Work Power & Energy-SPP.pptx

WORK-POWER &
ENERGY
Lecture21
CL-101 ENGINEERING MECHANICS
B. Tech Semester-I
Prof. Samirsinh P Parmar
Mail: samirddu@gmail.com
Asst. Professor, Department of Civil Engineering,
Faculty of Technology,
Dharmsinh Desai University, Nadiad-387001
Gujarat, INDIA
SPP-CL, DoCL, DDU- Nadiad, Gujarat, Bharatvarsh 1

Content of the presentation
• Forms of energy
• Energy – introduction
• Work- introduction
• Work done
• Zero work done
• Work done at constant force
• Kinetic energy
• Net work
• Potential energy
• Energy in springs
• Spring constant
• Work and Energy correlations
• Solved Problems
• Examples, Numericals

Forms of Energy
When work is done on an object the amount of energy the object has as well as the
types of energy it possesses could change. Here are some types of energy you
should know:
• Kinetic energy
• Rotational Kinetic Energy
• Gravitational Potential Energy
• Elastic Potential Energy
• Chemical Potential Energy
• Mass itself
• Electrical energy
• Light
• Sound
• Other waves
• Thermal energy

Energy
• Energy and Mechanical Energy
• Work
• Kinetic Energy
• Work and Kinetic Energy
• The Scalar Product of Two Vectors

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

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

Kinetic Energy
• Kinetic Energy is energy associated with the state of
motion of an object
• For an object moving with a speed of v
• SI unit: joule (J)
1 joule = 1 J = 1 kg m2/s2
2
2
1
mv
KE 

Kinetic Energy
Kinetic energy is the energy of motion. By definition, kinetic energy is
given by:
K = ½ mv2
The equation shows that . . .
. . . the more kinetic energy it’s got.
• the more mass a body has
• or the faster it’s moving
K is proportional to v2, so doubling the speed quadruples kinetic
energy, and tripling the speed makes it nine times greater.

Kinetic Energy Example
A 55 kg toy sailboat is cruising at
3 m/s. What is its kinetic
energy?
This is a simple plug and chug
problem:
K = 0.5 (55) (3)2 = 247.5 J
Note: Kinetic energy (along with
every other type of energy) is a
scalar, not a vector!

Energy Units
The formula for kinetic energy, K = ½ m v2, shows that its units are:
kg·(m/s)2 = kg·m2 / s2 = (kg·m / s2)m = N·m = J
So the SI unit for kinetic energy is the Joule, just as it is for work.
The Joule is the SI unit for all types of energy.
One common non-SI unit for energy is the calorie. 1 cal = 4.186 J.
A calorie is the amount of energy needed to raise the temperature of
1 gram of water 1 C.
A food calorie is really a kilocalorie. 1 Cal = 1000 cal = 4186 J.
Another common energy unit is the British thermal unit, BTU, which the energy
needed to raise a pound of water 1 F. 1 BTU = 1055 J.

Work - Energy Theorem:
The net work done on an object equals its change in
kinetic energy.
Wnet = Fnet x = max = mx(v/t)
= m(x/t)v = mv(vf -v0)
= m[½ (vf + v0)](vf -v0)
= ½ m(vf
2 -v0
2) = ½ mvf
2 -½ mv0
2
= Kf - K0 = K
Here’s a proof when Fnet is in line with the displacement, x. Recall
that for uniform acceleration, average speed = v = ½ (vf + v0 ).

Work – Energy- Example 1
Schmedrick takes his 1800 kg pet rhinoceros, Gertrude, ice skating on a frozen pond.
While Gertrude is coasting past Schmedrick at 4 m/s, Schmedrick grabs on to her tail
to hitch a ride. He holds on for 25 m. Because of friction between the ice and
Schmedrick, Gertrude is slowed down. The force of friction is 170 N. Ignore the
friction between Gertrude’s skates and the ice. How fast is she going when he lets go?
Friction, which does negative work here, is the net force, since weight and normal
force cancel out. So,Wnet= -(170N) (25 m) = -4250 J. By the work-energy theorem this
is the change in her kinetic energy, meaning she loses this much energy. Thus,
-4250 J = K = ½ mvf
2 - ½ mv0
2 = ½ m(vf
2 - v0
2)
= ½ (1800 kg) [vf
2 - (4 m/s)2]  vf = 3.358 m/s
You should redo this problem using the 2nd Law &
kinematics and show that the answer is the same.

