Mechanics

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Chapter No. 05
Mohd Yousuf Soomro ( Lecturer )

Applications Of Newton’s Laws Institute of Physics

5-1 Force laws:
To understand the effects of forces a detailed microscopic understanding of interaction
of objects with their environment is necessary.
All of known forces in the universe can be grouped into four types.

1- The Gravitational forces
2- The Electromagnetic forces
3- The Weak Nuclear forces
4- The Strong Nuclear forces.

1- The Gravitational forces:
It originates with the presence of matter. The gravitational force between two protons
just touching their surfaces is about 10-38 of the strong force between them.

The principal difference between gravitation & other forces is that, on the practical
scale, gravity is cumulative and infinite in range. Gravity is the weakest of the four
fundamental forces, yet it is the dominant force in the universe for shaping the large
scale structure of galaxies, stars, etc.

The gravitational force between two masses m1 and m2 is given by the relationship:

This is often called the "universal law of gravitation" and G the universal gravitation
constant. It is an example of an inverse square law force. The force is always attractive
and acts along the line joining the centers of mass of the two masses. The forces on the
two masses are equal in size but opposite in direction, obeying Newton's third law.
Viewed as an exchange force, the mass less exchange particle is called the graviton.

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The gravity force has the same form as Coulomb's law for the forces between electric
charges, i.e., it is an inverse square law force which depends upon the product of the
two interacting sources.

2- The Electromagnetic forces:
Electromagnetism is important in the structure & the interaction of fundamental particle.
It is also responsible for the binding of atoms & the structure of solids. the
electromagnetic force manifests itself through the forces between charges.

The electromagnetism force between two neighboring protons is about 10-2 of the strong
force. The electromagnetic force is a force of infinite range which obeys the inverse
square law, and is of the same form as the gravity force.

3- The Weak Nuclear forces:
The weak force is responsible for certain radioactive process & other similar decay
processes involving fundamental particles. The weak force between two neighboring
protons is about 10-7 of strong force between them. One of the four fundamental forces,
the weak interaction involves the exchange of the intermediate vector bosons, the W and
the Z. Since the mass of these particles is on the order of 80 GeV, the uncertainty
principle dictates a range of about 10-18 meters which is about 0.1% of the diameter of a
proton.

It is crucial to the structure of the universe in that

1. The sun would not burn without it since the weak interaction causes the transmutation
p ≥ n so that deuterium can form and deuterium fusion can take place.
2. It is necessary for the buildup of heavy nuclei.

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It was in radioactive decay such as beta decay that the existence of the weak interaction
was first revealed.

4- The Strong Nuclear force:
A force which can hold a nucleus together against the enormous forces of repulsion of
the protons is strong indeed. However, it is not an inverse square force like the
electromagnetic force and it has a very short range.

It is responsible for the binding of the nuclei, is the dominant force in the reaction &
decays of most of the fundamental particles.
The relative strength b/w two neighboring protons is 1. the strong force is 1038 times
greater than gravitational force.

Fundamental Forces

Unification for forces:
In 19th century James Clerk Maxwell united the separate electric & magnetic fields into
a single electromagnetism force.

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In 1967 a theory was proposed by “Dr. Abdus Salam”, “Stephen Weinberg”.
According to this theory the weak & electromagnetism forces could be regarded a part
of a single force called electroweak force.

Theories called GUTs (Grand Unification Theories) predicts about the unification of
strong and electroweak forces.

Other theory called TOE ( Theory of everything ) which unified the Strong,
electroweak & gravitational forces.

Prediction of TOE:
One prediction of these theories is that proton should not be a stable particle but should
decay on a time scale greater than 1031 years.
One way to test this theory is to which collection of 5.3 X 10 33 protons from a year to if
one of proton decays.

Electroweak Unification
The discovery of the W and Z particles, the intermediate vector bosons, in 1983 brought
experimental verification of particles whose prediction had already contributed to the
Nobel prize awarded to Weinberg, Salam, and Glashow in 1979. The photon , the
particle involved in the electromagnetic interaction, along with the W and Z provide the
necessary pieces to unify the weak and electromagnetic interactions. With masses
around 80 and 90 GeV, respectively, the W and Z were the most massive particles seen
at the time of discovery while the photon is mass less.

The discovery of the W and Z particles in 1983 was hailed as a confirmation of the
theories which connect the weak force to the electromagnetic force in electroweak
unification

Grand Unification
Grand unification refers to unifying the strong interaction with the unified electroweak
interaction.

One prediction of the grand unified theories is that the proton is unstable at some level.

In the 1970's, Sheldon Glashow and Howard Georgi proposed the grand unification of
the strong, weak, and electromagnetic forces at energies above 1014 GeV. If the ordinary
concept of thermal energy applied at such times, it would require a temperature of 1027K
for the average particle energy to be 1014 GeV.

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The unification of the strong force is well beyond our reach at the present time, and the
unification of gravity with the other three is out of reach for earthbound experiments.
This has led to greater cooperation between high-energy particle physicists and
astrophysicists as each group realizes that some of their answers can only come from the
other.

Ordinary Mechanical Systems:
Ordinary mechanical system involves only two forces.

(1)Gravity
(2) Electromagnetism

The gravitational force is apparent property in the Earth’s attraction for objects, which
gives them weight.
All other forces are electromagnetism in origin.

