CHAPTER 5 WORK AND ENERGY

CHAPTER 5.0 :
WORK and ENERGY

(6 HOURS)

1

Learning Outcomes:

5.1 Work (2 hours)

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

 Define work done by a constant force.

 Explain the physical meaning of the dot
product, W  F  s

 Use the equation for work done by a constant
force, W  Fscos

 Determine work done from the force- 2
displacement (F-s) graph.

5.1 Work, W

Work done by a constant force

 is defined as the product of the component of the force parallel
to the displacement times the displacement of a body.

OR

is defined as the scalar (dot) product between force and
displacement of a body.

Equation : where


W Fs
W  F cosθs  Fs cosθ F : magnitude of force

s : displacement of the body

θ : the angle between F and s

3

 It is a scalar quantity.
 The S.I. unit of work is kg m2 s2 or joule (J).
 The joule (1 J) is defined as the 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 m2 s2

4

Applications of work’s equation

Case 1 :

 Work done by a horizontal force, F on an object (Figure 5.1).
F W  Fs cosθ and θ  0

s W  Fs

Figure 5.1

Case 2 :

Work done by a vertical force, F on an object (Figure 5.2).



F W  Fs cosθ and θ  90

Figure 5.2  W 0J
s

5

Case 3 :

Work done by a horizontal forces, F1 and F2 on an object
(Figure 5.3).  
F1 W1  F1s cos 0

F2  W2  F2s cos 0
s
Figure 5.3

W  W1  W2  F1s  F2s
  W  F1  F2 s and Fnett  F1  F2

  W  Wnett  Fnett s

6

Case 4 :

Work done by a force, F and frictional force, f on an object

(Figure 5.4). 

 F 
f s

Figure 5.4

 Wnett  Fnett s and Fnett  F cos θ  f  ma

Wnett  F cos  f s OR Wnett  mas

7

 Caution :

Work done on an object is zero when F = 0 (Figure 5.5) or
s = 0 (Figure 5.6) and  = 90 (Figure 5.7).

Figure 5.5 Figure 5.6 Figure 5.7

8

Sign for work. 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.

9

Example 5.1 :

You push your 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 1.60 N F  5.00 N
s  1.50 m
a. Use work’s equation of constant force,

WF  Fs cosθ and θ  0 10

WF  5.001.50cos0

WF  7.50 J

Solution :

b. W f  fs cosθ and θ  180

W f  1.601.50cos180

W f  2.40 J

c. W  WF  Wf
W  7.50   2.40

W  5.10 J

OR W  5.00 1.601.50

W  Fnett s

W  F  f s

W  5.10 J 11

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.

12

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 mgcos 25
25

W  mg

a. Consider the work done along inclined plane, thus

i. WF  Fxs cosθ where : θ  0
 WF  250cos 25 3.80cos 0

WF  861 J 13

 Solution :

a. ii. Wg  mg sin 25 s cosθ where : θ  180

 Wg  209.81sin 25 3.80cos180

Wg  315 J

iii. WN  Ns cosθ where θ:  90

WN  0 J

iv. W f  fk s cosθ where : θ  180

W f  μk N s cos180

 W f  μk F sin 25  mg cos 25 s
 Wf  0.300 250sin 25  209.81cos 25 3.80

W f  323 J

14

Solution : 15

a. v. W  WF  Wg  WN  W f
W  861  315 0   323

W  223 J

b. Given u  0

By using equation of work for nett force,

W  mas

223  20a3.80

a  2.93 m s2

Hence by using the equation of linear motion,

v2  u 2  2as

v2  0  22.933.80

v  4.72 m s1

Example 5.3 :

F (N)

5

0 3 5 6 7 s(m)
4

Figure 5.8

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 5.8. 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

W  1 6  5  35  1 7  6 4 16

22
W  18 J

Exercise :

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.

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

17

Exercise : y
F3
2.
35

F1 x

 50
F2

Figure 5.8

Figure 5.8 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

18

Work done by a variable force

F/N

FN

Figure 5.9

F4 s4 sNs2 s
F1 W1

0 s1s1

W  F1s1  F2s2  ...  FN sN

19

Figure 5.9 shows a force, F whose magnitude changes with the
displacement, s.

For a small displacement, s1 the force remains almost constant
at F1 and work done therefore becomes W1=F1 s1 .

