Higher Secondary Physics Class Work, energy and power

1. By
Jenoy Mathew
A.K.M.H.S.S,Kottoor

2. Work
 The work done by a force is measured
by the dot product of force and
displacement. If a force F acting on a
body displaces it through a
displacement d,then the work done by
the force is given by W=F.d

3. F
dd
Work done(W)=F.d

4. Ө
FsinӨ
F
FcosӨ
d
Work done(W)=FcosӨ.d

5. Work is a scalar
 The body is not displaced in the
perpendicular direction. So the work
done by FsinӨ component of force is
zero. Work done is the dot product of
two vectors. So work done is a scalar
quantity.

6. Types of Work

7. F
Angle d F.d=FdcosӨ
between F
and d is zero =Fd cosO
=Fd
Work done is +ve

8. W=F.d=F.d cosӨ
=F.d cos 900
=zero
F
Ө =90o
d
Work done is Zero

9. W= F.dcosӨ
Ө=1800 =F.dcos1800
= -Fd
F d
The work done is -ve

10. Kinetic Energy

11. Kinetic Energy
It is the energy possessed
by virtue of its motion

12. Regarding with K.E.
 K.E. is always a +ve scalar quantity.
 K.E. depends on the frame of reference.
 K.E.related to its momentum as
K.E.=p2/2m,where m is the mass and p is
the momentum

13. Work energy principle
 Work done by a force in displacing a
body measures the change in
kinetic energy of the body

14. U=O V
m
S
By equation of
motion, v2= u2+2aS
U=O , So v2= 2aS
2
W = F.S = ma X v /2a S = v2/2a
=1/2mv2

15. This work done appears as the
K.E. Of the
body, i.e.,K.E.=1/2mv 2

16. Work done by a variable force

17. B
Work done for the
Force displacement
dx, dw=F.dx
A
D C Xf
O Xi dx Displacement
The total work done on the body from A to B,
W= = Area ABCD

18. Work energy theorem for a
variable force

19. The time rate of change of K.E. is, dK/dt=d/dt(1/2 mv2)
=1/2m(2v).dv/dt=m dv/dt.v=F.v=F.dx/dt
=
i.e,Ki – Kf = =W
dK= F.dx

20. Problems

21. Calculate the work done in lifting a mass of
5kg vertically through 8m
 Force on the mass = weight of the body =
5 X 9.8 N
 Vertical lift = 8m
 Work done = Force X vertical lift = 5 X
9.8 X 8 = 392 J

22. A man pushes a roller with a force of 50 N through
a distance of 20m.Calculate the work done if the
handle of the roller is inclined at an angle of 60o
with the ground?
 W = F X S cosϴ=50 X 20X cos60 = 500J

23. Calculate the work done in raising a stone of mass 5kg
and specific gravity 3 lying at the bed of a lake through
a height of 5m?
 Loss of weight of stone in water =
 = kg wt.
 weight of stone in water = kg wt = kg wt
 Force,F = kg wt = X 9.8 N
W= X 9.8 X 5 = 163.3 J

24. A bullet weighing 10g is fired with a velocity of
800m/s.After passing through a mud wall 1m thick, its
velocity decreases to 100m/s.Find the average
resistance offered by the mud wall?
 W = 1/2mv2 – 1/2mu2
 F.d = 1/2mv2 – 1/2mu2
F= { v2 - u2}= -3150N

25. Potential energy

26. Potential energy
 The energy possessed by a body by virtue
of its position (or configuration) in a
field or state of strain is called Potential
energy.
 e.g. Water stored in the reservoir of a
dam, an apple hanging on an apple tree.

27. Gravitational potential energy
 The energy possessed by a body due
its position in the gravitational field
of earth is called gravitational
potential energy.

28. h
180o W=F.h
mg = F h cosϴ
= mgh cos 1800
h
= - mgh
P.E. = V = mgh
Earth

29. Potential energy of a spring
When a spring is stretched or
compressed from its normal
position(x = 0),a restoring force is
set up inside the spring

30. The same spring is stretched or
compressed as shown below. In which case
does the force exerted by the spring have
the largest magnitude?

31.  This restoring force is directly proportional to
the displacement from the equilibrium
position.
 i.e., F α - x
 F = -kx,where k is a constant called spring
constant.
 –ve sign indicates that restoring force is
opposite to the displacement.

32.  The external force required to keep the spring
at a displacement x is Fext = -F = +kx
 Therefore the total work done to stretch the
spring from x = 0 to x = x is, W = =
= 1/2 kx2

33. From the graph
Force
kx
x Displacement

34.  For a particular displacement x,the
area of the graph = area of ∆OAB
 area of ∆OAB = 1/2 x × kx = 1/2 kx2 = P.E.

35. Conservative forces
and Non-conservative forces

36. Conservative forces
 if the amount of work done against a
force depends only on the initial and
final positions of the body moved and
not the path followed,then such a force
is called conservative force.
 e.g. gravitational force,electrostatic
force,elastic force,Magnetic
force,Lorentz force

37. Non-conservative
 If the workdone against a force depends
on the path followed,such a force is
called non-conservative force.
 e.g. Frictional force,Viscous
force,Damping force,A force that
generally depends on velocity

38. Conservation of mechanical energy
 Energy can neither be created nor
destroyed.It can only be
transformed from one form into
another without gain or loss.i.e. the
total energy of a closed system
remains constant.