I. Contact forces
II. Viscous forces
III. Tensile forces
IV. Elastic forces
I. Contact forces:
There are two kinds of contact forces:

a. Normal Force
b. Frictional Force

a. Normal force:
Normal force is produced in bodies when one body pushes on another.

The normal force is defined as the net force compressing two parallel surfaces together;
its direction is perpendicular to the surfaces.

b. Frictional force:
Frictional force Produced b/w bodies when one surface rubs against another.

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II Viscous forces:
Air resistance.

III Tensile forces:
Tensile forces are produced in a stretched rope or string.

Tension is a reaction force applied by a stretched string (rope or a similar object) on the
objects which stretch it. The direction of the force of tension is parallel to the string, away
from the object exerting the stretching force. So if an object hangs from a rope due to
gravity, the gravitational force on the object points downward, and there is an equal
tension force in the rope point upward, making the net force on the object equal to zero.

Tension exists also inside the string itself: if the string is considered to be composed of
two parts, tension is the force which the two parts of the string apply on each other. The
amount of tension in the string determines whether it will break, as well as its vibrational
properties, which are used in musical instruments.

IV Elastic forces:
Elastic forces are that kind of forces arising from the deformation of a solid body which
depends only on the body's instantaneous deformation and not on its previous history,
and which is conservative. These are produced in a spring

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6.2 FRICTIONAL FORCES:
When one body slides over the surface of another body, an opposing force is set up to
resist the motion. This force which opposes the motion is called “force of friction” or
friction.

Friction is the resistive force acting between bodies that tends to oppose and damp out
motion. Friction is usually distinguished as being either

1. static friction (the frictional force opposing placing a body at rest into motion)
and
2. Kinetic friction (the frictional force tending to slow a body in motion).
3. In general, static friction is greater than kinetic friction.

The friction force is electromagnetic in origin: atoms of one surface "stick" to atoms of
the other briefly before snapping apart, causing atomic vibrations, and thus transforming
the work needed to maintain the sliding into heat.

The study of friction is called tribology.

The frictional forces on each body are in a direction opposite to its motion relative to the
other body.
Frictional forces automatically oppose this relative motion and never aid it even when
there is no relative motion, frictional forces may exist b/w surfaces.

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Factors on which force of friction depends:
1- Normal reaction forces
2- Nature of surface
1- Normal reaction ( R ):
Force of friction is directly proportional to normal reaction “ R “ which acts in upward
direction against the weight of the body sliding on a surface.

The normal force is defined as the net force compressing two parallel surfaces together;
its direction is perpendicular to the surfaces.

In the simple case of a mass resting on a horizontal surface, the only component of the
normal force is the force due to gravity, where FN=mg. In this case, the magnitude of the
friction force is the product of the mass of the object, the acceleration due to gravity,
and the coefficient of friction. However, the coefficient of friction is not a function of
mass or volume; it depends only on the material. For instance, a large aluminum block
will have the same coefficient of friction as a small aluminum block. However, the
magnitude of the friction force itself will depend on the normal force, and hence the
mass of the block.

If an object is on a level surface and the force tending to cause it to slide is horizontal,
the normal force N between the object and the surface is just its weight, which is equal
to its mass multiplied by the acceleration due to earth's gravity, g.

If the object is on a tilted surface such as an inclined plane, the normal force is less,
because less of the force of gravity is perpendicular to the face of the plane. Therefore,
the normal force, and ultimately the frictional force, is determined using vector analysis,
usually via a free body diagram. Depending on the situation, the calculation of the
normal force may include forces other than gravity.

2-Nature of surface:
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Force of friction also depends on the nature of two surfaces in contact with each other.
And it is constant for various pair of surfaces in contact and its value is known as
Co-efficient of friction.
Friction is very important in our daily lives.

Advantages of frictional forces:
1- In the absence of frictional forces b/w ground & our feet we can not walk.
2- The frictional force b/w the roads & the driving wheels propel an automobile the
frictional force prevents the tier from slipping & provides the necessary propelling
force.
3- Proper forces of frictional are maintained in the joints of the body to facilitate
running or rapid movement of joints in the absence of friction we can not stand
upright.

Disadvantages of frictional forces:
In a car engine oil under pressure is supplied continuously to all bearing surfaces.
Absence of oil will allow metal to metal contact & the resulting friction will raise the
temperature and cause the bearing and position to seize up.

Force Law for Frictional Forces:
Now if we want to know the force law for frictional forces or we want to know how to
express frictional forces in terms of the properties of the body and its environment.
The force law for dry, sliding friction is empirical in character & approximate in their
predictions. They do not have the elegant simplicity and accuracy as for the
gravitational force law & for the electrostatic force law.
Consider a block at rest on a horizontal table ( a ) attach a spring to it to measure the
horizontal force ( F ) required to set the block in motion.

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We find that the block will not move even though we apply a small force fig ( b ).

It means that applied force is balanced by an opposing frictional force “ f “ exerted on
the block by the table, acting along the surface of contact.

As we increase the applied force fig (c) (d) then there is a definite force at which the
block will “brake away “from the surface & begins to accelerate. Fig ( e ).

By reducing the force once motion has started, it is possible to keep the block in
uniform motion without acceleration. Fig ( f ).

