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Chapter 6:
Conservation of Energy
Introduction
 Energy

– ability to perform work.
Unit of energy (and of work) – Joules (J)
1 J = 1 N- m = 1 kg- m2 / s2
Forms of energy:
Light, chemical, nuclear, mechanical,
electrical, sound, heat, kinetic, elastic,
magnetic etc.
Each type of energy is calculated
differently.
Law of Conservation of Energy

In any natural process, total energy is always
“conserved”, i.e. energy can not be created
nor destroyed.
Can be transformed from one form to
another.
Can be transferred from one system to
another.
In science, any law of conservation is a very
powerful tool in understanding the physical
universe.
Work by Constant Force
 Work
– transfer of energy as a result of
force.

Constant force – its magnitude and
direction unchanged.

The force acts on the object throughout
the process.
 Only
component of force parallel to
direction of motion performs work!
Work by Constant Force
Work done by force F in moving the object
at constant speed through displacement
Dx :
W = (F cos q) Dx
F cos q = component of F parallel to Dx.
Work is a scalar quantity.
Unit = Joule (J).
f
1 J = 1 N-m
F
N
q
W
Dx
Work by Constant Force
W = (F cos q) Dx
1. If Dx = 0, then W = 0
If you held a 50 kg bag on your head
without moving for 3 hours, would you
have done work?
W = (F cos q) Dx
2. If q = 0, cos 0 = 1 ie F and Dx are parallel
and W = F Dx
F
Dx
W = (F cos q) Dx
3. If q = 90o, cos 90o = 0, ie F and Dx are
perpendicular to each other
and W = 0
f
F
N
q
Dx
W
Work done by the normal force N is zero
W = (F cos q) Dx
4. If q = 180o, cos 180o = -1, ie F and Dx are
antiperpendicular to each other
and W = negative.
f
F
N
q
Dx
W
Work done by the frictional force f is - F Dx
Example:
A 30 N chest was pulled 5 m across the floor
at a constant speed by applying a force of
50 N at an angle of 30o. How much work
was done by tension T?
W = F Dx cos q
= (50 N) (5 m) cos (30)
= 217 Joules
50 N
30o
N
f
mg
T
Example:
A 30 N chest was pulled 5 m across the floor at
a constant speed by applying a force of 50 N at
an angle of 30o. How much work was done by
gravity?
T
N
50 N
f
30o
f
mg
mg
W = mg Dx cos q
= 30 x 5 cos(90) = 0
Dx
mg
90o
Example:
A 30 N chest was pulled 5 m across the floor at
a constant speed by applying a force of 50 N at
an angle of 30o. How much work was done by
50 N
friction f?
f
mg
To find the magnitude of f:
Consider x-component of forces acting.
Since constant velocity: Fnet = 0
So Tcos30o – f = 0
Thus f = 43.3 N.
W = 43.3 x (cos180o) x 5 = -217 J
T
N
f
mg
f
Dx
180o
Work by Variable Force
W = Fx Dx
Work = area under F vs x
plot
Force
F
Dx
x
Force
Spring F = k x
Area = ½ k (x)2 = Wspring
x
Example
A dart gun with a spring constant
k = 400.0 N/m is compressed
8.0 cm. How much energy will it
transfer to a dart when the
spring is released?
W = ½ k(Dx)2
Example
As you ride in an elevator going upward at
constant speed, work done on you by
the normal force from the floor of the
elevator
(A)is negative.
(B) is zero.
(C) is positive.
(D) is zero, negative or positive depending on the
speed of the elevator.
Kinetic Energy
Suppose a constant force F is applied
horizontally on a small rigid object of mass m.
Its velocity will change, say from vo to v as it
moves through a distance Dx (without rotation).
Work done on the object
W = Fx Dx [F parallel to Dx]
= m ax Dx [Recall: v2 = vo2 + 2a(x-xo)]
= ½ m (v2 – v02) = ½ mv2 - ½ mv02
The quantity ½ . mass . (velocity)2 is called
kinetic energy of the object.
An object of mass m, moving with
velocity v, has kinetic energy
K = ½ mv2
K  m: If m is doubled, K will double.
K  v2 If v is doubled, K will
quadruple.
Can K ever be zero?
Can K ever be negative?
W = DK
Can W ever be negative?
Kinetic Energy
F
vo
v
Dx
If more than one force acts on the object,
the change in kinetic energy is the total
work done by all the forces acting on
the object.
Total work Wtotal = DK
This is called the Work- Energy Principle
Total Work
Total work Wtotal = DK = change in K.E.
Wtotal = First find work done by each force ,
then add them.
= first add all the forces and then
then calculate work done by the net force
Q: What is the total work done on an
object moving with constant velocity?
(A) Positive (B) Negative (C) Zero
Example 1:
A 30 N chest was pulled 5 m across the floor
at a constant speed by applying a force of
50 N at an angle of 30o. How much work
was done by all the forces acting on it?
WT = F Dx cos q = (50 N)(5 m) cos (30) = 217 J
WN = 0, Wmg = 0 Wf = -217J
Wtotal = 217 + 0 +0 +(-217) = 0
T = 50 N
30o
N
f
mg
T
Example 2:
A 30 N chest was pulled 5 m across the
floor by applying a force of 50 N at an
angle of 30o. How much work was
done by all the forces acting on it if its
speed changed from 2 m/s to 5 m/s in
the process?
Example
FN
You are towing a car up a hill with constant
velocity. The work done on the car by the
normal force is:
1. positive
2. negative
correct
3. zero
• Work done by
gravity?
• Work done by
tension?
V
T
W
Since the direction of the force is
positive the value of work will be
positive.
it's negative because it's trying to
slow down the car. The normal
force does no work because it
acts in a direction perpendicular
to the displacement of the car.
Example: Block w/ friction

