehicles, to gas
for cooking and heating and, of course, electrical energy. A source of energy that may be used
by a country or individuals within that country is commonly referred to as an energy resource.
Examples of energy resources are Oil, Fuel, Nuclear, Natural gas, Solar, Geothermal …etc
Energy resources are very precious commodities, perhaps the most obvious being oil.
Selecting which energy resources to use is often a very difficult decision. There are lots of
factors to consider, chief among them being the availability of the resource, the economics
involved and the subsequent environmental impact (more on this later).

We can‘t use each directly as it is. We need a suitable device to convert the naturally existing
energy resources into useful forms of energy.

Wastage
Energy

Energy from natural Energy Useful work done or

source supplied to Converter energy converted
energy converter

Fig. 4.3.8 Energy converter

Wind mill, Hydroelectric power generator, solar power …etc are examples of energy convertors.
Energy resources are often used to generate electricity.
Electricity is exceptionally useful as it is quite simple
to transfer a vast amount ofenergy from one place to
another (all you need is a suitable wire!) and it can be
easily transformed into most other forms of energy.
Most methods of electricity generation involve a rotating
turbine. This turbine turns a generator (a magnet or series
of magnets inside coils of wire). This generator converts
kinetic energy into electrical energy.
Fig. 4.3.9 A simple generator

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Unit 4 Work, Energy and Power
Globally the most common method for generating electricity involves the burning of fossil fuels
such as coal, oil and natural gas. The chemical energy contained within these fuels is released as
heat (through burning), this heat is used to turn water into steam, this steam then
turns a turbine to generate electricity.
Large fossil fuel power stations
can generate up to 4 billion joules
per second! However, such a global
reliance on fossil fuels is
problematic for two main
reasons.

 Fossil fuels are a finite energy Fig. 4.3.10 An example of a coal-fired

resource. Eventually we will run power station
out of coal, oil and natural gas.
 Burning fossil fuels produces several atmospheric pollutants, including sulphur dioxide
and perhaps more worryingly, carbon dioxide. Carbon dioxide (CO2) is a powerful
greenhouse gas. It is thought the increase in CO2 output is a significant factor in
manmade global warming, heating up the entire planet and leading to dramatic changes
to weather and climate.
Ethiopia has few proven fossil fuel resources. However, some people estimate that there is
considerable potential for oil and natural gas exploration in the future. In a nuclear power station
uranium is used as a fuel. Inside the reactor there is a complex nuclear reaction (fission –
splitting the atom). This process generates heat, which is used to turn water to steam, etc. The
only real difference between a nuclear power station and a coal-fired one is the method for
generating the heat. In a nuclear reactor a great deal of heat can be produced per kg of uranium,
and so nuclear plants can generate vast amounts of electricity. As no fuel is ‗burnt‘ there are no
greenhouse gases produced; however, this process produces radioactive waste. This waste will
remain dangerous for millions of years.
Renewable and non-renewable energy resources
Resources that do not involve a fuel that will eventually run out are referred to as renewable.
Table 4.3.1 includes a selection of some of the forms of renewable energy resources. This is not
a definitive list; other forms include tidal (energy from tidal movements), wave (energy from
water waves) and biomass (burning organic matter specifically grown for the task).

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Unit 4 Work, Energy and Power
Table 4.3.1 Comparison of some renewable energy resources

Energy in Ethiopia

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Unit 4 Work, Energy and Power
In 2008 as a country we generated just over 1 × 1016 J (10 000 000 000 000 000 J!!) of electrical
energy. At the time of writing around nearly all of our electricity generation comes from
hydroelectric power.

The location of hydroelectric power plants This graph shows the amount of electricity
generated per resource.

As part of the country‘s general development plan, with the aim of expanding the Electric Power
generation capacity, the GERD, Gilgel Gibe III, Tekeze, Genale-Dawa (GD-3), Gilgel Gibe II
and Tana Beles power plants with respective generating capacities of 6450MW, 1870MW,
300MW, 254MW, 184MW and 460MW became operational in 2009 and 2010. Reliance on
hydroelectric power has advantages and disadvantages, as listed in Table 4.3. Ethiopia can
diversify its electricity sources by exploiting its geothermal (> 9000 MW) and wind (>10
000MW ) electricity generating Potential. Figure 4.42 shows the location of several hydroelectric
power plants. Ethiopia is among only a few African countries with the potential for significant
energy generation to come from geothermal wind power.
Non-renewable energy resources
Fossil fuels such as Coal, gas and oil are energy sources which are results of
ancient remains of plants and animals. The energy from such supplies is used by
burning then. Once you burn them the energy is used up forever. You cannot
replace the energy from such sources.
The sources which give energy that cannot be replaced are known as non-
renewable energy sources.
Nuclear energy is also another form of non-renewable energy source which