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TYPES OF FRICTION:
1- Static friction
2- Kinetic friction
1- Static friction:
The frictional forces existing b/w surfaces at rest with respect to each other are called
forces of static friction.

A motionless body is subject to static friction. The direction of the static friction force
can be visualized as directly opposed to the force that would otherwise cause motion,
were it not for the static friction preventing motion. In this case, the friction force
exactly cancels the applied force, so the net force given by the Vector sum, equals zero.
It is important to note that in all cases, Newton's first law of motion holds.

2-Kinetic friction:
The forces acting b/w surfaces in relative motion are called forces of kinetic friction.
The friction force is directed in the opposite direction of the Resultant force acting on a
body.

In the case of kinetic friction, the direction of the friction force may or may not match
the direction of motion: a block sliding atop a table with rectilinear motion is subject to
friction directed along the line of motion; an automobile making a turn is subject to
friction acting perpendicular to the line of motion (in which case it is said to be 'normal'
to it).

LAWS OF Static FRICITION:
These laws were discovered experimentally by Leonardo da vinci. (1452- 1519)
Two laws of friction about the maximum force of Static friction b/w any pair of dry un-
lubricated surfaces.

1. It is approximately independent of the area of contact over wide limits.
2. It is proportional to the normal force.

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Normal force / loading force:
It is the force which arises from the elastic properties of the bodies in contact.
For a block resting on a horizontal table or sliding along it, the normal force is equal in
magnitude to the weight of the block.

Laws of Kinetic Friction:
Two laws of friction about of kinetic friction fk b/w dry un-lubricated surfaces:

1. It is approximately independent of the area of contact over wide limits.
2. It is proportional to the normal force

Coefficient of friction
The coefficient of friction is a dimensionless quantity used to calculate the force of
friction (static or kinetic).

While it is often stated that the coefficient of friction (COF) is a "material property," it
is better categorized as a "system property." Unlike true material properties (such as
conductivity, dielectric constant, yield strength), the COF for any two materials depends
on system variables like temperature, speed, atmosphere, as well as on geometric
properties of the interface between the materials.

For example, a copper pin sliding against a thick copper plate can have a COF that
varies from 0.6 at low speeds (metal sliding against metal) to below 0.2 at high speeds
when the copper surface begins to melt due to frictional heating. The latter speed, of
course, does not determine the COF uniquely; if the pin diameter is increased so that the
frictional heating is removed rapidly, the temperature will drop, the pin remains solid
and the COF rises to that of a 'low speed' test.

1. Coefficient of Static friction
2. Coefficient of kinetic friction

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1. Coefficient of Static friction
The coefficient of static friction is defined as the ratio of the maximum static
friction force (F) between the surfaces in contact to the normal (N) force.

Co-efficient of Static friction: ( μs ):

The ratio of magnitude of the maximum force of static friction to the magnitude of the
normal force is called the co-efficient of static friction for the surfaces involved.

f S ≤ µS N −−−−−− →(1)

fS
µS =
N
Where
f S = magnitude of the force of static friction
µS = co-efficient of static friction
N = magnitude of normal force

NOTE :
The equality sign hold only when fS has its maximum value.

substance
wood on wood 0.25-0.50
steel on steel 0.58
glass on glass 0.9-1.0

Muhammad Yousuf Soomro (Lecturer) Institute of Physics

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2. Coefficient of kinetic friction
The ratio of the magnitude of the force of kinetic friction to the magnitude of the normal
force is called the co-efficient of kinetic friction.

f K = µ K N − − − − − − → (2)
fK
µK =
N
Where
f K = magnitude of the force of kinetic friction
µK = co-efficient of kinetic friction
N = magnitude of the normal force

NOTE :
The force of the kinetic friction is also reasonably impendent of the relative speed with
which the surfaces move over each other.

Some values for common substances are given in the following table

substance T( )
hickory on dry snow -- 0.08
ice 0 0.020
ice -10 0.035
brass on ice 0 0.020

NOTE :
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Equations (1) & equation (2) are relations b/w the magnitude only of the normal &
frictional forces these forces are always directed perpendicularly to one another

IMPORTANT :
1- Both µS , µK ,are dimensionally constant , each being the ratio of ( magnitude)
two forces.
2- Usually for a given pair of surfaces µS > µK

3- The actual values of µS & µK depend on the nature of both the surfaces in
contact.
4- Both µS , µK can exceed unity, although commonly they are less than 1.

NOTE :
1. Both static and kinetic coefficients of friction depend on the pair of surfaces in
contact.
2. Their values are usually determined experimentally.
3. For a given pair of surfaces, the coefficient of static friction is larger than that of
kinetic friction.

A free-body diagram of a block resting on a rough inclined plane, with its weight (W),
normal force (N) and friction (F) shown.

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Sliding Coefficient of Friction
when the tangential force F overcomes the frictional force between two surfaces
then the surfaces begins to slide relative to each other. In the case of a body
resting on a flat surface the body starts to move. The sliding frictional
resistance is normally different to the static frictional resistance.

The coefficient of sliding friction is expressed using the same formula as the
static coefficient and is generally lower than the static coefficient of friction..

