Third ppt

What is Work?
• In physics, work is the scalar product of force and
displacement. A force is said to do work if, when acting,
there is a displacement of the point of application in the
direction of the force.
Work=Force*displacement
The SI unit of work is the joule (J).
• Example: Moving a chair, Throwing a stone etc.

Solved Examples (Work)
Example: A box is dragged across
a floor by a 100N force directed
60o above the horizontal. How
much work does the force do in
pulling the object 8m?
• Solution: F = 100N
• θ = 60
• S = 8m
• W = (F*Cosθ)* S
=(100*Cos60)* 8
= 100x1/2x 8 = 400 J

Example: A horizontal force F pulls a 10 kg carton across
the floor at constant speed. If the coefficient of sliding
friction between the carton and the floor is 0.30, how much
work is done by F in moving the carton by 5m?
Answer: The carton moves with constant speed.
We look at all the forces acting on the carton:
F is the applied force to the right.
P is the frictional force to the left.
W is the weight of the carton acting downwards.
N is the normal reaction by the floor on the carton acting
upwards.
Since there is no movement in the vertical direction,
Also since the carton is moving with constant speed,
Thus F = μN= μmg =0.3 x 10 x 9.8
= 29.4 N
Therefore work done W = FScos0
=(29.4 Cos0)*5
=147 J

Work done as Area under Curve
• The area under the force-displacementcurve represents
the work done. It is the distance (s) of the object
displaced by the application of the force (f). In this case
force (f) is presented on the y-axis and displacement (s)
on the x-axis. The work done (w) is a scalar quantity.

Kinetic energy
In physics, the kinetic energy of an object is the energy
that it possesses due to its motion. It is defined as the
work needed to accelerate a body of a given mass from
rest to its stated velocity.
Having gained this energy during its acceleration, the
body maintains this kinetic energy unless its speed
changes.
Example: An airplane has a large amount of kinetic
energy in flight due to its large mass and fast velocity.
SI Unit of Kinetic Energy is Jule.

Types of Kinetic Energy
• rotational kinetic energy is the energy possessed by a
body which is moving in circles, e.g. planets revolving
around the sun have rotational kinetic energy;
• vibrational kinetic energy is the energy possessed by
an object due to vibration, e.g. vibrating phone has a
vibrational kinetic energy;
• translational kinetic energy is the energy possessed
by an object moving from one point to another.
Translational kinetic energy can be easily observed in
our everyday life.
• Kinetic energy depends upon two things, i.e. mass (m)
and velocity (v) as the formula of kinetic energy depicts;
Kinetic Energy=(1/2)*m*v^2
Where ‘m’ is the mass of the object that is in motion and
‘v’ is the velocity.

HydroPower Plant
• When the moving water, possessing some kinetic
energy, hits the turbine present in the dam, the kinetic
energy of the water gets converted into mechanical
energy. This mechanical energy moves the turbines and
then, ultimately, it leads to the production of electrical
energy.
• Other Examples: wind mill, moving cars, flying planes etc

What are the different types of
kinetic energy?
• Mechanical energy
• Electrical energy: Electrical energy can be supplied
easily to any desired place through wires. We get
electrical generators. These electric generators are run
by hydropower, thermal power or nuclear power.
• Light energy: Light is a form of energy. The sun is the
major source of light for us. Light helps plants to make
food in the process of photosynthesis.
• Sound energy: A sound is a form of energy. It is
produced when a body vibrates; such as vibrating
diaphragm of a drum
• Thermal energy: Heat is a form of energy given out by
hot bodies. A large amount of heat is obtained by burning
fuel. Heat is also produced when motion is opposed
by frictional forces.

Work-Energy Principle
• The work-energy principle states that an increase in
the kinetic energy of a rigid body is caused by an equal
amount of positive work done on the body by the
resultant force acting on that body. Conversely, a
decrease in kinetic energy is caused by an equal
amount of negative work done by the resultant force.
• Examples, 1) A car is accelerating constantly. Here
velocity and kinetic energy are also increasing. So
positive work is done.
2) A motor bike applies break and comes into rest after
few second. Here work done by the break is negative.
Because speed decreases, for which K.E also
decreases.