Work-Energy - Example 2
A 62 pound upward force is applied to a 50 pound can of Spam. The
Spam was originally at rest. How fast is it going if the upward force
is applied for 20 feet?
Spam
62 lb
50 lb
Wnet = K
Fnet x = Kf - K0
(12 lb) (20 ft) = ½ mvf
2 - 0
240 ft · lb = ½ (mg)vf
2 /g
240 ft · lb = ½ (50 lb)vf
2 / (32.2 ft/s2)
vf
2 = 309.12 ft2 /s2 vf = 17.58 ft /s
multiply &
divide by g
mg is the
weight 9.8 m  32.2 ft
continued on next slide

Work-Energy Sample 2 check
vf
2 - v0
2 = 2ax vf
2 = v0
2 + 2ax = 2ax
= 2(Fnet / m)x = 2Fnet g /(mg) ·x
= 2 (12 lb)(32.2 ft /s2) / (50 lb) · (20 ft)
= 309.12 ft2 /s2
vf = 17.58 ft/s
Let’s check our work with “old fashioned” methods:
This is the same answer we got using energy methods.

Why ?
2
2
1
mv
KE 

Work
The simplest definition for the amount of work a force does on an object
is magnitude of the force times the distance over which it’s applied:
W = F x
This formula applies when:
• the force is constant
• the force is in the same direction as the displacement of the object
F
x

Work Example
Tofu Almond
Crunch
50 N
10 m
The work this applied force does is independent of the presence of any
other forces, such as friction. It’s also independent of the mass.
A 50 N horizontal force is applied to a 15 kg crate of granola bars over
a distance of 10 m. The amount of work this force does is
W = 50 N · 10 m = 500 N · m
The SI unit of work is the Newton · meter. There is a shortcut for
this unit called the Joule, J. 1 Joule = 1 Newton · meter, so we can say
that the this applied force did 500 J of work on the crate.

Negative Work
7 m
fk = 20 N
A force that acts opposite to the direction of motion of an object does negative work.
Suppose the crate of granola bars skids across the floor until friction brings it to a
stop. The displacement is to the right, but the force of friction is to the left.
Therefore, the amount of work friction does is -140 J.
Friction doesn’t always do negative work. When you walk, for example, the friction
force is in the same direction as your motion, so it does positive work in this case.
v
Tofu Almond
Crunch

When zero work is done
7 m
N
mg
As the crate slides horizontally, the normal force and weight do no work at all,
because they are perpendicular to the displacement. If the granola bar were moving
vertically, such as in an elevator, then they each force would be doing work.
Moving up in an elevator the normal force would do positive work, and the weight
would do negative work.
Another case when zero work is done is when the displacement is zero. Think about
a weight lifter holding a 200 lb barbell over her head. Even though the force applied
is 200 lb, and work was done in getting over her head, no work is done just holding
it over her head.
Tofu Almond
Crunch

Net Work
The net work done on an object is the sum of all the work done on it by the individual
forces acting on it. Work is a scalar, so we can simply add work up. The applied
force does +200 J of work; friction does -80 J of work; and the normal force and
weight do zero work.
So, Wnet = 200 J - 80 J + 0 + 0 = 120 J
FA = 50 N
4 m
fk = 20 N
N
mg
Note that (Fnet )(distance) = (30 N) (4 m) = 120 J.
Therefore, Wnet = Fnet x
Tofu Almond
Crunch

Tofu Almond
Crunch
When the force is at an angle
x
F

When a force acts in a direction that is not in line with the displacement, only
part of the force does work. The component of F that is parallel to the
displacement does work, but the perpendicular component of F does zero
work. So, a more general formula for work is
W = Fxcos
Fcos
Fsin This formula assumes
that F is constant.

Work: Incline Example


k
F
A box of tiddlywinks is being dragged across a ramp at a
toy store. The dragging force, F, is applied at an angle 
to the horizontal. The angle of inclination of the ramp is ,
and its length is d. The coefficient of kinetic friction
between the box and ramp is k. Find the net work done
on the tiddlywinks as they are dragged down the ramp.