Rolling Friction
When a cylinder rolls on a surface the force resisting motion is termed rolling
friction. Rolling friction is generally considerably less than sliding friction. If W is
the weight of the cylinder converted to force, or the force between the cylinder and the
flat surface, and R is radius of the cylinder and F is the force required to overcome the
rolling friction then.

W
F =f ×
R

“ f “ is the coefficient of rolling friction and has the same unit of length as the radius R.

Typical values for f are listed below

Surfaces Rolling Friction f
Steel on Steel 0,0005m
Wood on Steel 0,0012m
Wood on Wood 0,0015m
Iron on iron 0,00051m
Iron on granite 0,0021m
Iron on Wood 0,0056m
Polymer of steel 0,002m
Hardrubber on Steel 0,0077m
Hardrubber on Concrete 0,01 -0,02m
Rubber on Concrete 0,015 -0,035m

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Factors affecting the friction between
surfaces
Dry surfaces
1. For low surface pressures the friction is directly proportional to the
pressure between the surfaces. As the pressure rises the friction factor
rises slightly. At very high pressure the friction factor then quickly
increases to seizing
2. For low surface pressures the coefficient of friction is independent of
surface area.
3. At low velocities the friction is independent of the relative surface velocity.
At higher velocities the coefficient of friction decreases.

Well lubricated surfaces
1. The friction resistance is almost independent of the specific pressure
between the surfaces.
2. At low pressures the friction varies directly as the relative surface speed
3. At high pressures the friction is high at low velocities falling as the velocity
increases to a minimum at about 0,6m/s. The friction then rises in
proportion the velocity 2.
4. The friction is not so dependent of the surface materials
5. The friction is related to the temperature which affects the viscosity of the
lubricant

University of Sindh Jamshoro

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6-3 The Dynamics Of Uniform Circular Motion
If a particle moves at constant speed in a circular path both the velocity and acceleration
are constant in magnitude but both change their direction continuously this motion is
called uniform circular motion.

Examples of uniform circular motion:

1- In atoms, the electrons move in circular orbits around the nucleus.
2- The earth and others plants keep as moving around the sun in fixed orbits.
3- Computer disks.

If a body is moving at uniform speed “ v ” in a circle or a circular arc of radius “r” it
v2
experience a centripetal acceleration “a” whose magnitude is .
r
The direction of centripetal acceleration is always radially inward towards the center of
circle. This centripetal acceleration is a vector because even though its magnitude
remains constant, its direction changes continuously as the motion progresses.

If a body undergoing uniform circular motion, then a net force must be acting on it &
the magnitude of net force

mv 2
∑ F = ma = r

The direction of net force acting at any instant must be the direction of centripetal
acceleration “a“ at that point, radially inward.

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Example :

A disk is moving with uniform circular motion & it moving on the end of a string in a
circle on a frictionless horizontal table as shown in figure.

The net force on the disk is provided by the tension “T“in the string. It string accelerates
the disk by constantly changing the direction of its velocity so that the disk moves in s
circle.

The direction of T is always towards the pin at the center, and its magnitude must equal
mv 2
to
r

mv 2
T=
R

If the string were to be cut where it joins the disk, there would be no net force exerted
on the disk. The disk would then move with constant speed in a straight line along the
direction of the tangent to the circle at the point at which the string was cut.

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Centripetal Forces:
Forces responsible for uniform circular motion are called centripetal forces because they
are directed towards the center of the circular motion. So

1- for the revolving disk , the centripetal force is a tensile provided by the string.
2- For the Moon revolving around the earth the centripetal force is the
gravitational pull of the earth on the Moon.
3- For an electron circulating about an atomic nucleus the centripetal force is
electrostatic.

Examples of Centripetal forces:
The Conical Pendulum:

Consider a small body of mass “ m “ revolving in a horizontal circle with constant speed
“ v” at the end of a string of length “ L “. As the body swings around, the string sweeps
over the surface of an imaginary cone. This device is called a Conical Pendulum.
We want to find the time required for one complete revolution of the body.

If the string makes an angle θ with the vertical, the radius of the circular path is

R = L sin θ

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The forces acting on the body of mass “ m “ are its

( i ) Weight mg
( ii ) Tension T

Of the string.

Apply Newton’s second law of motion

∑F = T + mg
∑ F = mg ∴g = a

Or ∑F = ma

We can resolve T at any instant into a radial and vertical components

Tr = −T sin θ (Radial component)
Tr = T cos θ (Vertical component)

The radial component is –ve because if radial direction to be positive outward from axis.

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As there is no vertical acceleration in body so Z- component of Newton’s second law of
motion
∑ F = T − mg = 0
z z

Or T cos θ =mg −−−−−−→(1)

The radial acceleration is

V2
ar =−
R

It is –ve because it acts radially inward.
Radial acceleration is supplied by Tr which provides the centripetal force acting
on “m “.
From the radial component of Newton’s second law of motion

∑F r = Tr = mar

V2
∑ F = −T sin θ = m(− R )

mV 2
−T sin θ = −
R

mV 2
T sin θ = − − − − − − → (2)
R

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Dividing equation ( 2 ) by equation ( 1 )

T sin θ mV 2 R
=
T cos θ mg

sin θ
=V 2 R
cos θ g

V2
tan θ =
Rg

V 2 =tan θRg

V = Rg tan θ

This equation gives the constant aped of the body.

