Chapter III

Work and Energy

October 5, 2020
Chapter III Work and Energy

III.1. Work
III.1.1. Work done by a constant force
III.1.2. Work done by a forces system
III.1.3. Graphical determination of work
III.1.4. Work Done by a Variable Force
III.1.5. Work Done By a Spring
III.1.6. Work done by Gravitational Force

III.2. Power and efficiency

III.3. Kinetic Energy
III.4. Work-Kinetic Energy theorem
III.5. Applications of work-energy theory
October 5, 2020
 Introduction
• Previously, problems dealing with the motion of particles
were solved through the Newton laws of motion,
Current chapter introduces other method of analysis.

• Method of work and energy: directly relates force,

mass, velocity and displacement.

• Method of impulse and momentum: directly

relates force, mass, velocity, and time.(Chapter 4)

October 5, 2020
Energy

Why Energy?
• Why do we need a concept of energy?
• The energy approach to describing motion is
particularly useful when Newton’s Laws are
difficult or impossible to use
• Energy is a scalar quantity. It does not have a
direction associated with it

October 5, 2020
Energy

What is Energy?
• Energy is a property of the state of a system, not a
property of individual objects: we have to
broaden our view.
• Some forms of energy:
– Mechanical:
• Kinetic energy (associated with motion, within system)
• Potential energy (associated with position, within system)
– Chemical
– Electromagnetic
– Nuclear
• Energy is conserved. It can be transferred from
one object to another or change in form, but
cannot be created or destroyed
October 5, 2020
 III.1. Work
III.1.1. Work of a Constant Force in Rectilinear Motion

The work W done on a system by an agent exerting a constant force on

the system is the product of the magnitude of the force, the magnitude
of the displacement of the point of application of the force, and Cosθ,
where  is the angle between the force and displacement vectors:

W  F .S .cos W  F  r

Unit: Joule (J)

October 5, 2020
III.1.1. Work of a Constant Force in Rectilinear Motion

Work of a Constant Force in Rectilinear Motion

W12  F (cos  )x  F(cos  )( x2  x1 )


F
The directed effect on the motion along the path
2
W>0: F support motion 
W>0: F obstruct motion 1 x
W=0: F indirected effect to motion

October 5, 2020
III.1.1. Work of a Constant Force in Rectilinear Motion

Ex III.1 Mr. Clean

A man cleaning a floor pulls a vacuum cleaner with a force of
magnitude F=50.0 N at an angle of 30.0° with the horizontal (Fig. 3.4).
Calculate the work done by the force on the vacuum cleaner as the
vacuum cleaner is displaced 3.00 m to the right.

W  F r cos  50.0 N  3.00m  cos 30.0o  130J

Figure 3.4

October 5, 2020
III.1.1. Work of a Constant Force in Rectilinear Motion

Ex III.2. When Work is Zero

W  ( F cos  ) x

• A man carries a bucket of water

horizontally at constant velocity.
• The force does no work on the
bucket
• Displacement is horizontal
• Force is vertical
• cos 90° = 0

October 5, 2020
III.1.1. Work of a Constant Force in Rectilinear Motion

Forces which do not do work (ds = 0 or cos a = 0):

• reaction at frictionless pin supporting rotating body,

• reaction at frictionless surface when body in

contact moves along surface,

• reaction at a roller moving along its track

• weight of a body when its center of gravity moves

horizontally.

October 5, 2020
III.1.1. Work of a Constant Force in Rectilinear Motion

Ex III.3. Work Can Be Positive or Negative

• Work is positive when lifting the box

• Work would be negative if lowering the

box
– The force would still be upward, but
the displacement would be downward

October 5, 2020
III.1.1. Work of a Constant Force in Rectilinear Motion

Ex III.4. Work Can Be Positive or Negative

October 5, 2020
III.1.1. Work of a Constant Force in Rectilinear Motion

Note 1: Work is a scalar quantity

Note 2: the expression for work we have

developed apply when F is constant.
Note 3: We have made the implicit assumption
that the moving object is point-like

III.1.2. Work done by a forces

system
Method 1: First calculate the work done by each force: WA by force FA ,
WB by force FB , and WC by force FC . Then determine Wnet  WA  WB  WC

Method 2: Calculate first Fnet  FA  FB  FC ; Then determine Wnet  F  d

October 5, 2020
III.1.2. Work done by a forces system

N(20N)

W F 
i   F  .S  8.3  24( J )
i F(10N)

A B
Fms(2N)
P(20N)

W  Wi   ( Fi .S )  10.3  (2.3)  24( J ) AB=3m

W F 
i   F  .S  W   ( F .S )
i i i

October 5, 2020
 III.1.3. Graphical determination of work
F (N)

20

5 15 x (m)
W515  20.(15  5)  40 J F (N)

20

5 15 x (m)
The work is the area under the curve.