A block is sliding on a surface with an initial speed of 5
m/s. If the coefficient of kinetic friction between the block
and table is 0.4, how far does the block travel before
y
stopping?
N
Y direction: F=ma
N-mg = 0
N = mg
Work
WN = 0
Wmg = 0
Wf = f Dx cos(180)
= -mmg Dx
5 m/s
f
x
mg
W=DK
-mmg Dx = ½ m (vf2 – v02)
-mg Dx = ½ (0 – v02)
mg Dx = ½ v02
Dx = ½ v02 / mg
= 3.1 meters
POTENTIAL ENERGY
Potential Energy (U) = stored energy.
It is the energy an object possess due
to its position or its configuration.
There are different types of potential
energy:
Gravitational potential energy.
Elastic potential energy.
Electrical potential energy.
Gravitational Potential Energy
To move mass m from initial point to final
point at constant velocity, an external
force FEXT, equal but opposite to the
force of gravity must be applied.
WEXT = F cosq Dy = mgh
Yf = h
= (mg)(cos0)h
= mgh
Wg = mgcos(180o)(h)
= -mgh
FEXT
mg
Yo = 0
Gravitational Potential Energy
 The
quantity mgh is called gravitational
potential energy near the earth’s surface:
U = mgh
where U = 0 wherever we have chosen h = 0
• More general way of
writing gravitational
potential energy is
U = mg h
Gm1m2
U (r ) = r
FEXT
U = 0 at r = 
mg
U=0
The gravitational potential energy of an
object of mass m at a height h is
Ug = mgh
Ug = mgh and Wg = -mgh ie Ug = - Wg.
2. Ug  m
3. Ug  h
4. A reference level where h = 0 can be
chosen to be at any convenient location.
5. Only change in Ug is important.
DUg = mgDh = mg(hf – hi)
1.
Gravitational Potential Energy
Example 3: A block slid down a plane
If a block is slid down a plane.
Distance moved down the plane = Ds
Vertical height moved = h
How much work will the force of gravity
do?
W = F cosq Dx
Wg = mg . cosq . DS
But DS = Dy /cosq
h
q
mg
Wg = (mg)(cosq)(h/cosq) = mgh
Elastic Potential Energy
Elastic (Spring) force F = -kx
Work done by spring force W = ½ kx2
Energy stored in the spring when
compressed or stretched by distance x
Elastic Potential
Energy U = ½ kx2
Example
A dart gun with a spring constant
k = 400.0 N/m is compressed
8.0 cm.
(a) How much energy will it
transfer to a dart when the
spring is released?
(b) With what speed will the dart
(mass 200 g) leave the gun?
Example
Emil throws an orange straight up and
catches it at the same point it was thrown.
(a) How much work is done during the
orange’s free fall?
(b) If the orange was thrown upward from a
1.2 m height above the ground and falls to
the ground, how much work is done by
gravity? [Mass of orange = .25 kg]
Conservative and Non Conservative Forces
Sliding down a plane
h
Dropped
q
mg
Thrown
Work done by force of gravity, Wg = - mgh
Independent of path
Gravitational Potential Energy
To move mass m from initial point to final
point at constant velocity, an external
force FEXT, equal but opposite to the
force of gravity must be applied.
WEXT = F cosq Dy = mgh
Yf = h
= (mg)(cos0)h
= mgh
Wg = mgcos(180o)(h)
= -mgh
FEXT
mg
Yo = 0
1. As you ride an elevator going upward at
constant speed, the work done on you by the
normal force from the floor of the elevator is
A.
B.
C.
D.
93%
negative
zero
positive
Negative, zero or positive depending on
the speed of the elevator
4%
A.
2%
B.
0%
C.
34
D.
2. What is the total work done on an object
moving with constant velocity?
93%
A. Positive
B. Negative
C. Zero
7%
A.
0%
B.
C. 35
3. A passenger whose mass is 95 kg is seated
in a jet airliner flying 1,200 m above the
ground at a speed of 253 m/s. What is the
kinetic energy of the passenger?
89%
A.
B.
C.
D.
931 J
6.08 x 106 J
3.04 x 106 J
1.12 x 106 J
0%
A.
7%
4%
B.
C.
36
D.
4. A passenger whose mass is 95 kg is seated