Period of Motion:
Now if t represents the time for one complete revolution of the body
.

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2π R
V =
t
or
2π R
t =
V
2π R
t =
Rg tan θ

Squaring both sides

4π R 2
2
t 2
=
Rg tan θ

4π R
2
t 2
=
g tan θ

4π R2
t 2
=
g tan θ

R
t = π
2
g tan θ

As R =l sin θ

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L sin θ
t =π
2
g tan θ

L sin θ
t =π
sin θ
2
g( )
cos θ

L cos θ
t =π
2
g

Note :
This equation gives the relation b/w t, L, θ.
T is called period of motion and it does not depend on mass m.

6-4 Equations & Motion
Constant and non constant forces
Constant forces:

Forces that dose not depend on time, velocity or position.

Examples :

1. Object falling near the earth surface.
2. Braking car

If force is constant, then the acceleration is constant, and for constant acceleration
the solution in one dimensional for V(t) & x(t) are easily obtained.

Constant forces demonstrate the applications of Newton’s law and they are certainly
easier to work with than non constant forces.

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Many particle problems often include forces that remain constant under many
circumstances.

Examples :

1 Gravity near the earth surface
2 Frictional forces
3 Tension forces in a strings

Find V(t) for constant force:
Let “ a “ represent constant acceleration & an object starts with velocity “ Vo “ at
time t = 0 & at time “ t “ later it has velocity V.

V2 − V1 ∆V
a= =
t2 − t1 ∆t
Or

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∆V V − V0
a= =
∆t t −0

V − V0
a=
t

at = V − V0

V = V0 + at − − − − − − → (1)

This important result allows us to find the velocity at all later times. This equation
No. 01 shows the velocity as a function of time, and it can be written as V(t).

Note :
Equation 01 is in the form of Y= mx+b, which describes the graph a straight line.

Non-Constant forces:

Forces that depends on time, velocity & position.

Examples :

1 Swinging pendulum bob
2 Rocket blasting toward earth’s orbit
3 A rain drop falling against air resistance.
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When the force is not constant then we can not use constant acceleration formulae to
find V(t) & x(t).

Methods of solving problems:
1 Analytical method
2 Numerical method

1 Analytical method:

This method involving integral calculus.

Find V( t ):
As
V
a =
t
or
dv
a =
dt

a.dt =dv

dv = .dt
a

Take integration both sides with limits of velocity V as follows
Vo at time t = 0 &
V at time t = t

Limits of time from 0 to t

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V 1

∫ dv =∫a.dt
Vo 0

V 1

∫ dv =a ∫dt
Vo 0

[V ]Vo =a [ t ] 0
V 1

Appling limits

V − Vo = a (t − 0)

V − Vo = at

V = Vo + at
or
V (t ) = Vo + at

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Find x ( t ):
As

S
V = S =x
t
or
x
V =
t

dx
V =
dt

dx = V .dt

dx = (Vo + at )dt

dx = Vo.dt + a.t.dt

Integrate with limits from position xo at time 0 to position x at time t
x t t

∫ dx = ∫V
xo 0
0 dt + ∫ at dt
0

x t t

∫ dx = V ∫ dt + a ∫ tdt
x0
0
0 0

t
t 2 
[ x] x = V0 [ t ] 0 + a  
x t
0
 2 0

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Apply limits

( x − x0 ) = V0 (t − 0) + a( 1 t 2 )
2

x − x0 = V0 t + 1 at 2
2

x = x0 + V0 + 1 at 2
2

or
x(t ) = x0 + V0 t + 1 at 2
2

Physical phenomenon involving non-constant forces:
1. Forces depending on the time
2. Forces depending on the velocity
3. Forces depending on the position

2-Numerical methods:

In this method we use a computer to evaluate the integrals, obtaining not the analytic
functions V( t ) & x( t ) but instead the numerical values of V & x at any time t this
can be done to any desired level of precision.


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Time-Dependent Forces: (analytical methods)

When the forces is time dependent, we can obtain an acceleration a(t) using Newton’s
laws of motion.

Find V( t ):

Find the velocity by direct integration.

V
as a=
t
dv
a=
dt
dv = a (t ) dt
Integrate with limits from
Time t = 0 , when the initial velocity is Vo , to time t when velocity is V
V t

∫ dv= ∫a(t)dt
V0 0

t

[ V]V = ∫a(t)dt
V

0
0

t
V-V0 = ∫a(t)dt
0

t
V=V0 + ∫a(t)dt
0

or
t
V(t)=V0 + ∫a(t)dt
0

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Find x ( t ):

as V = dx
dt
dx = V dt
or dx = V (t ) dt

Integrate with limits time t = 0 when the particle is located at xo, to time t = t, when the
position is x .

x t

∫ dx= ∫V (t)dt
x0 0

t

[ x] x = ∫V (t)dt
x
0
0

t
x-x 0 = ∫V (t)dt
0

t
x=x 0 + ∫V (t)dt
0

or

t
x (t)=x 0 + ∫V (t)dt
0

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34
x t
x t

∫ dx = ∫ V (t )dt
x0 0
∫ dx =∫V (t )dt
x0 0

t t
[ x] x = ∫ V (t )dt
x
[ x ] x =∫V (t ) dt
x
0 0
0 0

t t
x − x0 = ∫ V (t ) dt x −x0 = ∫V (t ) dt
0 0

t t

x = x0 + ∫ V (t ) dt x = x0 +∫V (t ) dt
0 0

or or
t t

x (t ) = x0 + ∫ V (t ) dt x (t ) = x0 +∫V (t ) dt
0
0

Chapter No. 07 Mohd Yousuf Soomro

WORK & ENERGY Institute of Physics

Work:

When force acts upon a body and if it causes displacement also, then work is said to be
done by the force upon the body.
Or
Dot product of two vector quantities (force & displacement) is called work.
Work is a scalar quantity.
rr
W = F .d
Where
F = force
d = displacement moved by the body in the direction of force.