39. At A, A
K.E.=1/2 mu2 = 0
P.E.(V) = mgh
Total energy= K.E. +P.E. = mgh
x
At B,
P.E. = mg(h – x)
K.E. = ½ mv2 h
For a free falling body,v = B
√2gx
So,K.E. = ½ m X 2gx = mgx
Total energy = K.E.+P.E.
= mg (h- x) + mgx
C
= mgh

40. At C,
P.E. = O Energy
Here velocity at C(V)=√2gh
So K.E. = mgh
Total energy = mgh+ O = mgh T.E.
K.E
Variation of K.E.,P.E.
P.E.
And total energy
with height
Height

41. Problems

42. A ball is thrown vertically downward from a height of 20m with an
initial velocity u.After collision with the ground, it losses 40% of its
initial velocity and rebounds to the same height. Find out u ?
 Total energy of the ball when just thrown
downwards = 1/2mu2 + mgh.
 Energy after collision = (1/2mu2 + mgh)60/10o
 Energy of the ball at a height ‘h’ after collision = mgh
 By conservation of
energy, (1/2mu2 + mgh)60/10o =
mgh
 3/2mu2 = 2 mgh, i.e., u2 = 4/3 gh
 u = √4/3 9.8 20 = 16.17 m/s

43. A body of mass 1kg initially at rest is dropped
from a height of 2m on to a vertical spring having
force constant 490 N/m.Calculate the maximum
distance through which the spring will be
compressed
Loss in gravitational P.E. Of the body = Gain
in the P.E. Of the spring
Mg (h + x) = 1/2kx2
1 9.8(2+x) = ½ 490 x2
100 x2 – 4x – 8 = 0,on solving x = 0.303m

44. A rain drop of radius 2mm falls from a height of 250m above the
ground.What is the work done by the gravitational force on the
drop ?
 Volume,V = 4/3 πr3
 Mass, m = V d
 Work done by gravitational force(W) = mgh
= 4/3 πr3 d g h
 W = 4/3 π (2 10-3 )3 103 9.8 250 = 0.082 J

45. Different forms of energy

46. Light energy
 When an excited electron in an atom or
molecule makes a transition to a lower energy
level,this form of energy is produced
 Light transmitted as photons of energy E = hυ

47. Internal energy
 The molecules of a body are in random
motion.The molecules thus possess K.E.
 The intermolecular attraction produces P.E.
 The sum of these kinetic and potential
energies of the system is called internal
energy.

48. Heat or thermal energy
 Heat is a form of energy
 An object possesses heat energy due
to the molecules moving in it

49. Nuclear energy
 The binding energy of
nucleons(neutrons and protons) in the
nucleus is called nuclear energy
 During nuclear reactions like nuclear
fusion or fission, this energy is released

50. Electrical energy
 Electric charges develop an attraction or
repulsion among them
 A work is to be done to move a charge in
an electric field
 This work is stored as electrical energy

51. Mass energy equivalence
 When a mass ‘m’ is converted into energy E
 The corresponding energy released is E = mc2
, Where c is the velocity of light
 This equation is known as Einstein’s mass energy
equivalence
 If 1 kg of matter is converted as energy,then
 E = 1 ( 3 108 )2 = 9 1016 J
 The energy released during nuclear reaction is
due to the conversion of mass into energy

52. About 4 1010 kg matter per second is converted
into energy in the sun.What is the power
output of the sun?
 E = mc2 = 4 1010 ( 3 108 )2 = 3.6 1027 J
 This much of energy is liberated per second
 Power out put of sun = 3.6 1027 J/s = 3.6 1027 W

53. Power
 Power is defined as the rate at which work is done
 If W is the work in a time t,the average
power, P= W/t
 If dw is the work done in a small time interval
dt,then the instantaneous power, P= dw/dt
 Power is a scalar quantity
 P=W/t = F.S/t= F.v (since v=S/t)

54. Problems

55. A water pump driven by petrol raises water at a rate of
0.5 m3/min from a depth of 30m.If the pump is 70%
efficient,what power is developed by the engine
 Volume of water taken/s = 0.5/60 m3/s
 Mass of water taken/s = 0.5/60 X 103 kg/s
 Out put power=mgh/t=0.5/60 X 103 X9.8 X 30=2450W
 i.e , Input power = output power/efficiency
 2450/70/100=2450 X 100 /70 = 3500 W

56. A railway engine of mass 12000kg is moving at a
constant speed 5m/s up an inclined plane of
150.Calculate the power of the engine?
(Take g=9.8 m/s)
R F
Mg sinϴ
Mg cos ϴ
Mg
150

57.  Force (F) = mg sinϴ
 F = 12000 X 9.8 X sin 150 = 3.04 X 104 N
 P = F X v = 3.04 X 104 N X v
 P = F X v = 3.04 X 104 X 5 = 152000 W
= 152 kW

58. Collisions

59.  The abrupt change to the path of a
moving body(or bodies)due to its
interaction with other body(or bodies)is
called collision.
 The magnitude and direction of the
velocity of the colliding bodies may
change in a collision.

