Prepared by senior and educated professor . (education, physics, science, Work, power, energy)

1. PHYSICS
Work, Power & Energy

2. How work is done
Energy Force Function
Work is done

3. Work is the result of force moving an object

4. W = f • d
{Distance
Force
{Distance

5. WORKWORK
 When a force acts on a body in such a way thatWhen a force acts on a body in such a way that
body covers some displacement, it is called work.body covers some displacement, it is called work.
 *Work(W) is defined as a product of force(F) and*Work(W) is defined as a product of force(F) and
displacement(d)displacement(d)
 Work can be defined as transfer of energy.Work can be defined as transfer of energy.
 Work is scalar quantity.Work is scalar quantity.
 InSI unit of work is Joule( J ). J=N×mInSI unit of work is Joule( J ). J=N×m
 If force makes an angle with the direction of motion, thenIf force makes an angle with the direction of motion, then
for finding work done, that component of force will befor finding work done, that component of force will be
taken which acts in the direction of motion(Fcostaken which acts in the direction of motion(Fcosθθ ) .W=F) .W=F
coscosθθ ××dd
 W=F cosW=F cosθθ ××dd
d
F
θ
F
d
W= F×d
W = F× d

6. TYPES OF WORKTYPES OF WORK
 (A) When direction of force and(A) When direction of force and
displacement is same the work done isdisplacement is same the work done is
positivepositive
 (B) When direction of force and(B) When direction of force and
displacement is opposite the work done isdisplacement is opposite the work done is
negative.negative.
 (C) When force and displacement are(C) When force and displacement are
mutually perpendicular to each other themutually perpendicular to each other the
work done is zero.work done is zero.
F d
F d
F d

7. Perpendicular forcePerpendicular force
 Since cosSince cos ΘΘ = 0,= 0,
 F*d*F*d* ΘΘ = 0= 0
No work done!No work done!
REMEMBER!REMEMBER!
A vertical forceA vertical force
CANNOT causeCANNOT cause
horizontalhorizontal
displacement!displacement!
Cos Θ = 0 when Θ = 90 degrees

8. • Man does positive
work
lifting box
• Man does negative
work
lowering box
• Gravity does positive
work when box lowers
• Gravity does negative
work when box is
raised

9. Although it takes less force for car A to get the top of the
ramp ,but all cars do the same amount of work.

10. POWERPOWER
In unit time Work performed is termedIn unit time Work performed is termed
powerpower..
Power is a measure of how quickly workPower is a measure of how quickly work
can be donecan be done
P=W/t.(P=Power,W=Work,t=Time)P=W/t.(P=Power,W=Work,t=Time)
Unit of power is watt (J/s).Unit of power is watt (J/s).
Power is scalar quantity.Power is scalar quantity.
Other common unit of Power is horseOther common unit of Power is horse
power(hp)power(hp)
1hp= 746 watt.1hp= 746 watt.
P = W/ t But W = F . d
P = F .d /t But d/t = V
Then P = F . V
So power is product of
force & velocity.

11. ENERGY
Mechanical Non- Mechanical
Kinetic Potential
Linear
Rotational
Gravitational
Elastic
Electric
Magnetic
Light
Sound
Heat

12. ENERGYENERGY
Capability of doing work is called EnergyCapability of doing work is called Energy..
In SI units energy is measured in Joule(J).In SI units energy is measured in Joule(J).
Energy is scalar quantity.Energy is scalar quantity.
There are two basic kinds of energy.There are two basic kinds of energy.
Kinetic Energy(K.E).Kinetic Energy(K.E).
Potential Energy(P.E).Potential Energy(P.E).
Other forms of energy are as under:-Other forms of energy are as under:-
(Electrical,Mechanical,Light,Sound,(Electrical,Mechanical,Light,Sound,
Chemical,Thermal,Nuclear,Geo-Chemical,Thermal,Nuclear,Geo-
thermal,Tidal,Magnetic,Solar e.t.c.)thermal,Tidal,Magnetic,Solar e.t.c.)

13. POTENTIAL ENERGYPOTENTIAL ENERGY
 Definition=Capacity to perform work by a body dueDefinition=Capacity to perform work by a body due
to its position or state or condition is termedto its position or state or condition is termed
potential energy (P.Epotential energy (P.E))
 Examples=Bullet loaded in a gun & water stored in aExamples=Bullet loaded in a gun & water stored in a
dam have potential energydam have potential energy
 The P.E possessed by a body in the gravitationalThe P.E possessed by a body in the gravitational
fields is called gravitational P.E.fields is called gravitational P.E.
 Derivation of equation of Gravitational PotentialDerivation of equation of Gravitational Potential
energyenergy..
 W=FW=F××d. But F=W(Weight) & d=h (Height)d. But F=W(Weight) & d=h (Height)
 W=wW=w××h. But W=mgh. But W=mg
 W=mgh. ButW(work)=P.EW=mgh. ButW(work)=P.E
 P.E = mghP.E = mgh
 P.E=Potential energy( Gravitational)P.E=Potential energy( Gravitational)
 g=Acceleration due to gravityg=Acceleration due to gravity
 m= Mass, h= Heightm= Mass, h= Height
B
A
m
m
Height

15. Elastic Potential EnergyElastic Potential Energy
 Definition:-Definition:- It is stored as a result ofIt is stored as a result of
deformation of an elastic object, such asdeformation of an elastic object, such as
stretching of springstretching of spring..