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Unit 4 Work, Energy and Power
developed nations use to generate electricity. The world has a tendency of
replacing non-renewable energy sources by renewable sources.
4.4. Power
Consider Conceptual Example, which involved rolling a refrigerator up a ramp into a truck.
Suppose that the man is not convinced by our argument that the work is the same regardless of
the length of the ramp and sets up a long ramp with a gentle rise. Although he will do the same
amount of work as someone using a shorter ramp, he will take longer to do the work simply
because he has to move the refrigerator over a greater distance. While the work done on both
ramps is the same, there is something different about the tasks—the time interval during which
the work is done.

Figure 4.2.11 Mechanical power

method in this discussion, but keep in mind that the notion of power is valid for any means of
energy transfer. If an external force is applied to an object (which we assume acts as a particle),
and if the work done by this force in the time interval t is W, then the average power during
this interval is defined as
𝑊𝑜𝑟𝑘 𝑜𝑛𝑒
𝑝 =
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟 𝑙
𝑊
𝑝 =
𝑡
Thus, while the same work is done in rolling the refrigerator up both ramps, less power is
required for the longer ramp.

Power in terms of Force and Velocity

We already know that power is the time rate of Work done. So, at any given point in time, the
power can be defined as:

P=

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Unit 4 Work, Energy and Power
W is the work done in the time period of t. Since, we know that W = F.S and F is constant, we
can rewrite the above equation as:

P=F.
Here, F is the force, S is the displacement of the object and t is the time period of that
displacement. Since, we know that vava = S / t is the average velocity of the object, we can rewrite
the above equation as:

P=Fvave
For steady velocity, vave = v, then we can express the power in terms of force and velocity as

P=Fv
P = Fvcos𝜃
Where 𝜃 is the angle between F and V vectors..
When a body accelerates uniformly from Vi to Vf, the average velocity can be taken as

vave = then

P = Fvave
In general, power is defined for any type of energy transfer. Power is the rate at which energy is
crossing the boundary of the system by a given transfer mechanism.
The SI unit of power is joules per second (J/s), also called the watt (W) (after James Watt):
Another unit of power is the horsepower (hp):
1 hp = 746 W
A unit of energy (or work) can now be defined in terms of the unit of power.
One kilowatt-hour (kWh) is the energy transferred in 1 h at the constant rate of 1 kW =1 000 J/s.
The amount of energy represented by 1 kWh is
1 kWh = (103W)(3600 s) =3.60 X 106 J
Note that a kilowatt-hour is a unit of energy, not power. When you pay your electric bill, you are
buying energy, and the amount of energy transferred by electrical transmission into a home
during the period represented by the electric bill is usually expressed in kilowatt-hours.

Example 1:
a) How much work is done by a 500 W motor running for 30 minutes?
In joules:

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Unit 4 Work, Energy and Power
W=Pt
In this case the time taken is 30 minutes, which is 1800 s, and P = 500 W.
W = 500 W X1800 s
W = 900 000 J or 900 kJ

In kilowatt-hours:
W=Pt
In this case the time taken is 30 minutes, which is 0.5 hours, and P= 500 W, which is 0.5 kW.
W = 0.5 kWX0.5 h
W = 0.25 kWh

Example 2: Calculate the power required for a force of 10 N applied to move a 10 kg box at a
speed of 1 m/s over a frictionless surface.
Solution:
F=10N and V=1m/s
We can therefore calculate power from:
P=F·v
= (10N)(1m/s)
= 10W

Summery
A system is most often a single particle, a collection of particles or a region of space. A system
boundary separates the system from the environment. Many physics problems can be solved by
considering the interaction of a system with its environment.
The work W done on a system by an agent exerting a constant force F on the system is the
product of the magnitude ∆𝑥 of the displacement of the point of application of the force and the
component F cos 𝜃 of the force along the direction of the displacement ∆𝑥:
𝑊 = ∆𝑥𝐹𝑐𝑜𝑠𝜃

Frictional force always does negative work. Its effect is always to decrease the energy of the
body on which it is doing work.
= =
The kinetic energy of a particle of mass m moving with a speed v is
= 𝒎
The work–kinetic energy theorem states that if work is done on a system by external forces and
the only change in the system is in its speed, then

𝑊𝑛𝑒𝑡= 𝑓− 𝑖

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Unit 4 Work, Energy and Power
If a particle of mass m is at a distance y above the Earth‘s surface, the gravitational potential
energy of the particle–Earth system is
Ug = mgy

The elastic potential energy stored in a spring of force constant k is

= 𝑥
A reference configuration of the system should be chosen, and this configuration is often
assigned a potential energy of zero.
A force is conservative if the work it does on a particle moving between two points is
independent of the path the particle takes between the two points. Furthermore, a force is
conservative if the work it does on a particle is zero when the particle moves through an arbitrary
closed path and returns to its initial position. A force that does not meet these criteria is said to be
nonconservative.

