Note Chap 5 SP015 (Work, Energy and Power) Sitinorsham

Chapter 05 Physics

CHAPTER 5:
Work, Energy and Power

(F2F:6.5 Hours)

1

Overview: Chapter 05 Physics

Work

Energy Power

Kinetic Gravitational Elastic
energy potential potential energy
energy

Principle of Average and
conservation instantaneous

of energy power

2

Chapter 05 Physics

Learning Outcome:

5.1 Work (2.5 hours)
At the end of this chapter, students should be able to:

 Define work done by a constant force,

W =  • 
F s

 Apply work done by a constant force and
determine work done from a force-displacement
graph.

3

5.1 Work Chapter 05 Physics

5.1.1 Work, W

At the end of this lesson, students should be able to:

 Define work done by a constant force.

Work done by a constant force

 is a product of the component of the force parallel to the
displacement times the displacement of a body.

OR

is a scalar (dot) product between force and displacement of

a body. W = (F cos θ )s = Fs cos θ


W = F•s

where F : magnitude of force

s : displaceme nt of the body

θ : the angle betw een F and s

4

Chapter 05 Physics

 a scalar quantity.

 S.I. unit of work:

W = Fs

( )unit of W = kg m s−2 ( m )

= kg m 2 s−2 OR joule (J)

 The joule (1 J) is a work done by a force of 1 N which

results in a displacement of 1 m in the direction of the

force. 1 J = 1 N m = 1 kg m 2 s−2

5

Chapter 05 Physics

Work done by a constant force

At the end of this lesson, students should be able to:

 Apply work done by a constant force and determine work
done from a force-displacement graph.

Force/N

F

W = s2 Fds Work = Area
s1
= F (s2 − s1 ) s1 s2 Displacement/m
0

W = area under the force-displacement graph

6

Chapter 05 Physics

5.1.2 Applications of work’s equation

Case 1 :

 Work done by a horizontal force, F on an object.

  W = F s c os θ and θ = 0
F s
W = Fs
Figure 5.2

Case 2 :

 Work done by a vertical force, F on an object.

 W = F s c os θ and θ = 90 
F
W = 0 J
Figure 5.3 s

7

Chapter 05 Physics

Case 3 :

 Work done by a horizontal forces, F1 and F2 on an object.
F1  W1 = F1s cos 0
F2  W2 = F2 s cos 0
s
 ( )Figure 5.4
W = W1 + W2 = F1s + F2 s

 ( )W = F1 + F2 s and Fnett = F1 + F2
 ( )W = Wnett = Fnett s
Case 4 :

 Work done by a force, F and frictional force, f on an object.
F 
 s
f

Figure 5.5

( )( )Wnett = Fnett s and Fnett = F c os θ − f = ma 8

Wnett = F cos − f s OR Wnett = m as

Chapter 05 Physics

 Caution :

⚫ Work done on an object is zero when F = 0 or s = 0 and

 = 90.

9

⚫ Sign for work. Chapter 05 Physics

W = Fs cos

 If 0< <90 (acute angle) then cos > 0 (positive value)

therefore

W > 0 (positive)  work done on the system ( by
the external force) where
energy is transferred to the system.

 If 90< <180 (obtuse angle) then cos <0 (negative

value) therefore

W < 0 (negative)  work done by the system
where energy is transferred
from the system.

10

Chapter 05 Physics

Example 5.1 :

You push your physics reference book 1.50 m along a horizontal
table with a horizontal force of 5.00 N. The frictional force is 1.60
N. Calculate

a. the work done by the 5.00 N force,

b. the work done by the frictional force,

c. the total work done on the book.

Solution : F = 5.00 N

a. f = 1.60 N

s = 1.50 m

WF = Fs c os θ and θ = 0
=
(5.00)(1.50)cos 0

WF = 7.50 J 11

Chapter 05 Physics

Solution :

b. W f = fs cos θ and θ = 180 

= (1.60)(1.50)cos180

W f = −2.40 J

c. W = WF + W f

= 7.50 + (− 2.40)

W = 5.10 J

OR  W = (5.00 − 1.60)(1.50)

 W = Fnetts

 W = (F − f )s

W = 5.10 J 12

Chapter 05 Physics

Example 5.2 :
A box of mass 20 kg moves up a rough plane which is inclined to

the horizontal at 25.0. It is pulled by a horizontal force F of

magnitude 250 N. The coefficient of kinetic friction between the
box and the plane is 0.300.
a. If the box travels 3.80 m along the plane, determine

i. the work done on the box by the force F,

ii. the work done on the box by the gravitational force,
iii. the work done on the box by the reaction force,
iv. the work done on the box by the frictional force,
v. the total work done on the box.
b. If the speed of the box is zero at the bottom of the plane,
calculate its speed when it is travelled 3.80 m.