Solved Examples
• A body of mass 10kg at rest is acted
upon simultaneously by two
forces 4N and 3N at right angles to
each other. The kinetic energy of the
body at the end of 10s is
• A 50 J
• B 100 J
• C 125 J
• D 144 J

• Net force on body =root(4^2+3^2)=5N
∴a=F/m=5/10=1/2m/s^2
• Kinetic energy =1/2mv^2
=1/2m(at)^2
=125J
Example: Suppose that you push on the 30.0-kg package
in Figure with a constant force of 120 N through a
distance of 0.800 m, and that the opposing friction force
averages 5.00 N. Calculate the net work done on the
package?
Answer: The net force is the push force minus friction,
or Fnet = 120 N – 5.00 N = 115 N. Thus the net work is
• Wnet=Fnet*d
=(115 N)(0.800 m)
=9.20 N⋅m=92.0 J

Example: A car with a mass of 1,000 kg brakes to a stop
from a velocity of 20 m/s (45 mi/hr) over a length of 50
meters. What is the force applied to the car?
Answer: ∆KE = 0 – [(1/2)(1,000 kg)(20 m/s)2]
= –200,000 J
W = –200,000 Nm
= (F)(50 m);
F = –4,000 N

What is Spring?
• In classical physics, a spring can be seen as a device
that stores potential energy, specifically elastic potential
energy. A characteristic of a spring which is defined as
the ratio of the force affecting the spring to the
displacement caused by it.
F = − k*x, (Hooke's Law)
U = 1/2*k*x^2 (Potential energy)
knet = k1 + k2 (combining 2 springs in parallel)
knet = (1/k1) + (1/k2) (combining 2 springs in series)
where F is the force exerted by the spring on an object
attached to it. x is the spring's displacement from it's
equilibrium length. k is the spring constant, with SI units
of N/m.

Solved Example (Spring’s Potential Energy)
Example: Two carts connected by
a 0.25m spring hit a wall, compressing the
spring to 0.1m. The spring constant k is
100N/m.
What is the elastic potential energy
stored from the spring’s compression?
• Choose answer:
A -15.00J
B 1.125J
C -1.125J
D 2.250J

Potential Energy stored in spring is: U=(k*x^2)/2
Compared to the unstretched length, the spring
compresses,
x=0.1-0.25m=-0.15m
U=1/2*k*x^2=1/2*(100N/m)*(-0.15)^2
=1.125J
Example: A 2-kg mass is attached to a
spring. If the elongation of spring is 4 cm,
determine potential energy of elastic
spring. Acceleration due to gravity is 10
m/s^2.

Known :
Mass (m) = 2 kg
Acceleration due to gravity (g) = 10 m/s^2
Weight (w) = m g = (2)(10) = 20 N
Elongation (x) = 4 cm = 0.04 m
Wanted : Potential energy of elastic spring
Solution :
Formula of Hooke’s law :
F = k x
F = force, k = spring constant, x = the change in length of
spring
Spring constant :
k = w / x = 20 / 0.04 = 500 N/m
Potential energy of elastic spring :
PE = ½ k x^2 = ½ (500)(0.04)^2 = (250)(0.0016) = 0.4Joule

How to analyze a spring force vs.
displacement graph
• The area under the force in the
spring vs. displacement curve is
the work done on the spring.
Figure shows a plot of force on the
spring vs. displacement, where
displacement is 0 when the spring
is unstretched. The work done on
a spring stores elastic potential
energy Us, in the spring until the
spring goes back to its original
length. Therefore, Us is equal to
the work done and also to the area
under the curve.

Elastic Potential Energy
• Elastic potential energy is energy stored as a result of
applying a force to deform an elastic object. The energy
is stored until the force is removed and the object
springs back to its original shape, doing work in the
process. The deformation could involve compressing,
stretching or twisting the object.