60. Types of collisions
 Perfectly elastic collision
 The collision in which,both K.E. and
linear momentum of the colliding
bodies remain conserved is called
perfectly elastic collision.

61.  Inelastic collision
 The collision in which linear
momentum of the system remains
conserved but its kinetic energy is not
conserved is called inelastic collision.

62.  Perfectly inelastic collisions
 A collision is said to be perfectly
inelastic if after collision,the two bodies
stick together an move as a single system

63. Elastic collision in One Dimension
m1 m2
u1 u2
At the time of
collision
Before collision

64. After collision
V1 V2
m1 m2
The collision is elastic, the K.E. And
linear momentum remain conserved

65. By conservation of momentum,
m1 u 1+m2 u2 = m1 v 1+m2 v2
m1(u 1 - v 1 ) = m2 (v2 - u2) ------- (1)
By conservation of K.E.,
1/2 m1 u 1 2+1/2m2 u2 2 = 1/2m1 v 1 2+1/2 m2 v2 2
m1(u 12 - v 12 ) = m2 (v22 -u22)
m1(u 1 + v 1 ) (u 1 - v 1 ) = m2 (v2 + u2) m2 (v2 - u2)-----(2)
Dividing eq (2) by (1),
u 1 + v 1 = v2 + u2 -------(3)
i.e., u 1 - u2 = v2 - v 1 -------(4)

66. (u 1 - u2 ) is the relative velocity of approach
and (v2 - v 1 ) is the relative velocity of
separation

67. Velocities after collision
From equation ( 3) , v2 = u 1 + v 1 - u2
Substituting the value of v2 in Eq.(1),
m1(u 1 - v 1 ) = m2 (u 1 + v 1 - u2- u2)
m1 u 1 - m1 v 1= m2 u 1 +m2 v 1- 2m2 u2
m1 u 1- m2 u 1+ 2m2 u2 = v 1( m1+ m2 )

68. There fore,
v 1 ={ ( m1- m2 ) u 1 / ( m1+ m2 ) }+ {2 m2 u2/ ( m1+ m2 )}
Similarly we can see that,
v 2 ={ ( m2- m1 ) u 2 / ( m1+ m2 ) }+ {2 m1 u1/ ( m1+ m2 )}

69. Special cases
 When the two bodies have equal masses,i.e., m1 = m2 = m
v 1 = 2m u2/2m = u2 and v 2 = u 1 . So if the two bodies
have equal masses after elastic collision in one
dimension,exchange their velocities
 If the second body is at rest before collision.then, u2 =
O, also m1 = m2 = m ,after collision v 1 = O and v 2 = u 1

70.  If m1>> m2 ,we can neglect m2 and u2 = O,then
v 1 = m1 u 1 / m1 = u 1 , v 2 = 2m1 u 1 / m1 = 2 u 1 .i.e.,If a
heavy body collides with a light body,there is no
change in the velocity for the heavy body and the
light body moves with twice the velocity of the
heavy body

71.  If m1<< m2 , here we can neglect m1. Also if u2 = O
then, v 1 = - m2 u 1 / m2 = u 1 , v 2= 2m1 u 1 / m2 = O.
i.e,if a light body collides with a heavy body at
rest,then it rebounds with its initial velocity,while
the heavy body remains at rest

72. Inelastic collision in one dimension
 Consider a perfectly in elastic collision.After
collision,the two bodies stick together and
move.
 By conservation of
momentum, m1 u
1+m2 u2 = ( m1+ m2 )V ,where V is the velocity of
the combined mass after collision.
 V = m1 u 1+m2 u2 / ( m1+ m2 )

73.  The loss in K.E. is
∆K = 1/2 m1 u 1 2+1/2m2 u2 2 - ½(m1+ m2 )v2
 ∆K = 1/2 m1 u 1 2+1/2m2 u2 2 - ½{(m1 u 1+m2 u2 )2}/ (m1+ m2 )
 i.e., ∆K = m1 m2 (u 1- u2 )2 /2(m1+ m2 )
 ∆K is +ve and hence there is always a loss in K.E.

74. Elastic collision in 2 dimension
v1
ϴ1 X
m1 m2
ϴ2
u1
v2

75. m1 v 1 sin ϴ1
ϴ1
m1 v 1 cos ϴ1
By conservation of momentum along X direction,
m1 u 1= m1 v 1 cos ϴ1 + m2 v 2 cos ϴ2

76. Conservation of momentum along Y direction,
O = m1 v 1 sin ϴ1- m2 v 2 sin ϴ2
During elastic collision,K.E. Is conserved,
m1 u 1 2= m1 v 1 2+ m2 v2 2