13

Chapter 05 Physics

Solution :

m = 20 kg; F = 250 N ; μk = 0.300; s = 3.80 m
a
N Fx

Fy 25 F 
s
mg sin 25
y
x fk 25 25 mgcos 25
W = mg

a. i. WF = Fs cos θ and θ = 25

= (250)(3.80)cos 25

WF = 861 J

14

Chapter 05 Physics

Solution :

a. ii. W g = (m g )s cos θ and θ = 115

= (20)(9.81)(3.80)cos115

Wg = −315 J

iii. W N = Ns cos θ and θ = 90 
WN = 0 J

iv. W f = f k s cos θ and θ = 180 

= (μk N )s cos180

( )= − μk Fy + mg cos 25 s

( )= −(0.300) 250 sin 25 + (20)(9.81)cos 25 (3.80)

W f = −323 J

15

Solution : Chapter 05 Physics

a. v. W = WF + Wg + WN + W f

= 861 + (− 315)+ 0 + (− 323)

W = 223 J

b. Given u = 0

W = mas

223 = (20 )a(3.80 )

a = 2.93 m s−2

v2 = u 2 + 2as

v 2 = 0 + 2(2.93)(3.80 )

v = 4.72 m s−1

16

Chapter 05 Physics

Example 5.3 : F (N)

5

0 3 5 6 7 s(m)
−4

Figure 5.6

A horizontal force F is applied to a 2.0 kg radio-controlled car as it
moves along a straight track. The force varies with the
displacement of the car as shown in Figure 5.6. Calculate the
work done by the force F when the car moves from 0 to 7 m.

Solution : W = area under the F − s graph

= 1 (6 + (5 − 3))5 + 1 (7 − 6)(− 4) 17

22
W = 18 J

Chapter 05 Physics

Exercise 5.1 :

1. A block of mass 2.50 kg is pushed 2.20 m along a frictionless
horizontal table by a constant 16.0 N force directed 25.0 below
the horizontal. Determine the work done on the block by
a. the applied force,
b. the normal force exerted by the table, and
c. the gravitational force.
d. Determine the total work on the block.

(Given g = 9.81 m s−2)

ANS. : 31.9 J; (b) & (c) U think; 31.9 J

18

Chapter 05 Physics

2. y
F3

F1 35

x

 50 
F2

Figure 5.7

Figure 5.7 shows an overhead view of three horizontal forces
acting on a cargo that was initially stationary but that now
moves across a frictionless floor. The force magnitudes are

F1 = 3.00 N, F2 = 4.00 N and F3 = 10.0 N. Determine the total

work done on the cargo by the three forces during the first

4.00 m of displacement.

ANS. : 15.3 J

19

Chapter 05 Physics

Learning Outcome:

5.2 Energy and conservation of energy (3 hours)
At the end of this chapter, students should be able to:
 State the principle of conservation of energy.
 Apply the principle of conservation of energy

(mechanical energy and heat energy due to friction).
 State and apply the work-energy theorem,

Wnett = K = Kf − Ki

20

Chapter 05 Physics

5.2 Energy and conservation of energy

5.2.1 Energy

 is a system’s ability to do work.
 The S.I. unit for energy is same to the unit of work (joule, J).
 is a scalar quantity.

Forms of Description
Energy

Chemical Energy released when chemical bonds between atoms
and molecules are broken.

Electrical Energy that is associated with the flow of electrical charge.

Heat Energy that flows from one place to another as a result of
Internal a temperature difference.

Total of kinetic and potential energy of atoms or molecules
within a body.

21

Chapter 05 Physics

Forms of Description
Energy Energy released by the splitting of heavy nuclei.

Nuclear

Mass Energy released when there is a loss of small amount

of mass in a nuclear process. The amount of energy
can be calculated from Einstein’s mass-energy

equation, E = mc2

Radiant Heat Energy associated with infra-red radiation.

Sound Energy transmitted through the propagation of a series
of compression and rarefaction in solid, liquid or gas.