NOTE :
Sometimes the force and displacement may not be in the same direction.

Examples:
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35
A wooden block is pulled by means of a rope, it moves in the horizontal direction, but
the force acts along the rope. If the force makes an angle “ θ “ with the direction of
motion then work done can be found by the production of the component of the force
along the direction of motion and the displacement “ S “ moved by the block.

We can express work done by the force analytically as

W = F . d cosθ
Where r
F = magnitude of the vector F
r
d = magnitude of the vector d
r r
θ = angle b/w F and d

Work is an algebraic quantity.
It can be positive, negative , zero depending upon the value of angle b/w force ( F ) and
displacement ( d ).

Cases of work done:
( i ) case:
When the components of force is in the same direction of the displacement ( θ = 0 ) then
the work is positive.

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36
r r
W = F . d cos θ

if θ =0

W = F .d cos 0

W = F .d (1)
cos00 = 1
W = F .d

Example :
When a spring is stretched, then work done by the stretched ( stretching force ) is
positive.

( ii ) case:
When the direction of the force is opposite to the direction of the displacement ( θ =1800 )
the work done is negative.

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37
rr
W = F .d cos θ

if θ = 1800

W = F .d cos1800 cos1800 = − 1

W = F .d (− 1)

W = − F .d

Example :
Work done by the “gravitational force “on the body being lifted is negative.
When a body is lifted against gravity at very slow speed , the angle b/w gravity and
displacement is 1800.

( iii ) case:
When the force acts at right angle to the displacement ( θ = 900 ) THEN THE WORK
done is zero.
It means that force does not produce work.
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38

rr
W = F .d cos θ

if θ =0

W = F .d cos 900

W = F .d X 0 cos 900 = 0

W =0

Example :
work done by the centripetal force is zero because the centripetal force is always at right
angles to the direction in which the body moving.

Units:
SI System: (MKS System)
In the S.I system the unit of work is Newton-meter which is commonly known as
Joule(J ).
1 Newton. meter = 1 Joule

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39

1N. m = 1 J
The work of 1 joule is said to be done if a force of 1 Newton acts on a body and if it
moves through a distance of 1 meter in the direction of force.

CGS system: ( centimeter , gram , second system )
In CGS system the unit of work is erg.

1 erg = 1 dyne 1 cm

1 erg = 10-7 J

The work of 1 erg is said to be done if a force of 1 dyne acts on a body and if it moves
through a distance of 1 centimeter in the direction of force.

FPS System ( foot, pound , second system ) / British Engineering
System :

In FPS System the unit of work is Foot . Pound ( ft-lb ).

1 ft. lb = 1.355 joule

Relationships b/w units:
1 Joule = 105 dynes X 102 cm
1 Joule = 107 ( dynes .cm )
1 Joule =107 Joule

1055 Joule = 1 Btu (Btu= British thermal unit)

NOTE :
In the physics of molecules , atoms , and elementary particles a much smaller unit is
used , it is called electron volt ( e.V )

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40

1eV = 1.60 X 10−19 J

Multiples of the electron volts:
1 million eV = 1 M eV = 106 eV
1 billion eV = 1 B eV = 1012 eV

Muhammad Yousuf Soomro Lecturer Institute of Physics
University of Physics Jamshoro

Physical dimension of Work:

As
Work = force. displacement

W = F . d − − − − − −− → (1)

Since F = ma
Or
F = mass × acceleration
L
F = M × ( ) M = mass
T2
a = acceleration
L = length
T = time

So equation ( 1 ) becomes

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41
L
W = M× ( 2
)( L)
T

ML2
W = 2
T

W = M L2 T −2

7.1 Constant Force on a Particle:
Suppose a particle of mass “m “is subjected to a constant force “ F “. We assume the
motion of a particle along x- axis.

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42

W = F .d
or
W = F .dx − − − −− → (∗)
As we know from Newton’s second law of motion
F = ma
dv
F =m
dt
So equation ( * ) becomes

dv
F . dx = m .dx − − − − − (1)
dt
Suppose the particle is displaces from “Xo” to “X”.
Let the velocity at “ Xo” be “ Vo” and velocity at “ X “ be “ V “.

