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2. Work
The force acting through a distance.
When a force acts on the body and point of
application of the force is displaced in the
direction of force.

3. Formula and Unit
Work = Force X Displacement
W = F. s (or) F. d (or) F. x
The S.I base units is Kg m2/ s2
The unit of work is joule (J)
1 joule :
When a force of 1N acting on body displacing
it by distance of 1m.

4. Work involves force
The more force is applied the more work is
done.
Work involves distance
The more distance travelled the more work is
done.
So work is
Force acting upon an object in the direction the
object moves or displaced.

5. Work done
1.Work is done on an object when force
acts on it in the direction of motion.
2.Work is not done when there is no
motion or when the force is perpendicular to the
motion.
3. If the displacement is against a force,
the work is done against the force.

6. W = F . d
For constant force in the direction of Motion.
W = F (cos Ө) d
For constant force with a component in
direction of motion.
i.e Work by a constant force W= F. d cos Ө
Ө  angle between F & d

9. The condition need to be satisfied for work to
be done :
* A force should act on an object
* The object must be displaced

22. Energy

24. Energy
The capacity to do work or in other words,
Energy as the ability to do work done.
The unit of energy is the same as that of
work. i.e joule
1 joule
The energy required to do 1 joule of work.

25. Commercial unit of energy
1 kwh is the energy used in one
hour at the rate of 1000 J/s
1 kwh = 1 kw X 1 h
= 1000 w X 60 X 60
= 3600000
= 3.6 X 106 J

26. Forms of Energy
The energy in many forms like mechanical
energy, thermal energy, electric energy, light
energy, nuclear energy, etc …

28. Mechanical Energy
The energy possessed by a body due to its
position or due to its motion.
The mechanical energy of a body consists
of potential energy and kinetic energy.

29. Kinetic Energy
The energy possessed by an object due to its
motion.
The K.E of an object increases with its
speed.

31. Expression for Kinetic Energy
v2 – u2 = 2as  s = v2-u2/ 2a
W = mg X h
= ma X (v2-u2 ) / 2a
= m ( v2 – u2 ) / 2
= ½ m ( v2 – u2 )
(Initial velocity u = 0)
W = ½ mv2
The K.E possessed by an object of mass m
and velocity with a uniform velocity v.
E k = ½ mv2

32. Alternative Method:
E K = F . s
= ma X s
To Find s :
v2 = u2 + 2as
v2 = 0 + 2as
s = v2 / 2a
i.e
E K = ma X v2/2a
E K = ½ mv2

36. Potential Energy
The P.E of a body is the energy stored in
the body by virtue of its position.
P.E is given by the amount of work done
by the force acting on the body, when the body
moves from its given position to some other
position.

38. Expression for Potential Energy
when the body is taken vertically up through a
height then work done
W = Force X Displacement
= mg x h
This work done is stored as potential energy
in the body.
E p = mgh

40. Kinetic Energy Potential Energy
1. Energy at work 1. It is stored energy, energy ready to go.
2. Kinetic is motion 2. Potential is energy of position
3. Increases weight, mass & Speed 3. Increases weight, mass & Height
4. E k = ½ mv2 4. E p = mgh
5. Example: Water falling over the fall, A
race car speeding around a corner, a
bicycle cursing down a hill, and students
running a home from school.
5. Example: water stored in reservoir, A
race car at the starting line of a race, a
bicycle on top of a hill, and students
waiting to go home from school

41. In physics, the law of conservation of
energy states that the total energy of an isolated
system remains constant—it is said to be
conserved over time.
Energy cannot be created or destroyed. It
only can be transformed from one form to
another.
If a body or system of bodies is in motion
under a conservative system of forces, the sum
of its kinetic energy and potential energy is
constant.
Law of Conservation of Energy

42. Law of Conservation of Energy
P.E  E P = 0
v2 = u2 + 2as
v2 = 0 + 2gh
K.E  E K = ½ mv2
= ½ m (2gh)
E K = mgh
T.E  E T = EP + EK
= 0 + mgh
E T = mgh

43. Problem:
1. m = 15 kg v = 4 m/s K.E = ?????
2. m = 24 kg v = 8 m/s K.E = ?????
3. m = 10 kg v = 6 m/s K.E = ?????
4. m = 28 kg v = ???? K.E = 686 J
5. m = ???? v = 3 m/s K.E = 27 J

44. Answer :
1.120 J
2.768 J
3.180 J
4.7 m/s
5.6 kg

45. Problems
1. m = 10 kg g= 9.8 m/s^2 h= 6m P.E=????
2. m = ???? g= 9.8 m/s^2 h= 10m P.E=294 J
3. m = 5 kg g= 9.8 m/s^2 h= ??? P.E= 196 J

46. Answer :
1. 588 J
2. 3 kg
3. 4 m

47. Problems
1.m= 3 kg, v= 6 m/s, h=15m, g=9.8m/s^2, T.E=??
2.m= 7 kg v= 4 m/s h= 5m g=9.8m/s^2 T.E=??
3.m= 2 kg v= 5 m/s h=10m g=9.8m/s^2 T.E=??
4. m=?? v= 2 m/s K.E=12J g=9.8m/s^2 h=??
P.E= 117.6 J T.E=??
5. m= 5kg v=??? K.E= 62.5J g=9.8m/s^2 h=10m
P.E=??? T.E=???

48. THANK YOU