Mechanical

a. Kinetic Energy associated with the motion of a body.

b. Gravitational Energy associated with the position of a body in a

potential gravitational field.

c. Elastic Energy stored in a compressed or stretched spring.

potential

Table 5.1 22

Chapter 05 Physics

5.2.2 Kinetic energy, K

 is an energy of a body due to its motion.

K = 1 mv 2
2

where K : kinetic energy of a body

m : mass of a body
v :speed of a body

Work-kinetic energy theorem

 states “work done by the nett force on a body equals the
change in the body’s kinetic energy”.

Wnett = K = Kf − Ki

23

Chapter 05 Physics

Example 5.4 :

A stationary object of mass 3.0 kg is pulled to the north by a
constant force of magnitude 50 N on a smooth surface. Determine
the speed of the object when it is travelled 4.0 m away.

Solution : m = 3.0 kg ; F = 50 N; s = 4.0 m; u = 0
F
The nett force acting on the object is given by

Fnett = F

Fnett = 50 N

By applying the work-kinetic energy theorem, thus
Wnett = K f − Ki
 1
sF Fnett s = 2 mv2 − 0

Top view (50)(4.0) = 1 (3.0)v2 24

2
v = 11.6 m s−1

Chapter 05 Physics

Example 5.5 :

F (N)

10

0 4 67 10 s(m)

−5

Figure 5.9

An object of mass 2.0 kg moves along the x-axis and is acted on
by a force F. Figure 5.9 shows how F varies with distance
travelled, s. The speed of the object at s = 0 is 10 m s−1.
Determine
a. the speed of the object at s = 10 m,
b. the kinetic energy of the object at s = 6.0 m.

25

Chapter 05 Physics

Solution : m = 2.0 kg; u = 10 m s −1

a. W = area under the F − s graph from 0 m to 10 m

= 1 (6 + 4)10 + 1 ((10 − 6)+ (10 − 7))(− 5)

22
W = 32.5 J

W = K f − Ki

= 1 mv2 − 1 mu2
22

32.5 = 1 (2.0)v2 − 1 (2.0)(10)2

22

v = 11.5 m s−1

26

Chapter 05 Physics

Solution :

b. W = area under the F − s graph from 0 m to 6 m

= 1 (6 + 4)10

2
W = 50 J

W = K f − Ki

= K f − 1 mu2
2
1 (2.0)(10)2
50 = K f −
2
K f = 150 J

27

Chapter 05 Physics

Exercise 5.2 :

Use gravitational acceleration, g = 9.81 m s−2

1. A bullet of mass 15 g moves horizontally at velocity of
250 m s−1.It strikes a wooden block of mass 400 g placed at rest
on a floor. After striking the block, the bullet is embedded in the
block. The block then moves through 15 m and stops. Calculate
the coefficient of kinetic friction between the block and the floor.

ANS. : 0.278
2. A parcel is launched at an initial speed of 3.0 m s−1 up a rough

plane inclined at an angle of 35 above the horizontal. The
coefficient of kinetic friction between the parcel and the plane is
0.30. Determine
a. the maximum distance travelled by the parcel up the plane,
b. the speed of the parcel when it slides back to the starting

point.
ANS. : 0.560 m; 1.90 m s−1

28

Chapter 05 Physics

5.2.3 Potential Energy
 is an energy stored in a body or system because of its

position, shape and state.
Gravitational potential energy, U
 is an energy stored in a body or system because of its

position.

U = mgh

where U : gravitatio nal potential energy

m : mass of a body
g : acceleration due to gravity
h : height of a body from the initial position

 depends only on the height of the object above the surface
of the Earth.

29

Chapter 05 Physics

Work-gravitational potential energy theorem
 states “ the change in gravitational potential energy as the

negative of the work done by the gravitational force”.

W = −U

 For calculation, use W = U = Uf −Ui

Uf : final gravitational potential energy

Ui : initial gravitational potential energy
W : work done by a gravitational force

30

Chapter 05 Physics

Example 5.6 :


F 20.0 m

Figure 5.11

In a smooth pulley system, a force F is required to bring an
object of mass 5.00 kg to the height of 20.0 m at a constant
speed of 3.00 m s−1 as shown in Figure 5.11. Determine
a. the force, F
b. the work done by the force, F.