16. Explanation:-Explanation:-The force that is applied on object is stored as itsThe force that is applied on object is stored as its
elastic potential energy, which helps the object to restore its initialelastic potential energy, which helps the object to restore its initial
position.position.
According to Hooke’s law F = k XAccording to Hooke’s law F = k X
Average force required for displacement( X ) is F/2.Average force required for displacement( X ) is F/2.
Hence work done for displacement isHence work done for displacement is
W (work ) = F/2 × X But F = kXW (work ) = F/2 × X But F = kX
EP = W = kX/2 × X = ½ kX2EP = W = kX/2 × X = ½ kX2
EP = ½ k X2EP = ½ k X2
EP= Elastic P.E, K= Spring constant, X= Extension of spring.EP= Elastic P.E, K= Spring constant, X= Extension of spring.

17. KINETIC ENERGYKINETIC ENERGY
 Definition=Energy produced by a body due to its motion is calledDefinition=Energy produced by a body due to its motion is called
Kinetic energy(K.EKinetic energy(K.E))
 Examples=Bullet fired from a gun& Water flowing from dam haveExamples=Bullet fired from a gun& Water flowing from dam have
Kinetic energy.Kinetic energy.
 Derivation of equation of kinetic energy.Derivation of equation of kinetic energy.
 As 2as=VAs 2as=Vff
22
– V– VII
22
.But V.But VII = o m/s & V= o m/s & Vff = V= V
 Then 2as =VThen 2as =V 22
 Or ” S” = VOr ” S” = V 22
As ” F “= m aAs ” F “= m a
 2a2a
 Put value of “S” & “F “ in below equation.Put value of “S” & “F “ in below equation.
 As W = F × SAs W = F × S
 W = m. a × VW = m. a × V 22
= m V= m V 22
. But W = K.E So K.E = mV. But W = K.E So K.E = mV22
 2a 2 22a 2 2
F
Sm m
V i V f

20. LAW OF CONSERVATION OFLAW OF CONSERVATION OF
ENERGYENERGY
 Statement=Energy is neither created nor destroyed, but oneStatement=Energy is neither created nor destroyed, but one
form of energy is converted into an other form of energy, soform of energy is converted into an other form of energy, so
total energy of the system remains constant.total energy of the system remains constant.
 ExplanationExplanation= In an isolated system from the motion of the= In an isolated system from the motion of the
pendulum above law of conservation of energy is proved inpendulum above law of conservation of energy is proved in
following way.following way.
P.E=100J
K.E= OJ
P.E= O J
K.E = 100 J
A
B
C
h

21. At point AAt point A
 Suppose: m=2Kg , g=10m/ sSuppose: m=2Kg , g=10m/ s22
,h= 5m,h= 5m
 As P.E = mghAs P.E = mgh
 P.E = 2P.E = 2××1010 ×× 5 = 1005 = 100
 P.E = 100 J(At point A )P.E = 100 J(At point A )

 Suppose: V = O m/ sSuppose: V = O m/ s 22
 As K.E = mVAs K.E = mV 22
= 2= 2×× 00 ×× 0 = 00 = 0
 2 22 2
 K.E = 0 J ( At point A )K.E = 0 J ( At point A )
 Total energy at point A = 100 J(P.E ) + 0 J ( K.E )= 100 JTotal energy at point A = 100 J(P.E ) + 0 J ( K.E )= 100 J

22. At point BAt point B
 Suppose: m = 2mg , g = 10 m / sSuppose: m = 2mg , g = 10 m / s 22
,h = 0m.,h = 0m.
 As P.E = mghAs P.E = mgh
 P .E = 2P .E = 2 ×× 1010 ×× 0 = 00 = 0
 P.E = 0JP.E = 0J
 Suppose: m = 2 Kg , V = 10 m/ sSuppose: m = 2 Kg , V = 10 m/ s 22
 K.E = m VK.E = m V 22
= 2= 2 ×× 1010 ×× 10 = 100 J10 = 100 J
 2 22 2
 K .E = 100 JK .E = 100 J
 Total energy at point B = 0J (P.E )+ 100 J ( K .E )= 100JTotal energy at point B = 0J (P.E )+ 100 J ( K .E )= 100J
 Hence it is proved that at point A and B total energyHence it is proved that at point A and B total energy
is 100J (Remains constant)is 100J (Remains constant)
 though energy from one form to anotherthough energy from one form to another
changes(P.E toK.E )changes(P.E toK.E )

25. KE=Kinetic Energy, EE= Energy Efficiency, GP= Gravitational potential Energy.

29. Thanks *****