The total mechanical energy of a system is defined as the sum of the kinetic energy and the
potential energy:
Emech = K +U
If a system is isolated and if no nonconservative forces are acting on objects inside the system,
then the total mechanical energy of the system is constant:

Kf + Uf =Ki + Ui

If nonconservative forces (such as friction) act on objects inside a system, then mechanical
energy is not conserved. In these situations, the difference between the total final mechanical
energy and the total initial mechanical energy of the system equals the energy transformed to
internal energy by the nonconservative forces.

The average power during this interval is defined as

𝑊𝑜𝑟𝑘 𝑜𝑛𝑒 𝑊
𝑝 = =
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟 𝑙 𝑡
The instantaneous power P is defined as the time rate of energy transfer. If an agent applies a
force F to an object moving with a velocity v, the power delivered by that agent is
P=F.V

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Unit 4 Work, Energy and Power

End of unit Questions and problems

Questions
1. Cite two examples in which a force is exerted on an object without doing any work on the
object.
2. One bullet has twice the mass of a second bullet. If both are fired so that they have the same
speed, which has more kinetic energy? What is the ratio of the kinetic energies of the two
bullets?
3. (a) If the speed of a particle is doubled, what happens to its kinetic energy? (b) What can be
said about the speed of a particle if the net work done on it is zero?
4. An elevator is pulled by its cables at constant speed. Is the total work done on the elevator
positive, negative or zero?
5. Explain why the total energy of a system can be either positive or negative, whereas the
kinetic energy is always positive.
6. Are there any cases where a frictional force can increase the mechanical energy of a system?
If so give examples.
7. If only one external force acts on a particle, does it necessarily change the particle‘s (a) kinetic
energy? (b) velocity?
8. Discuss the changes in mechanical energy of an object–Earth system in (a) lifting the object,
(b) holding the object at a fixed position, and (c) lowering the object slowly. Include the
muscles in your discussion.
9. Can the average power over a time interval ever be equal to the instantaneous power at an
instant within the interval? Explain.
10. As a simple pendulum swings back and forth, the forces acting on the suspended object are
the gravitational force, the tension in the supporting cord, and air resistance.
(a) Which of these forces, if any, does no work on the pendulum?
(b) Which of these forces does negative work at all times during its motion?
(c) Describe the work done by the gravitational force while the pendulum is swinging.

Problems
1) How much energy is stored on a 2000N/m spring when it is compressed by a 4cm?
2) By what amount will it be stretched in order to store a 5.4J of energy on a 1500N/m spring?
3. Suppose a student stretches a spring with a spring constant of 1600N/ in increasing the force
applied uniformly from 0 to 80N. a) How much energy does he store in the spring?
b) How much work is done on the spring? c) How much work is done by the spring?

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Unit 4 Work, Energy and Power
4. a) Calculate the energy stored in a spring when it is compressed 8mm by an average force of
120N? b) What is the spring constant of the spring?
5) A 1000kg car starting from rest accelerates uniformly through a distance of 800m along a
straight road. The net force causing this acceleration is 810N. Determine the final speed of the
car.
6) A boy kicks a ball of mass 80g and the ball leaves his foot with a velocity of 20m/s. What is
the kinetic energy of the ball just when it left the foot of the boy?
7) A student uses a catapult to shoot a stone of mass 30g by stretching its rubber and storing an
elastic potential energy of 6J. If all the EPE is converted into kinetic, with what speed will the
stone leave the Catapult?
8. A box of mass 6kg starts from rest at A and slides down a rough slope length 10m which
makes an angle of 370above the horizontal as shown below. Calculate the
a) work done on the box by the gravitational force
b) change in kinetic energy of the box
c) work done by the net force
d) magnitude of the frictional force acting on the box
e) speed of the box at the bottom of the plane.
9. A 50kg crate is pulled across a horizontal floor at a constant
velocity with a force ―F‖ at 370 above the horizontal.
A 24N frictional force opposes the motion of the crate as it moves through a distance of 5m.
i) What is the work done on the crate by the a) applied force b) frictional force
c) gravitational force d) normal force ii) what os the total work done on the crate
10. A 0.09kg arrow is fired horizontally. The bowstring exerts an average force of 80N on the
arrow over a distance of 0.9m. With what speed does the arrow leave the bow?
11. Hailue and Negasa are running with the same speed but Hailu has 3times the mass as
Negasa. If Hailu has a kinetic energy of 420J, what will be the kinetic energy of Negasa?
12. Two identical baseballs A and B are hit by the same batter but ball A has 5times the velocity
as ball B. If ball B has a kinetic energy of 1.2x103J, what is the kinetic of ball A?
13.If five a 60W and two a 100W electric lamps are operating for 5hours per a day, how much
electric energy is consumed by the lamps per a month?
14. A body has a kinetic energy of 840J. What will be the new kinetic energy of the body if its
speed is
i) doubled? ii) reduced to one fourth of its initial value?
15. A block falling through the air by 60m does a work equal to -7200J. What is the mass of
the block?
16. A crane lifts a 400kg load vertically to a height of 10m from the ground in 5seconds.
What is the power developed by the crane?
17.A boy shoots a 0.04kg stone using a catapult by stretching its rubber to 0.4m with an average
force of 20N. What will be the speed of the stone as it leaves the Catapult?