Work: Incline Example (cont.)
First we break F into components  and // to the ramp. N is the
difference between F and mg cos.


k
F
F// = F cos ( + )

F = Fsin ( + )
mg cos
fk
N
Wnet = Fnet d = [F// + mgsin - fk]d
= [Fcos( +) + mgsin - k{mgcos - Fsin( + )}]d
Only forces // to the
ramp do any work.

Work: Circular Motion Example
A ‘69 Thunderbird is cruising around a
circular track. Since it’s turning a
centripetal force is required. What type
of force supplies this centripetal force?
answer:
friction
How much work does this force do?
None, since the centripetal
force is always  to the
car’s motion.
answer:
r
v

Work W
• Start with Work “W”
• Work provides a link between force and energy
• Work done on an object is transferred to/from it
• If W > 0, energy added: “transferred to the object”
• If W < 0, energy taken away: “transferred from the object”
x
F
mv
mv x


2
0
2
2
1
2
1

Definition of Work W
• The work, W, done by a constant force on an
object is defined as the product of the component
of the force along the direction of displacement
and the magnitude of the displacement
• F is the magnitude of the force
• Δ x is the magnitude of the
object’s displacement
•  is the angle between
x
F
W 
 )
cos
( 
and 
F x

Work Unit
• This gives no information about
• the time it took for the displacement to occur
• the velocity or acceleration of the object
• Work is a scalar quantity
• SI Unit
• Newton • meter = Joule
• N • m = J
• J = kg • m2 / s2 = ( kg • m / s2 ) • m
x
F
W 
 )
cos
( 
x
F
mv
mv 

 )
cos
(
2
1
2
1 2
0
2


Work: + or -?
• Work can be positive, negative, or zero. The
sign of the work depends on the direction of the
force relative to the displacement
• Work positive: W > 0 if 90°>  > 0°
• Work negative: W < 0 if 180°>  > 90°
• Work zero: W = 0 if  = 90°
• Work maximum if  = 0°
• Work minimum if  = 180°
x
F
W 
 )
cos
( 

Example: When Work is Zero
• 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
x
F
W 
 )
cos
( 

Example: 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

Work Done by a Constant Force
• The work W done on a system by
an agent exerting a constant force
on the system is the product of the
magnitude F of the force, the
magnitude Δr of the displacement
of the point of application of the
force, and cosθ, where θ is the
angle between the force and
displacement vectors:

cos
r
F
r
F
W 






F

II
F

III
r


F

I
r


F

IV
r


r


0

I
W

cos
r
F
WIV 

r
F
WIII 

r
F
WII 



Work and Force
• An Eskimo pulls a sled as shown. The total mass of
the sled is 50.0 kg, and he exerts a force of 1.20 ×
102 N on the sled by pulling on the rope. How much
work does he do on the sled if θ = 30°and he pulls
the sled 5.0 m ?
J
m
N
x
F
W
2
2
10
2
.
5
)
0
.
5
)(
30
)(cos
10
20
.
1
(
)
cos
(









Work Done by Multiple Forces
• If more than one force acts on an object, then the
total work is equal to the algebraic sum of the
work done by the individual forces
• Remember work is a scalar, so
this is the algebraic sum
r
F
W
W
W
W F
N
g
net 



 )
cos
( 
 
net by individual forces
W W

Work and Multiple Forces
• Suppose µk = 0.200, How much work done on the
sled by friction, and the net work if θ = 30° and
he pulls the sled 5.0 m ?


sin
0
sin
,
F
mg
N
F
mg
N
F y
net






J
m
N
s
m
kg
x
F
mg
x
N
x
f
x
f
W
k
k
k
k
fric
2
2
2
10
3
.
4
)
0
.
5
)(
30
sin
10
2
.
1
/
8
.
9
0
.
50
)(
200
.
0
(
)
sin
(
)
180
cos
(

























J
J
J
W
W
W
W
W g
N
fric
F
net
0
.
90
0
0
10
3
.
4
10
2
.
5 2
2












Kinetic Energy
• Kinetic energy associated with the motion of an
object
• Scalar quantity with the same unit as work
• Work is related to kinetic energy
2
2
1
mv
KE 
x
F
mv
mv net