October 5, 2020
 III.1.4. Work Done by a Variable Force

If the force takes on several successive constant values:

October 5, 2020
III.1.4. Work Done by a Variable Force

We can then approximate a continuously varying force by a

succession of constant values.

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III.1.4. Work Done by a Variable Force

Ft

0 s1 s2 s

The work is the area under the curve.

October 5, 2020
III.1.4. Work Done by a Variable Force

* To get the total work done along a path

xf

lim  Fx x   Fx dx
xf
dr
x0
xi
xi ds

* Notice that

F
2 2
W12   F (cos  )ds   Ft ds
1 1

* General
2 2
W12   F  dr W12   ( Fx dx  Fy dy  Fz dz )
1 1
October 5, 2020
III.1.4. Work Done by a Variable Force

EXAMPLE III.5. Calculating total work done from a

graph
A force acting on a particle varies with x as shown in Figure.
Calculate the work done by the force on the particle as it
moves from x=0 to x=6.0 m.

WAB  5.0 N  4.0m=20J

1
WBC  5.0 N  2.0m=5.0J Figure 3.6
2
WAC  WAB  WBC  20J+5.0J=25J

October 5, 2020
October 5, 2020
 III.1.6. Work done by Gravitational Force
* In limited space near the ground

Ag  A  B  mgd cos180  mgAB B

Ag  B  A   mgd cos0  mgAB

A
Ag  A  B   mg ( yA  yB )
 III.1.5. Work Done By a Spring

• Spring force
Fx  kx

October 5, 2020
III.1.5. Work Done By a Spring

Spring at Equilibrium

• F=0

October 5, 2020
III.1.5. Work Done By a Spring

Spring Compressed

October 5, 2020
III.1.5. Work Done By a Spring

Spring undeformed

F  kx
x
 1
F
2
x
W12    kxdx ( 21 kx22  21 kx12 )
x2
1

F   12 k ( x2  x1 )( x2  x1 )
x
x1 x2 x

F1

F2
F  kx
III.1.5. Work Done By a Spring
Measuring k for a spring
The spring is hung vertically and an object of mass m is attached to its
lower end. Under the action of the “load” mg, the spring stretches a
distance d from its equilibrium position.
a. If a spring is stretched 2.0 cm by a suspended object having a mass of
0.55 kg, what is the force constant of the spring?
b. How much work is done by the spring on the object as it stretches
through this distance?

mg 0.55kg  9.80m/s2
k  2
 2.7  10 2
N/m
d 2.0  10 m
1 2 1
 
2
Ws  0  kd    2.7 10 N/m  2.0 10 m  5.4 102 J
2 2
2 2
* Work done by gravitational force

Wg  F  r  mg  d cos  mgd
 0.55 kg  9.80m/s 2  2.0 102 m=1.1 101 J
?????
 III.2. Power and efficiency

Power is the rate at which work is done.

W
Pavg 
t
dW
Instanstaneous Power P
dt

Unit of P : Watt. 1W = 1 J/1s

The kilowatt-hour The kilowatt-hour (kWh) is a unit of work. It is defined
as the work performed by an engine of power P = 1000 W in a time t = 1 hour
W  Pt  1000  3600  3.60 106 J The kWh is used by electrical utility
companies (check your latest electric bill)
III.2. Power and efficiency

dW F cos  dx dx
P   F cos   Fv cos 
dt dt dt

P  Fv cos   F  v

v
III.2. Power and efficiency

Efficiency

work out power out

 
work in power in
 III.3. Kinetic Energy

dv dv dx dv
F  ma  m  m  mv
dt dx dt dx

f f f
dv
Wi  f   Fdx   mv dx   mvdv
i i
dx i

Wi f  mv  mv
1
2
2
f
1
2
2
i
III.3. Kinetic Energy

Wi f  mv  mv
1
2
2
f
1
2
2
i

mv 2
K
2

We define a new physical parameter to describe the state of motion

of an object of mass m and speed v - kinetic energy K

• Units of work and kinetic energy are the same:

2
m  m
K  mv  kg     kg 2  m  N  m  J
1
2
2

 s   s 

Wi  f  K f  K i K i  Wi  f  K f
 III.4. Work – kinetic energy theorem

Wnet  K f  Ki  K K f  K i  Wnet
“The work done on a particle by a “The final kinetic energy of an
net force acting on it equals the object is equal to its initial
change in kinetic energy of the kinetic energy plus the change
particle” due to the net work done on it”