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Unit 4 Work, Energy and Power
18. A car travel at 15m/s against a force of friction of 5600N between its fire and the road. What
is the power developed by the engine of the car?

19. In a mass-spring oscillating system, the 2kg mass on the free end of the spring oscillates
freely back and forth b/n the two amplitude positions A and B as shown below. Given that
x=4cm.

A x C x B

i) Calculate the mechanical energy at

a) the amplitude positions. b) the equilibrium position.
C) a distance of 2cm from the equilibrium position.
ii) Calculate the speed of the mass at a distance of 2cm from the equilibrium position.
20. In a vertical mass-spring oscillating system, the 2kg mass on the free end of the spring
oscillates freely up and down b/n the two amplitude positions A and B as shown below.
Given that x=4cm and the spring constant 2000N/m.
i) At position A or B, calculate the
A
KEA=________________
GPEA=_______________
x
EPEA= _______________ m
MEA= ________________ C
VA= _________________
ii) At the equilibrium position, calculate x
KEC=________________ GPEC=_______________ B
EPEC= _______________ MEC= ________________
Vc= _________________ Ground

21. A bullet, mass 50 g, is shot vertically up in the air with a muzzle velocity of 200 m·s−1. Use
the Principle of Conservation of Mechanical Energy to determine the height that the bullet will
reach. Ignore air friction.
22. A skier, mass 50 kg, is at the top of a 6.4 m ski slope.
(a) Determine the maximum velocity that she can reach when she skies to the bottom of the slope
(b) Do you think that she will reach this velocity? Why/Why not?

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Unit 4 Work, Energy and Power
23. In the oscillation of simple pendulum, there is a
transformation of energy from one form to another form.
What will be the transformation of energy when the mass
moves from:- 𝜃
a) C to A? b) C to B?
c) A to C? d) B to C?
B A

24. A 2 kg metal ball is suspended from a rope. If it is released from point

A and swings down to the point B (the bottom of its arc):
(a) Show that the velocity of the ball is independent of it mass.
(b) Calculate the velocity of the ball at point B.
25. A tennis ball, of mass 120 g, is dropped from a height of 5 m. Ignore air
friction.
(a) What is the potential energy of the ball when it has fallen 3 m?
(b) What is the velocity of the ball when it hits the ground?
26. A pendulum bob of mass 1.5 kg, swings from a height A to the bottom of its arc at B. The
velocity of the bob at B is 4 m·s−1. Calculate the height A from which the bob was released.
Ignore the effects of air friction.
27. In the diagram below, calculate the work done by the applied force, the frictional force, the
normal force and by the net force if = = 5m and 100N
a) F= 40N
b) F= 130N
F 370

28. What power is required to pull a 5kg block at a steady speed of 2m/s? The coefficient of
friction is 0.3.
29. Determine the work done by each of the following force.
a) F= (8 , 0)N and = ( 5, 0) m
b) F= (0 ,12)N and = ( 0, 10) m
c) F= ( 8, 6) N and = ( 3, 4) m

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Unit 4 Work, Energy and Power
30. A Winch lifts a 150kg crate 3m upwards with Winch
acceleration of 2m/s2.
a) How much work is done by the Winch?
b) How much work is done by the gravity?
a
31. A 7500W engine is propelling a boat at 72km/h.
What force is the engine exerting on the boat?
m

Grade 9 Page 33