Elastic Potential Energy(Continued)

Law of Conservation of Energy
• The law of conservation of energy is a law of science
that states that energy cannot be created or destroyed,
but only changed from one form into another or
transferred from one object to another. In a closed
system, i.e., a system that isolated from its surroundings,
the total energy of the system is conserved.
The amount of energy in any system, then, is
determined by the following equation:
Ut=Ui+W+Q
– Ut is the total internal energy of a system.
– Ui is the initial internal energy of a system.
– W is the work done by or on the system.
– Q is the heat added to, or removed from, the
system.

• The equation expressing conservation of energy is:
KEi+PEi=KEf+PEf. If you know the potential energy for
only some of the forces, then the conservation of
energy law in its most general form must be used:
KEi + PEi + Wnc + OEi = KEf + PEf + OEf,
where OE stands for all other energies
Example: Water can produce electricity. Water falls from
the sky, converting potential energy to kinetic energy.

What is Conservation of Energy
Total energy of an isolated system remains constant
irrespective of whatever internal changes may take place
with energy disappearing in one form reappearing in
another.
Solved Examples
Example: Consider a person on a sled sliding down a 100
m long hill on a 30° incline. The mass is 20 kg, and the
person has a velocity of 2 m/s down the hill when they're
at the top. How fast is the person traveling at the bottom
of the hill? All we have to worry about is the kinetic
energy and the gravitational potential energy; when we
add these up at the top and bottom they should be the
same, because mechanical energy is being conserved.

Solution: At the top: PE = mgh = (20) (9.8) (100sin30°)
= 9800 J
KE = ½ *m*v^2 = ½*(20)*(2)^2 = 40 J
Total mechanical energy at the top = 9800 + 40 = 9840 J
At the bottom: PE = 0, KE = ½*m*v^2
Total mechanical energy at the bottom = ½*m*v^2
If we conserve mechanical energy, then the mechanical
energy at the top must equal what we have at the
bottom. This gives:
½*m*v*2 = 9840, so v = 31.3 m/s.

Work as transfer of energy
• Work is the transfer of energy. In physics we say
that work is done on an object when you transfer
energy to that object. If you put energy into an object,
then you do work on that object (mass). If a first object is
the agent that gives energy to a second object, then the
first object does work on the second object. Energy is
the ability to do work. The energy transfer from a
source is equal to the amount of work done. Like
work, energy is measured in joules.

Thermal Energy from friction
• Thermal energy is due to the two surfaces rubbing
against each other, which is basically friction. So the
work done by friction is W-Fk*d (Fk is the force
of friction) and it is equal to the thermal energy.
Friction converts mechanical energy into thermal energy.
Friction force - against the direction of movement - can
be computed by multiplying the pressure force applied
with the friction coefficient. Multiplying the friction force
with the distance moved against it gives you the thermal
energy generated through friction.

What is Thermal Energy?
• Thermal energy (also called heat energy) is produced
when a rise in temperature causes atoms and molecules
to move faster, vibrate and collide with each other.
Thermal energy refers to the energy contained within a
system that is responsible for its temperature. As thermal
energy comes from moving particles, it is a form of
kinetic energy. Example: The warmth from the sun, A
cup of hot chocolate

Work/Energy problem with friction
• work done by friction will be negative because the force
is applied in the opposite direction as the motion, and will
decrease the initial energy, giving us a Final Energy
result that is less than the initial energy because
energy is lost to friction.
The thermal energy is usually expressed by Q. It is
directly proportional to the mass of the substance,
temperature difference and the specific heat.
The SI unit of thermal energy is Joule(J).
The thermal energy(Q)=m*c* ΔT
Where
Q = thermal energy,
m = mass of the given substance,
c = specific heat, and
ΔT = temperature difference.