If a force F is applied to an object of mass m it can accelerate it and

increase its speed v and kinetic energy K. Similarly F can decelerate m
and decrease its kinetic energy.
We account for these changes in K by saying that F has transferred energy
W to or from the object. If energy it transferred to m (its K increases) we
say that work was done by F on the object (W > 0). If on the other hand.
If on the other hand energy its transferred from the object (its K decreases)
we say that work was done by m (W < 0)
III.4. Work-Kinetic Energy Theorem

m m K  K f  Ki  Wnet

Change in the kinetic  net work done on 

energy of a pareticle    the particle 
   

The work-kinetic energy theorem holds for both positive and negative values of Wnet
If Wnet  0  K f  Ki  0  K f  Ki
If Wnet  0  K f  Ki  0  K f  Ki
 III.5. Applycations of work-energy theory
III.5.1. Work the pendulum problem in the text.
• Wish to determine velocity of
pendulum bob at A2. Consider work
& kinetic energy.
K1  W12  K 2
1 mg 2
0  mgl  v2
2 g
v2  2 gl
• Velocity found without determining expression for acceleration
and integrating.

• All quantities are scalars and can be added directly.

• Forces which do no work are eliminated from the problem.

III.5.1. Work the pendulum problem in the text.

• Principle of work and energy cannot be

applied to directly determine the
acceleration of the pendulum bob.
• Calculating the tension in the cord
requires supplementing the method of
work and energy with an application of
Newton’s second law.
• As the bob passes through A2 ,
F n  m an
2
mg v
P  mg  2
g l
v2  2 gl mg 2 gl
P  mg   3mg
g l
Advantage of work and energy methods
• Velocity found without determining
expression for acceleration and integrating.

• All quantities are scalars and can be added

directly.

• Forces which do no work are eliminated

from the problem.

37
 III.5.2. A car is moving on the incline

SOLUTION:
• Evaluate the change in kinetic
energy.

• Determine the distance required

for the work to equal the kinetic
energy change.
An automobile weighing 4000 N
is driven down a 5o incline at a
speed of 88 m /s when the brakes
are applied causing a constant
total breaking force of 1500 N.
Determine the distance traveled
by the automobile as it comes to
a stop. 38
III.5.2. A car is moving on the incline

• Evaluate the change in kinetic

energy.

v1  88 m s
 4000 / 9.8188
2
K1  12 mv12  1
2  1578797 N  m
v2  0 K2  0

• Determine the distance required for the

work to equal the kinetic energy change.
W12   1500 N  x   4000 N  sin 5  x
  1151N  x
K1  W12  K 2
1578797 N  m  1151N  x  0
x  1371.67 m 39
 III.5.2. A car is moving on the incline

III.5.3. Example 3
SOLUTION:
• Apply the principle of work and
energy separately to blocks A and
B.

• When the two relations are

combined, the work of the cable
Two blocks are joined by an inextensible forces cancel. Solve for the velocity.
cable as shown. If the system is released
from rest, determine the velocity of block
A after it has moved 2 m. Assume that the
coefficient of friction between block A and
the plane is mk = 0.25 and that the pulley
is weightless and frictionless.
III.5.3. Example 3

SOLUTION:
• Apply the principle of work and energy
separately to blocks A and B.
 
WA   200 kg  9.81m s 2  1962 N
FA  k N A  kWA  0.25 1962 N   490 N
K1  W12  K 2 :
0  FC  2 m   FA  2 m   12 mAv 2
FC  2 m    490 N  2 m   1
2  200 kg  v 2

 
WB   300 kg  9.81m s 2  2940 N
K1  W12  K 2 :
0  Fc  2 m   WB  2 m   12 mB v 2
 Fc  2 m    2940 N  2 m   1
2  300 kg  v 2
III.5.3. Example 3

• When the two relations are combined, the work of

the cable forces cancel. Solve for the velocity.
FC 2 m   490 N 2 m   12 200 kg v 2

 Fc 2 m   2940 N 2 m   12 300 kg v 2

2940 N 2 m   490 N 2 m   12 200 kg  300 kg v 2

4900 J  12 500 kg v 2

v  4.43 m s
 III.5.4. Example 4
SOLUTION:
• Apply the principle of work and energy
between the initial position and the
point at which the spring is fully
compressed and the velocity is zero.
A spring is used to stop a 60 kg package The only unknown in the relation is the
which is sliding on a horizontal surface. friction coefficient.
The spring has a constant k = 20 kN/m
and is held by cables so that it is initially • Apply the principle of work and energy
compressed 120 mm. The package has a for the rebound of the package. The
velocity of 2.5 m/s in the position shown only unknown in the relation is the
and the maximum deflection of the spring velocity at the final position.
is 40 mm.
Determine (a) the coefficient of kinetic
friction between the package and surface
and (b) the velocity of the package as it
passes again through the position shown.
43
III.5.4. Example 4