To find the total work done by a variable force, W when

the displacement changes from s=s1 to s=s2, we can divide

the displacement into N small successive displacements :

s1 , s2 , s3 , …, sN

Thus

When N  , s  0, therefore W  s2 Fds
s1

20

W  s2 Fds Work done =
s1 Area under
F/N the graph

0 s1 s2 s/m

21

Example 5.4 :

A force , F acting on a particle varies with the displacement

x as shown in figure below.
Calculate the work done by the force as the particle moves
from x=0 to x=6 m.

F (N) Solution :
5
The Work done = area under graph
4
x (m) =½(4+6)5
6 = 25 J

22

Learning Outcome:

5.2 Energy and Conservation of energy (3 hours)

At the end of this chapter, students should be able to:
a) Define :

(i) kinetic energy
(ii) gravitational potential energy
(iii) elastic potential energy
b) Use :
(i) kinetic energy , K = ½ mv 2
(ii) gravitational energy, U = mgh
(iii) elastic potential energy, Us = ½ kx 2
c) State the principle of conservation of energy

d) Apply the principle of conservation of energy

23

5.2 Energy and Conservation of energy

Energy

o is defined as the system’s ability to do
work.

o The S.I. unit for energy is same to the unit

of work (joule, J).

o is a scalar quantity.

24

Potential Energy

 is defined as the energy stored in a body or system
because of its position, shape and state.

Gravitational potential energy, U

 is defined as the energy stored in a body or system because of
its position.

 Equation : U : gravitational potential energy

U  mgh where m : mass of a body
g : acceleration due to gravity

h : height of a body from the initial position

 The gravitational potential energy depends only on the height
of the object above the surface of the Earth.

25

 Work-gravitation potential energy
◦For calculation, use

W  U  U f Ui

where

U f : final gravitational potential energy

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

26

Example 5.5 :


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.

27

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 0

Fnett

F  mg
 F  5.009.81
mg
Constant F  49.05 N

Fs b. From the equation of work,

speed W  Fs cosθ and θ  0

 W  49.120.0
mg
W  981 J

OR
W  Fs cosθ and θ  0

W  U  mgh

W  981 J

28

Elastic potential energy, Us

 is defined as the energy stored in in elastic materials as the
result of their stretching or compressing.

 Springs are a special instance of device which can store
elastic potential energy due to its compression or
stretching.

 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”

where Fs  x Fs  kx

Fs : the restoring force of spring

k : the spring constant or force constant

x : the amount of stretch or compression (x f -xi ) 29

 Negative sign in the equation indicates that the direction of Fs is

always opposite to the direction of the amount of stretch or

compression (extension), x.

 Case 1:
The spring is hung vertically and its is stretched by a

suspended object with mass, m as shown in Figure 5.12.

The spring is in equilibrium

Figure 5.12 
Initial position Fs
Final position
x

thus Fs  W  mg 
W  mg

30

 Case 2:

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

compressed by a force, F as shown in Figure 5.13.

Fs is negative Fs 
x is positive F

The spring is in
x equilibrium, hence

x0 
Fs  F
Fs  0
x0

(Equilibrium position)

 x 0
F Fs Fs is positive
x is negative

x

Figure 5.13 31

 Caution:
For calculation, use : Fs  kx  F where F : applied force
◦ The unit of k is kg s2 or N m1

 From the Hooke’s law (without “ ve” sign), a restoring force, Fs
against extension of the spring, x graph is shown in Figure 5.14.

Fs W  area under the Fs  x graph
F

W  1 Fx1 W  1 kx1 x1
2 2

x W  1 kx12 Us
2
0 x1

Figure 5.14

32

 The equation of elastic potential energy, Us for compressing

or stretching a spring is

Us  1 kx2  1 Fs x
2 2

33

Example 5.6 :

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 :
Given F  800 N; x  0.200 m

From the Hooke’s law Fs  F  kx

a. Given x=0.300 m 800  k0.20 k  4 103 N m1

1  Us1 0.3002
Us  2 kx 2  2 4 103

U s  180 J 34

Solution :

b. Given m=60 kg. When the spring in

equilibrium, thus

 Fnett  0
Fs Fs  mg
kx  mg
 x 4 103 x  609.81

x  0.147 m

Therefore Us  1 kx2
2
  
mg 1 0.1472
W 2
 Us  4 103

U s  43.2 J

35

Kinetic energy, K

 is defined as the energy of a body due to its motion.