Solved Examples (Thermal Energy)
Example: Determine the thermal energy of a substance
whose mass is 6 kg and specific heat is 0.030 J/kg°c.
The temperature difference of this system is given as
20°c.
Solution:
Given:
m = 6kg,
c = 0.030J/kg°c,
ΔT = 20°c
The thermal energy formula is given by,
Q = mcΔT
= 6 x 0.030 x 20
Q = 3.6 J

Solved Examples (Work with Friction)

What is Power?
• In physics, power is the amount of energy transferred or
converted per unit time. In the International System of
Units, the unit of power is the watt, equal to one joule per
second. In older works, power is sometimes called
activity. Power is a scalar quantity. For example, when a
powerful car accelerates rapidly, it does a large amount
of work and consumes a large amount of fuel in a short
time.
• P = ∆W/∆t
• P = Fv cos θ
• The SI unit of power is Joules per Second (J/s), which is
termed as Watt.

Solved Examples(Power)
Example: A garage hoist lifts a truck up 2 meters above the
ground in 15 seconds. Find the power delivered to the
truck. [Given: 1000 kg as the mass of the truck]
Answer: First we need to calculate the work done, which
requires the force necessary to lift the truck against
gravity:
F = mg = 1000 x 9.81 = 9810 N.
W = F*d = 9810N x 2m = 19620 Nm = 19620 J.
The power is
P = W/t = 19620J / 15s
= 1308 J/s = 1308 W.

Solved Examples (Power)
Example: When doing a chin-up, a physics student lifts her
42.0-kg body a distance of 0.25 meters in 2 seconds.
What is the power delivered by the student's biceps?
Answer: To raise her body upward at a constant speed, the
student must apply a force which is equal to her weight
(m•g).
The work done to lift her body is
W = F * d = (411.6 N) * (0.250 m)
W = 102.9 J
The power is the work/time ratio which is (102.9 J) / (2
seconds) = 51.5 Watts (rounded)

Conservative and Non-Conservative
Forces

Elastic and Non-Elastic Collisions
• A perfectly elastic collision is defined as one in which
there is no loss of kinetic energy, momentum and total
energy in the collision and Forces involved during the
interaction are of conservative nature and The
mechanical energy is not converted into another form of
energy such as thermal energy or sound energy.
• Example: the collisions between two glass or preferably
ivory balls may be taken as elastic collisions.

Elastic Collision of two objects
with different and same masses

Non-elastic Collision
An inelastic collision is one in which part of the kinetic
energy is changed to some other form of energy in the
collision but total energy is conserved. In both an elastic
and non-elastic collision momentum is conserved.

Difference and Similarities between
Elastic and Non-Elastic Collision
Elastic Collision
1) No deformation of
Objects and Heat is not
generated.
2) Momentum & total energy
is same after Collision.
3) The forces involved are
conservative.
4) Mechanical energy is not
transformed into sound
energy or thermal energy.
Non-Elastic Collision
1) Deformation of Objects
and Heat is generated.
2) Momentum & total
energy is not same after
Collision.
3) The forces involved are
non-conservative.
4) Mechanical energy is
transformed into sound
energy or thermal
energy.

Difference and Similarities between
Elastic and Non-Elastic Collision
Elastic Collision
5) Kinetic energy is
conserved.
6) Linear momentum is
conserved.
7) Example: Collision of
atoms
Non-Elastic Collision
5) Kinetic energy is not
conserved.
6) Linear momentum is
not conserved.
7) Example:
automobiles’ collision

Solved Example( Elastic & Non-
Elastic Collision)
Example: Two billiard balls collide. Ball 1
moves with a velocity of 6 m/s, and ball 2
is at rest. After the collision, ball 1 comes
to a complete stop. What is the velocity of
ball 2 after the collision? Is this collision
elastic or inelastic? The mass of each ball
is 0.20 kg.

System of particles and
rotational motion

What is Center of Mass?
• In physics, the center of mass of a distribution of mass in
space is the unique point where the weighted relative
position of the distributed mass sums to zero. It is the
average position of all the parts of the system, weighted
according to their masses. This is the point to which a
force may be applied to cause a linear acceleration
without an angular acceleration. This is also point in a
body or system of bodies at which the whole mass may
be considered as concentrated. For Example: For simple
rigid objects with uniform density, the center of mass is
located at the centroid.

Center of Mass in different Situations
it is the unique position at which the weighted
position vectors of all the parts of a system sum up to
zero. If the object is subjected to an unbalanced force at
some other point, then it will begin rotating about the
center of mass.