31

Chapter 05 Physics

Solution :m = 5.00 kg; s = h = 20.0 m; v = constant = 3.00 m s−1
a. Since the object moves at the constant
F speed, thus
Fnett = 0

 F = mg
mg = (5.00)(9.81)
 
Constant F s F = 49.1 N
speed
 b. W = F s c os θ and θ = 0
mg
= (49.1)(20.0)cos 0

W = 982 J

OR

W = U = Uf −Ui 32
= mgh − 0

= (5.00)(9.81)(20.0)

W = 982 J

Chapter 05 Physics

Elastic potential energy, Us

 is an energy stored in in elastic materials as the result of
their stretching or compressing.

 Hooke’s Law states “the restoring force, Fs of spring is
directly proportional to the amount of stretch or
compression (extension or elongation), x if the limit of
proportionality is not exceeded”

OR Fs = −kx

Fs  − x

where Fs : the restoring force of spring

k : the spring constant or force constant

x : the amount of stretch or compressio n ( x f -xi )

 Negative sign : direction of Fs is always opposite to the

direction of the amount extension/compression, x. 33

Chapter 05 Physics

Case 1:

The spring is hung vertically and its is stretched by a suspended

object with mass, m.

Figure 5.12 
Initial position Fs
Final position
x

The spring is in equilibrium, thus 
W = mg
 Fy = 0
34
Fs − mg = 0

Fs = mg

Chapter 05 Physics

Case 2:

The spring is attached to an object and it is stretched and

compressed by a force, F. 
Fxs is negative Fs F
is positive

x The spring is in equilibrium,

x=0 hence  
Fs = F
Fs = 0
x=0

(Equilibrium position)

 x =0
F Fs Fs is p o s itiv e
x is ne ga tive

x

Figure 5.13

35

Chapter 05 Physics

 Caution:

⚫ For calculation, use : Fs = kx = F where F : applied force

⚫ The unit of k is :

k = Fs = ma unit of k = N = N m−1 OR kg s−2
xx m

 From the Hooke’s law (without “−” sign), a restoring force, Fs
against extension of the spring, x graph is shown.

Fs W = area under the Fs − x graph
F 1
W = 2 Fx1 W = 1 (kx1 )x1
0 Figure 5.14 2
1
W = 2 k x12 =Us

x1 x

36

Chapter 05 Physics

 The equation of elastic potential energy, Us for compressing or
stretching a spring is

Us = 1 kx2 = 1 Fs x
2 2

 The work-elastic potential energy theorem,

W = U s

OR

W = Usf − U si = 1 kxf2 − 1 kxi2
2 2

37

Chapter 05 Physics

 Work-energy theorem states the work done by the nett
force on a body equals the change in the body’s total

energy”. Kf +Uf Ki +Ui

 Wnett = E = Ef − Ei

Wnett = ( Kf +Uf ) − ( Ki +Ui )
= ( Kf − Ki ) + (Uf −Ui )

Wnett = K + U

38

Chapter 05 Physics

Example 5.7 :

A force of magnitude 800 N caused an extension of 20 cm on a
spring. Determine the elastic potential energy of the spring when

a. the extension of the spring is 30 cm.

b. a mass of 60 kg is suspended vertically from the spring.

Solution : F = 800 N ; x = 0.200 m

a. Fs =F = kx
800 =
k (0.20 )

k = 4 103 N m −1

Given x=0.300 m. 1
2
Us = k x2

( )U s 1 (0.300 )2
= 2 4 103 U s = 180 J

39

Chapter 05 Physics

Solution :

b. Given m=60 kg. When the spring in equilibrium, thus

 Fnett = 0
Fs
Fs = mg
x k x = mg

( )4 103 x = (60)(9.81)

x = 0.147 m

Us = 1 k x2
2
( )= 1 4 103 (0.147)2
  2
W = mg
U s = 43.2 J

40

Chapter 05 Physics

5.2.4 Principle of conservation of energy

At the end of this lesson, students should be able to:

 State and apply the principle of conservation of energy.
 states “in an isolated (closed) system, the total energy of

that system is constant”.

 we get

The initial of total energy = the final of total energy
OR

 Ei =  Ef

Conservation of mechanical energy

 In an isolated system, the mechanical energy of a system is the
sum of its potential energy, U and the kinetic energy, K of the
objects are constant.