 Equation : where

K  1 mv2 K : kinetic energy of a body
2 m : mass of a body

v :speed of a body

36

Work-energy theorem

 Consider a block with mass, m moving along the horizontal

surface (frictionless) under the action of a constant nett

force, Fnett undergoes a displacement, s in Figure 5.15.



mFnett

Figure 5.15 
s

 F  Fnett  ma (1)

By using an equation of linear motion: v2  u2  2as

a  v2 u2 (2)
2s
37

By substituting equation (2) into (1), we arrive

m v 2 u 2 
2s
Fnett 

Fnett s  1 mv2  1 mu2  Kf  Ki
2 2

Therefore Wnett  K

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

38

Example 5.7 :

A stationary object of mass 3.0 kg is pulled upwards by a
constant force of magnitude 50 N. Determine the speed of
the object when it is travelled upwards through 4.0 m.

Solution : m  3.0 kg ; F  50 N; s  4.0 m; u  0

 The nett force acting on the object is given by
F
Fnett  F  mg  50  3.09.81

mg Fnett  20.6 N

sF By applying the work-energy theorem, thus

 Wnett  K f  Ki
mg 1
Fnett s  2 mv 2  0

20.64.0  1 3.0v2 v  7.41 m s1
2
39

Example 5.8 :

A block of mass 2.00 kg slides 0.750 m down an inclined
plane that slopes downward at an angle of 36.9 below the
horizontal. If the block starts from rest, calculate its final
speed. You can ignore the friction.

m  2.00 kg ; s  0.750 m; u  0


N

a

mg sin 36.9 y

mg cos36.9  x
s
36.9 
mg

36.9 40

Solution : m  2.00 kg ; s  0.750 m; u  0

Since the motion of the block along the incline surface thus nett

force is given by

Fnett  mg sin 36.9

Fnett  2.009.81sin 36.9

Fnett  11.8 N

By using the work-energy theorem, thus

Wnett  K f  Ki
1
Fnett s  2 mv2  0

11.80.750  1 2.00v2

2

v  2.98 m s1

41

Example 5.9 :

F (N)

10

0 4 67 10 s(m)

5

Figure 5.16

An object of mass 2.0 kg moves along the x-axis and is acted on

by a force F. Figure 5.16 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.

42

Solution : m  2.0 kg; u  10 m s1

a. W  area under the F  s graph from 0 m to10 m

W  1 6  410  1 10  6 10  7 5

22
W  32.5 J

By using the work-energy theorem, thus

W  K f  Ki
W  1 mv2  1 mu2

22

32.5  1 2.0v2  1 2.0102

22
v  11.5 m s1

43

Solution :

b. W  area under the F  s graph from 0 m to 6 m

W  1 6  410

2
W  50 J

By using the work-kinetic energy theorem, thus

W  K f  Ki

W  Kf  1 mu 2
2

50  Kf  1 2.0102

2

K f  150 J

44

Exercise :

1. 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

45

Principle of conservation of energy

 states “in an isolated (closed) system, the total energy of that
system is constant”.

 According to the principle of conservation of energy, we get

 Ei   E f

The initial of total energy = the final of total energy

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 Ki Ui  K f U f

46

Example 5.10 : 30 cm

A 1.5 kg sphere is dropped from a height of x
30 cm onto a spring of spring constant,
Before After
k = 2000 N m1 . After the block hits the
Figure 5.17
spring, the spring experiences maximum

compression, x as shown in Figure 5.17.

a. Describe the energy conversion
occurred after the sphere is
dropped onto the spring until the
spring experiences maximum

compression, x.

b. Calculate the speed of the sphere just
before strikes the spring.

c. Determine the maximum compression, x.

47

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 kx2
1 2 48
E2  mgh1  2 mv 2 E3  mgh2 

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 mv2
2
mg h0 mgh0  mgh1 h0  h1 
 1 mv 2 and h 
 h1  2

v  2gh

v  29.810.30

v  2.43 m s1

49

Solution : m 1.5 kg; h  0.30 m; k  2000 N m1

c. Applying the principle of conservation of energy involving the
situation (2) and (3),

 E2  E3
1 1
mgh1  2 mv 2  mgh2  2 kx 2

 mgh1 1 1
 h2 2 mv 2  2 kx 2 and x  h1  h2

1.59.81x  1 1.52.432  1 2000x2

22

1000x2 14.7x  4.43  0

x  7.43102 m

50