• Apply principle of work and energy between

initial position and the point at which spring is
fully compressed.
 60 kg  2.5 m s 
2
K1  12 mv12  1
2  187.5 J K2  0

W12  f   k mg x

 
  k  60 kg  9.81m s 2  0.640 m     377 J  k

Pmin  kx0   20 kN m  0.120 m   2400 N

Pmax  k  x0  x    20 kN m  0.160 m   3200 N
W12 e   12  Pmin  Pmax  x
  12  2400 N  3200 N  0.040 m   112.0 J

W12  W12  f  W12 e    377 J  k  112 J

K1  W12  K 2 :
k  0.20
187.5J-  377 J  k  112 J  0
III.5.4. Example 4

• Apply the principle of work and energy for the

rebound of the package.
K2  0 K 3  12 mv32  12  60kg  v32

W23  W23  f  W23 e    377 J  k  112 J

 36.5 J
K2  W23  K3 :
0  36.5J  12  60 kg  v32 v3  1.103 m s

45
 III.5.5. Example 5
SOLUTION:
• Apply principle of work and energy to
determine velocity at point 2.
• Apply Newton’s second law to find
normal force by the track at point 2.
A 2000 N car starts from rest at
point 1 and moves without friction • Apply principle of work and energy to
down the track shown. determine velocity at point 3.
• Apply Newton’s second law to find
Determine:
minimum radius of curvature at point
a) the force exerted by the track 3 such that a positive normal force is
on the car at point 2, and exerted by the track.
b) the minimum safe value of the
radius of curvature at point 3.
III.5.5. Example 5

SOLUTION:
• Apply principle of work and energy to
determine velocity at point 2.
1 mg 2
K1  0 K 2  12 mv22  v2
2 g
W12   mg  40 m 
1 mg 2
K1  W12  K 2 : 0  mg  40 m   v2
2 g

v22  2  40 m  g  2  40 m  9.81m s 2  v2  28.01m s

• Apply Newton’s second law to find normal

force by the track at point 2.
mg v22 mg 2  40 m  g
W  N  m an  
g 2 g 20 m
N  10000N N  5 mg
III.5.5. Example 5

• Apply principle of work and energy to

determine velocity at point 3.
1 mg 2
K1  W13  K 3 0  mg  25 m   v3
2 g

v32  2  25 m  g  2  25 m  9.81m s 2 
v3  22.15 m s
• Apply Newton’s second law to find minimum
radius of curvature at point 3 such that a
positive normal force is exerted by the track.
mg  m an
mg v32 mg 2  25 m  g
  3  50 m
g 3 g 3
 Sample Problem

SOLUTION:
Force exerted by the motor
cable has same direction as
the dumbwaiter velocity.
Power delivered by motor is
D
equal to FvD, vD = 8 m/s.

The dumbwaiter D and its load have a • In the first case, bodies are in
combined weight of 600 N, while the uniform motion. Determine force
counterweight C weighs 800 N. exerted by motor cable from
conditions for static equilibrium.
Determine the power delivered by the
electric motor M when the dumbwaiter • In the second case, both bodies are
(a) is moving up at a constant speed of accelerating. Apply Newton’s
8 m/s and (b) has an instantaneous second law to each body to
velocity of 8 m/s and an acceleration of determine the required motor
2.5 m/s2, both directed upwards. cable force.
 Sample Problem
• In the first case, bodies are in uniform motion (a = 0).
Determine force exerted by motor cable from
conditions for static equilibrium.

Free-body C:
   Fy  0 : 2T  800 N  0 T  400 N

Free-body D:
   Fy  0 : F  T  600 N  0
F  600 N  T  600 N  400 N  200 N

Power  Fv D  200 N 8 m s 

 1600 N  m s

Power  1600 N  m s 
 Sample Problem
• In the second case, both bodies are accelerating. Apply
Newton’s second law to each body to determine the
required motor cable force.
a D  2.5 m s 2  aC   12 a D  1.25 m s 2 

Free-body
800
C:
  Fy  mC aC : 800  2T  1.25 T  100 N
10
Free-body
600
D:
  Fy  m D a D : F  T  600  2.5
10
F  100  600  150 F  650 N
Power  Fv D  650 N 8 m s   5200 N  m s

Power  5200 N  m s 