E = K + U = constant OR Ki +Ui = Kf +Uf

41

Chapter 05 Physics

Example 5.8 :

A 1.5 kg sphere is dropped from a height of 30 cm 30 cm
onto a spring of spring constant, k = 2000 N m−1 .
After the block hits the spring, the spring x
experiences maximum compression, x as shown
in Figure 4.15.

a. Describe the energy conversion

occurred after the sphere is

dropped onto the spring until the

spring experiences maximum Before After
compression, x.
Figure 4.15

b. Calculate the speed of the sphere just

before strikes the spring.

c. Determine the maximum compression, x. 42

Chapter 05 Physics

Solution :
a.

h = 30 cm

h0 v

x

h1
h2

(1) (2) (3)

The spring is not stretched The spring is not stretched The sphere is at height h2

hence Us = 0. The sphere is hence Us = 0. The sphere is above the ground after
compressing the spring by x.
at height h1 above ground

with speed, v just before
at height h0 above ground The speed of the sphere at

therefore U = mgh0 and it is this moment is zero. Hence

stationary hence K = 0.

  E1 = mgh0
strikes the spring. Therefore 1
1 2
E2 = mg h1 + 2 mv 2 E3 = mg h2 + k x2

43

Chapter 05 Physics

Solution : m = 1.5 kg; h = 0.30 m; k = 2000 N m −1

b. Applying the principle of conservation of energy involving the

situation (1) and (2),

 E1 = E2 + 1 mv 2
2
mg (h0 mg h0 = mg h1 (h0 − h1 )
= 1 mv 2 and h =
− h1 ) 2

v = 2gh

= 2(9.81)(0.30)

v = 2.43 m s−1

44

Solution : Chapter 05 Physics

c. Applying the principle of conservation of energy involving the

situation (2) and (3),

 E2 = E3
1 1
mg h1 + 2 mv 2 = mg h2 + 2 k x2

1 mv 2 1 k x2 and
2 2
( )mg (h1
− h2 ) + = x = h1 − h2

(1.5)(9.81)x + 1 (1.5)(2.43)2 = 1 (2000 )x2
22

1000 x2 − 14.7 x − 4.43 = 0

x = 7.43 10−2 m

45

Chapter 05 Physics

Example 5.9 :

m1 u1 m1 + m2

m2 h

Figure 5.16

A bullet of mass, m1=5.00 g is fired into a wooden block of mass,
m2=1.00 kg suspended from some light wires as shown in Figure
5.16. The block, initially at rest. The bullet embeds in the block,

and together swing through a height, h=5.50 cm. Calculate

a. the initial speed of the bullet.

b. the amount of energy lost to the surrounding. 46

Chapter 05 Physics

Solution : m1 = 5.00  10 −3 kg; m2 = 1.00 kg; h = 5.50  10 −2 m

a.

v12 = 0

u2 = 0 u12 m1 + m2

m1 u1 m2 m1 + m2 h

(1) (2) (3)

Applying the principle of conservation of energy involving the

situation (2) and (3),  E2 = E3

(m11 K = U
=
( )2 )(u12 )2 (m1 +
+ m2 = m2 )gh
u12 2gh = 2(9.81)
5.50 10 −2

u12 = 1.04 m s−1 47

Chapter 05 Physics

Solution : m1 = 5.00  10 −3 kg; m2 = 1.00 kg; h = 5.50  10 −2 m
Applying the principle of conservation of linear momentum

involving the situation (1) and (2),

  
p1 = p2

( ) ( )m1u1 = (m1 + m2 )u12

5.00  10 −3 u1 = 5.00  10 −3 + 1.00 (1.04 )

u1 = 209 m s−1

b. The energy lost to the surrounding, Q is given by

 Q = E1 − E2

= 1 m1 u12 − 1 (m1 + m 2 )(u12 )2
2 2
( ) ( )= 1 5.00 10−3 (209)2 − 1 5.00 10−3 + 1.00 (1.04)2
22

Q = 109 J

48

Example 5.10 : Chapter 05 Physics

Smooth
pulley

Q

2m

P

Figure 5.17

Objects P and Q of masses 2.0 kg and 4.0 kg respectively are
connected by a light string and suspended as shown in Figure
5.17. Object Q is released from rest. Calculate the speed of Q at
the instant just before it strikes the floor.

49

Chapter 05 Physics

Solution : mP = 2.0 kg; mQ = 4.0 kg; h = 2 m; u = 0

Smooth Smooth
pulley pulley

Q vP

2m 2m Q v

P

Initial Final

 Ei = E f
81)m(2Q)Ug=Qh(2==.0Um)(PP9g+.8hK1+)P(212+)m+KP1Qv
2 + 1 mQv 2
2
(4. )(9. (2.0)v2 + 1 (4.0 )v
0 2

v = 3.62 m s−1 2 2

50