in a jet airliner flying 1,200 m above the
ground at a speed of 253 m/s. What is the
potential energy of the passenger?
93%
A.
B.
C.
D.
931 J
6.08 x 106 J
3.04 x 106 J
1.12 x 106 J
2%
A.
2%
2%
B.
C.
37
D.
5. A spring has an elastic constant of 920 N/m.
By how much should it be stretched in order to
store 120 J of energy?
77%
A.
B.
C.
D.
E.
0.36 m
0.51 m
0.26 m
7.7 m
1.96 m
9%
7%
5%
2%
A.
B.
C.
D.
38
E.
Conservative Forces
Forces which perform same amount of work
independent of the path taken are called
conservative forces. Eg: gravity, elastic force,
electric force.
1. Work done by a conservative force depends
only on the initial and final position and not on
the path taken to get there.
2. Potential energy only goes with conservative
forces. There is potential energy associated
with gravity, elastic force, electric force.
3. WC = - DU
[ Ug = mgh and Uspring = ½ kx2]
Non Conservative Forces
1. Amount of work done by them
depend on the path taken. Eg:
friction, tension, push/pull.
2. There is no potential energy
associated with non conservative
forces.
Work - Energy w/ Conservative Forces
Total work = Work done by all types of forces.
= Work by conservative forces + work
by non-conservative forces
WTotal = WC + WNC
But WTotal = DK and WC = - DU
So, DK = - DU + WNC ie, WNC = DK + DU
WNC = DK + DU
Work - Energy w/ Conservative Forces
WNC = DK + DU
When only conservative forces are acting; WNC = 0
0 = DK + DU
OR
0 = (Kf – Ki) + (Uf – Ui)
OR Kf + Uf = Ki + Ui
The quantity K + U is called mechanical energy E
Thus Ef = Ei (Conservation of mechanical energy)
Skiing Example (no friction)
A skier goes down a 78 meter high hill with a
variety of slopes. What is the maximum speed she
can obtain if she starts from rest at the top?
[Assume
friction
is
negligible]
Conservation of energy:
Ki + Ui = Kf + Uf
½ m vi2 + m g yi = ½ m vf2 + m g yf
0 + g y i = ½ v f2 + g y f
vf2 = 2 g (yf-yi)
vf = sqrt( 2 g (yf-yi))
vf = sqrt( 2 x 9.8 x 78) = 39 m/s
Two similar objects are released from rest at the same
time to slide down two frictionless slopes A and B of
different inclines as shown in the figure below. Which
of these statements is true about the motion of the
two balls? They will reach the bottom with the same
(A)Acceleration (B) Speed (C) time duration
(D) Same acceleration, speed, time duration
A
B
h
30o
h
Example
Suppose the initial kinetic and potential energies
of a system are 75J and 250J respectively, and
that the final kinetic and potential energies of the
same system are 300J and -25J respectively.
How much work was done on the system by
non-conservative forces?
1. 0J
2. 50J
Initial = 75 J + 250 J = 325 J.
3. -50J
Final = 300 J - 25 J = 275 J.
4. 225J
Final -Initial = 275 - 325 = -50 J.
5. -225J
No energy goes into or comes out of
the system overall, so some nonconservative force has to be
accounted for.
Example
A dart gun with a spring constant k =
400.0 N/m is compressed 8.0 cm.
(a) How much energy will it transfer to
a dart of mass 200 g when the spring
is released?
(b) What would be the escape speed of
the dart?
(c) If the dart was aimed upward, how
high would it travel?
Uspring = ½ k(Dx)2 K = ½ mv2, Ug = mgh
Power
Average Power P = rate at which
work is being done.
= rate at which energy is being
transformed.
P = DW / Dt Units: J/s = Watt
DW/ Dt = [F Dx cos(q)]/ Dt
= F (v Dt) cos(q)
i.e., P = F v cos(q)
Example:
How much electrical energy will a
75-watt light bulb use if left lit for
5 hours? [1 hour = 3,600 s]
Example:
A hot plate used 225,000 J of
electrical energy in 5 minutes.
What is the power (wattage) of the
hot plate?