Center of Mass in different
Situations

Center of Mass in different Situations
We can extend the formula for the center of mass to a
system of particles. We can apply the equation
individually to each axis also. Meanwhile, the center of
gravity and the center of mass are only equal when the
entire system is subject to a uniform gravitational field.

Center of Mass(Continued)
Consider a system of "n" particles of masses m1, m2,….,mn whose
position vectors are r1, r2, r3,….,rn respectively.
The centre of mass of the system is defined as the point having the
position vector "r" and is equal to:
Where M is the total mass of the system.
If xc, yc, zc are the coordinates of the centre of mass
of a system of "n" particles.

Solved Problem (Center of Mass)
Example: The minute hand of a clock consists of an arrow
with a circle connected by a piece of metal with almost
zero mass. The mass of the arrow is 15.0 g. The circle
has a mass of 60.0 g. If the circle is at position 0.000 m,
the position of an arrow is at 0.100 m, then find out the
center of mass?
Answer: x=m1*x1+m2*x2/(m1+m2)
= (60*0.0 + 15*0.10)/(60 +15)
=(60×0.0+15×0.10)/75
= 1.50/75
x = 0.02 m
The center of mass will be at 0.020 m from the circle.

Example: A fisherman in a boat catches a great white
shark with a harpoon. The shark struggles for a while
and then becomes limp when at a distance of 300 m
from the boat. The fisherman pulls the shark by the
rope attached to the harpoon. During this operation, the
boat (initially at rest) moves 45 m in the direction of the
shark. The mass of the boat is 5400 kg. What is the
mass of the shark? Pretend that the water exerts no
friction.
Solved Example (Center of Mass)

Answer: This problem is very simple to solve because
friction is ignored. This means there are no external
forces, so the center of mass of the system (the shark
and boat) does not move at all.

Example: A 59-kg woman and a 73-
kg man sit on a seesaw, 3.5 m long.
Where is their center of mass?
Answer: In the case of this solution I will use the woman
as the origin.
Mwoman=59kg, Xwoman=0
Mman=73kg, Xman=3.5m
Xcom=Mwoman*Xwoman+Mman*Xman/(Mwoman+Mm
an)
Xcom=(59*0+73*3.5)/(59+71)
=(73*3.5)/(132)
=1.94m

Solved Example(Center of Mass
and Collisions on 2D)
Example: Suppose the following experiment is
performed. A 0.250-kg object (m1) is slid on a
frictionless surface into a dark room, where it
strikes an initially stationary object with mass of
0.400 kg (m2). The 0.250-kg object emerges
from the room at an angle of 45.0º with its
incoming direction.
The speed of the 0.250-kg object is originally
2.00 m/s and is 1.50 m/s after the collision.
Calculate the magnitude and direction of the
velocity (v'2 and θ2) of the 0.400-kg object after
the collision.

Pictorial Description of Problem

Solved Example(Center of Mass
and Collisions on 2D)

What is Center of gravity
• The center of gravity is the point through which the force
of gravity acts on an object or system. In most
mechanics problems the gravitational field is assumed to
be uniform. The center of gravity is then in exactly the
same position as the center of mass. center of
gravity of an object is the point from which you can
suspend an object at rest, and, no matter how the object
is oriented, gravity will not cause it to start rotating.

Relating Angular & Regular Kinematics
Variables

Solved Problem(Rotational and
Angular Kinematics)
Example: A deep-sea fisherman hooks a big fish that
swims away from the boat pulling the fishing line from his
fishing reel. The whole system is initially at rest and the
fishing line unwinds from the reel at a radius of 4.50 cm
from its axis of rotation. The reel is given an angular
acceleration of 110 rad/s2 for 2.00 s as seen in Figure 1.
(a) What is the final angular velocity of the reel? (b) At
what speed is fishing line leaving the reel after 2.00 s
elapses? (c) How many revolutions does the reel make?
(d) How many meters of fishing line come off the reel in
this time?

Solved Problem(Rotational and
Angular Kinematics)

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