Applying these conditions on equation (1) and integrate it.

x v
dv
∫ F .dx = ∫ m
x0 V0
dt
.dx

Since the force “ F “ and mass “ m “ are constants so we may take these constants out of
the integral sign.
x v
dv
F ∫ dx = m ∫ .dx
x0 V0
dt

x v
dx
F ∫ dx = m ∫ .dv
x0 V0
dt

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43
x v
dx
F ∫ dx = m ∫ Vdv V=
x0 V0
dt

2 v
v 
F [ x] x = m  
x
0
 2  v0
Appling the limits

2
v 2 v0
F ( x − x0 ) = m( − )
2 2

F ( x − x0 ) = 1 mv 2 − 1 mv0 − − − − → (2)
2 2
2

2
We call 1 mv as the Kinetic energy of the body. The R.H.S of equation ( 2 ) is the
2
change in the K.E of the particle when it changes its position from “xo” to “x”.
Thus
“The work done by a force is equal to the change in K.E of the particle. “

7.2 Work Done by a Variable Force:
The work done by a variable force can not be calculated by the direct use of formula
W= F.d for the entire displacement “ d” because the force is continuously changing with
the displacement.

So to calculate the work done by a variable force , we make two assumptions.

(1) The direction of force is along positive X-axis and its magnitude is changing with
displacement.
(2) The body is constrained to move in positive X-axis direction only.

Suppose a body of mass “ m “ is moving under a variable force “ f “.
The value of force “ f” is different at different positions.

43

44

We assume that the body moves from its initial position “ A “ to the final position “ B “
in the direction of the variable force.
If the body has moved from position “ A “ to the position “ B “ then we divide the total
displacement “ AB “ into small interval each of width “Δr “ .
During any short displacement interval “Δr “the force remains almost constant. Thus the
force for any one displacement interval = dot product of “F “and “Δr “.

The total work done for the displacement “ AB “

WA→ B = F1∆r1 + F2 ∆r2 + F3∆r3 + ........
or
B
W A→ B = ∑ F ( r ) ∆ r
A

The constant force is taken to be the force which acts at the beginning of each interval.
Let the total number of intervals from “A” to “ B “ be “ N “.
Now if the number of intervals are made infinite ( N → ∞ ) and the width “ Δr “ is made
so small that it tend to zero ( ∆r → 0 )

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45
B
WA→ B = ∆r → 0∑ F .∆r
A

Now
B B
∆r → 0 ∑ F .∆r = ∫ F .dr
A A

therefore the total force

B
WA→B = ∫ F .dr − − − −− → (1)
A

As we known that F=ma
dv dv
F=m a=
dt dt

B
dv
WA→B = ∫ m dr − − − −− → (2)
A
dt

Since “ m “ mass is constant so we may take it out of the ingral sign

B
dv
WA→B = m ∫ dr
A
dt
B
dr
WA→B = m ∫ dv
A
dt
B
WA→B = m ∫ Vdv − − − − → (3)
A

Integrate equation ( 3 ) with respect to ”v”

45

46
B
 2
V
=m  
 2 A
B
1 
= mV 2 
2 A
Applying limits

1 1
mVB2 − mVA
2

2 2
It means

K .E at B =K .E at A
1 1
mVB2 = mVA −− −− → (4)
2

2 2

Example :
An example of a variable force involving in the motion of spring.
We consider a spring that acts on a particle of mass “m “ . The particle moves in the
horizontal direction with the origin ( x = 0 ).

46

47
Let the particle be displaced a distance “ x “ from its original position x = 0 , as the
agent exerts a force “ F “ on the particle , the spring exerts an opposing force “ F “.

We know from the Hook’s law that
“Force acting on the body is directly proportional to the displacement of the
body”.

F = − kx
Where
K= force constant or spring constant.

Let us consider the work done on the particle by the spring when the particle moves
from initial position “ xi “ to “ xf”.

F = −k x

W = F .d

W = ( −kx ).dx

xf

W = ∫ ( −kx ).dx
xi

As k is constant so we may take it out of the integral sign.

xf

W = k ∫ (− x).dx − − − − → (1)
xi

Integrating equation ( 1 ) with respect to “ x”.

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48
xf
 x  2
W = k − 
 2 xi
Now applying limits

kx 2
kxi2
f
W =− +
2 2

2
kxi2 kx f
W= − − − − − → (2)
2 2

7-4 Work Energy & The Work Energy Theorem:
Work energy theorem:

The net work done by the forces acting on a particle is equal to the change in the Kinetic
energy of the particle.

Wnet = K f − Ki = ∆ K

Explanation :

Consider a net work done (W net) on a particle not by single force but all the forces that
act on the particle.

Ways to find Net Work:
There are two ways to find the net work.

1- Find the net force:
It means the vector sum of all the forces that acts on the particle.

Fnet = F1 + F2 + F3 + .......

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49
Then treat this force as a single force in calculating the net work by following equation.

xf

W = ∫ F ( x)dx (One dimension)
xi

f f

W = ∫ F .ds = ∫ F cos φ ds (Many dimensions)
i i


2- Calculate the work done by each of the forces that acts on the
particle:

W1 = ∫ F1.ds

W2 = ∫ F2 .ds

W3 = ∫ F3 .ds
Wn = ∫ Fn .ds

To find net work, add the work done by each of the individual forces.

Wnet = W1 + W2 + W3 + .........
We know that a net unbalanced force applied to a particle will change its state of
motion by accelerating it from initial velocity “ Vi “ to final velocity “ Vf “.

Under the influence of constant force , the particle moves from Xi to Xf & it
accelerates uniformly from Vi to Vf. The work done is

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50
Wnet =Fnet ( x f −xi )

Wnet =ma ( x f −xi )

Because acceleration is constant we use the equation

V 2 = V02 + 2a ( x − x0 )
or
V f2 = Vi 2 + 2a ( x f − xi )

1 1
Wnet = mv 2 − mv
f
2 2

It means:

“The result of the net work on the particle has been to bring about a change in the
value of quantity 1 mv 2 from point “ i “ to point “ f “. This quantity is called the Kinetic
2

energy K of the particle.

K = 1 mv 2
2
NOTE :

Work energy theorem holds for both constant forces as well as non constant forces:

The work energy theorem does not represent a new independent law of classical
mechanics. It is useful for solving problems in which the net work done on a particle by
external forces is easily computed and in which we are interested in finding the
particle’s speed at certain positions.

General Proof of Work-Energy Theorem:
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51
Let Fnet represent the net forces acting on the particle. The net work done by all the
external forces that act on the particle is

Wnet = ∫ Fnet dx − − − −− ⇒ (1)

Fnet = ma

As
dv dv
Fnet = m ∴a=
dt dt

By using chain rule

dv dx
Fnet = m
dx dt

dv
Fnet = m v
dx

dv
Fnet = mv − − − −− ⇒ (2)
dx
Put equation ( 2 ) in equation ( 1 )

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52

Wnet = ∫ Fnet dx
or
dv
Wnet = ∫ mv dx
dx

Wnet = ∫ mv dv

Integrate from initial velocity “Vi” to final Velocity “Vf”

vf

Wnet = ∫ mvdv
vi

vf

Wnet = m ∫ vdv
vi

2 vf
v 
Wnet = m  
 2  vi
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53
Applying limits

Wnet = m( 1 v 2 − 1 vi2 )
2 f 2

Wnet = 1 m(v 2 − vi2 )
2 f

or
Wnet = 1 mv 2 − 1 mvi2
2 f 2

This shows that the work – energy theorem holds also for non constant forces.

7-5 Power
“The work done in unit time called power”.

Or
“The rate of doing work is called power “.

F .d
Power =
t

W
Power = ∴ W = F .d
t

d
Power = F .V ∴ =V
t
Kinds of Power:
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54

1. Average power
2. Instantaneous power
1. Average Power:

If “Δw“ is the work done in a time “ Δt “ the power is called average power and it is
written as Pav.

∆W
P =
av
∆ t
2. Instantaneous power:
∆W
If ∆t → 0 the limiting value of is called Instantaneous power at time “ t “.
∆t

∆W
P = lim − − − −− → (1)
∆t →0 ∆t

As we know that

W = F .d

dr
P = lim F .
∆t →0 dt

P = F .V
It means power is equal to the dot product of force and velocity.

Units:
SI System:
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55
The S.U unit of power is Joule per Second (J/s) which is also called Watt (W).
This unit is named in honor of James Watt (1736-1819)

1 W = 1 J / s = 1 Kg.m2 / s3

Watt:
One watt is defined as:

“The rate of doing work or using energy at Joule per Second”.

Bigger Units:
The bigger units of power are Kilowatt ( kW) Mega watt ( MW) and Giga Watt ( Gw )

1 kW= 103 watt = 1000 w
1 Mw=106 watt=1000000w=1000 Kw
1 Gw= 109 watt=1000000000w

FPS System:
In British Engineering system the unit of power is ft-lb / sec (foot. Pound / second ).
Since this unit is quite small, therefore a bigger unit called Horse Power (hp) is used.

1 horsepower = 745.7 watts

The term "horsepower" was coined by the engineer James Watt (1736 to 1819) in 1782
while working on improving the performance of steam engines.

Horse Power:
“One horse power (hp) is the power of an agency which does work at the rate of
550 ft-lb/sec or 33000 ft-lb per minute.

550 ft-lb/sec = 33000 ft-lb / min
1 horsepower = 745.7 watts
Or
1 hp = 746 watt

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56

1 min = 60 seconds
1 ft = 0.3048 m and
1 lb = 0.45359237 kg

Relationship b/w Kilowatt hour & Joule:
The term kilowatt hour (KWh ) is originated from the unit of work.
One Kilo watt is defined as the work done in one hour by an agency working at the
constant rate of 1 Kw that is 1000 Joule per Second.

1 hour = 3600 seconds
1 Kwh = 1000 X 3600 = 3.6 X 106 Joules

Relationship b/w Horse power & Watt:

1 hp= 550 ft –lb / sec
We know that
1 ft = 0.3048 m
1 lb = 4.448 N
So 1 hp = 550 X 1.351 X 4.448
= 550 X 1.351 N.m / sec
= 550 X 1.351 Joules / sec

1 hp = 746 watt

Some Important Units of Energy
MECHANICAL ENERGY:

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57
Metric Units: SI: Joule (J) 1 J = 1 N-m
English Units: foot-pound (ft-lbs) 1 ft-lbs = 1.356 J

HEAT ENERGY:

Calorie (Cal) 1 Cal = 4.186 J
British Thermal Unit (BTU) 1 BTU = 1,055 J

ELECTRICAL & ATOMIC ENERGY:

Electron Volt (eV) 1 eV = 1.6x10-19 J
Kilowatt-hour (kWh) 1 kWh = 3.6x10+6 J


57

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