2
0
2
2
1
2
1
net f i
W KE KE KE
   

Work-Kinetic Energy Theorem
• When work is done by a net force on an object
and the only change in the object is its speed,
the work done is equal to the change in the
object’s kinetic energy
• Speed will increase if work is positive
• Speed will decrease if work is negative
2
0
2
2
1
2
1
mv
mv
Wnet 

net f i
W KE KE KE
   

Work and Kinetic Energy
• The driver of a 1.00103 kg car traveling on the interstate at 35.0 m/s
slam on his brakes to avoid hitting a second vehicle in front of him,
which had come to rest because of congestion ahead. After the breaks
are applied, a constant friction force of 8.00103 N acts on the car.
Ignore air resistance. (a) At what minimum distance should the brakes
be applied to avoid a collision with the other vehicle? (b) If the distance
between the vehicles is initially only 30.0 m, at what speed would the
collisions occur?

• (a) We know
• Find the minimum necessary stopping distance
N
f
kg
m
v
s
m
v k
3
3
0 10
00
.
8
,
10
00
.
1
,
0
,
/
0
.
35 





2
3
3
)
/
0
.
35
)(
10
00
.
1
(
2
1
)
10
00
.
8
( s
m
kg
x
N 





2
2
2
1
2
1
i
f
fric
N
g
fric
net mv
mv
W
W
W
W
W 





2
0
2
1
0 mv
x
fk 



m
x 6
.
76



• (b) We know
• Find the speed at impact.
• Write down the work-energy theorem:
N
f
kg
m
v
s
m
v k
3
3
0 10
00
.
8
,
10
00
.
1
,
0
,
/
0
.
35 





N
f
kg
m
s
m
v
m
x k
3
3
0 10
00
.
8
,
10
00
.
1
,
/
0
.
35
,
0
.
30 






x
f
m
v
v k
f 


2
2
0
2
2
2
2
1
2
1
i
f
k
fric
net mv
mv
x
f
W
W 





2
2
3
3
2
2
/
745
)
30
)(
10
00
.
8
)(
10
00
.
1
2
(
)
/
35
( s
m
m
N
kg
s
m
vf 




s
m
vf /
3
.
27


Work Done By a Spring
• Spring force
x
F kx
 

Spring at Equilibrium
• F = 0

Fig. 7.9, p. 173
0
lim
f
f
i
i
x
x
x x
x
x
x
F x F dx
 
 
 
max
0
2
1
2
f f
i i
x x
x
x x
x
W F dx kxdx
kxdx kx

  
  
 

2 2
1 1
2 2
f
i
x
i f
x
W kxdx kx kx
   

Work done by
spring on block

Measuring Spring Constant
• Start with spring at its
natural equilibrium length.
• Hang a mass on spring and
let it hang to distance d
(stationary)
• From
so can get spring constant.
0
x
F kx mg
mg
k
d
  


Scalar (Dot) Product of 2
Vectors
• The scalar product of
two vectors is written
as
• It is also called the dot
product
•
•  is the angle between
A and B
• Applied to work, this
means

A B
A B cos 
 
A B
cos
W F r 
   
F r

Dot Product
• The dot product says something about how
parallel two vectors are.
• The dot product (scalar product) of two
vectors can be thought of as the projection
of one onto the direction of the other.
• Components
A

B

x
A
A
i
A
AB
B
A







cos
ˆ
cos



z
z
y
y
x
x B
A
B
A
B
A
B
A 






B
A )
cos
( 
)
cos
( 
B
A

Projection of a Vector: Dot Product
• The dot product says something about how
parallel two vectors are.
• The dot product (scalar product) of two
vectors can be thought of as the projection
of one onto the direction of the other.
• Components
A

B

z
z
y
y
x
x B
A
B
A
B
A
B
A 





p/2
Projection is zero
x
A
A
i
A
AB
B
A







cos
ˆ
cos



1
ˆ
ˆ
;
1
ˆ
ˆ
;
1
ˆ
ˆ
0
ˆ
ˆ
;
0
ˆ
ˆ
;
0
ˆ
ˆ












k
k
j
j
i
i
k
j
k
i
j
i

Derivation
• How do we show that ?
• Start with
• Then
• But
• So
z
z
y
y
x
x B
A
B
A
B
A
B
A 





k
B
j
B
i
B
B
k
A
j
A
i
A
A
z
y
x
z
y
x
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ








)
ˆ
ˆ
ˆ
(
ˆ
)
ˆ
ˆ
ˆ
(
ˆ
)
ˆ
ˆ
ˆ
(
ˆ
)
ˆ
ˆ
ˆ
(
)
ˆ
ˆ
ˆ
(
k
B
j
B
i
B
k
A
k
B
j
B
i
B
j
A
k
B
j
B
i
B
i
A
k
B
j
B
i
B
k
A
j
A
i
A
B
A
z
y
x
z
z
y
x
y
z
y
x
x
z
y
x
z
y
x





















1
ˆ
ˆ
;
1
ˆ
ˆ
;
1
ˆ
ˆ
0
ˆ
ˆ
;
0
ˆ
ˆ
;
0
ˆ
ˆ












k
k
j
j
i
i
k
j
k
i
j
i
z
z
y
y
x
x
z
z
y
y
x
x
B
A
B
A
B
A
k
B
k
A
j
B
j
A
i
B
i
A
B
A









 ˆ
ˆ
ˆ
ˆ
ˆ
ˆ



Scalar Product
• The vectors
• Determine the scalar product
• Find the angle θ between these two vectors
ˆ
2
ˆ
and
ˆ
3
ˆ
2 j
i
B
j
i
A 






?

B
A





3
.
60
65
4
cos
65
4
5
13
4
cos
5
2
)
1
(
13
3
2
1
2
2
2
2
2
2
2
2




















AB
B
A
B
B
B
A
A
A y
x
y
x
4
6
-2
2
3
(-1)
2 








 y
y
x
x B
A
B
A
B
A



Gravitational Potential Energy
Objects high above the ground have energy by virtue of their height. This is potential
energy (the gravitational type). If allowed to fall, the energy of such an object can be
converted into other forms like kinetic energy, heat, and sound. Gravitational
potential energy is given by:
U = mg h
The equation shows that . . .
. . . the more gravitational potential energy it’s got.
• the more mass a body has
• or the stronger the gravitational field it’s in
• or the higher up it is

SI Potential Energy Units
From the equation U = mgh the units of gravitational
potential energy must be:
kg·(m/s2) ·m = (kg·m/s2) ·m = N·m = J
This shows the SI unit for potential energy is the Joule, as it is for
work and all other types of energy.

Reference point for U is arbitrary
Gravitational potential energy depends on an object’s height, but
how is the height measured? It could be measured from the floor,
from ground level, from sea level, etc. It doesn’t matter what we
choose as a reference point (the place where the potential energy is
zero) so long as we are consistent.
Example: A 190 kg mountain goat is perched precariously atop a
220 m mountain ledge. How much gravitational potential energy
does it have?
U = mgh = (190) (9.8) (220) = 409640 J
This is how much energy the goat has with respect to the ground
below. It would be different if we had chosen a different reference
point.

Reference point for U (cont.)
The amount of gravitation potential energy
the mini-watermelon has depends on our
reference point. It can be positive, negative,
or zero.
10 N
D
C
B
A
6 m
3 m
8 m
Reference
Point
Potential
Energy
A 110 J
B 30 J
C 0
D 60 J
Note: the weight of the object
is given here, not the mass.

Work and Potential Energy
If a force does work on an object but does not increase
its kinetic energy, then that work is converted into
some other form of energy, such as potential energy or
heat. Suppose a 10 N upward force is applied to our
mini-watermelon over a distance of 5 m. Since its
weight is 10 N, the net force on it is zero, so there is no
net work done on it. The work-energy theorem says
that the melon undergoes no change in kinetic energy.
However, it does gain gravitational potential energy in
the amount of U = mgh = (10 N) (5 m) = 50 J. Notice
that this is the same amount of work that the applied
force does on it: W = Fd = (10 N) (5 m) = 50 J. This
is an example of the conservation of energy.
10 N
FA = 10 N
mg = 10 N

Conservation of Energy
One of the most important principles in all of science is conservation of energy.
It is also known as the first law of thermodynamics. It states that energy can
change forms, but it cannot be created or destroyed. This means that all the
energy in a system before some event must be accounted for afterwards.
For example, suppose a mass is dropped from some
height. The gravitational potential energy it had
originally is not destroyed. Rather it is converted into
kinetic energy and heat. (The heat is generated due to
friction with the air.) The initial total energy is given by
E0 = U = mgh. The final total energy is given by Ef = K
+ heat = ½ mv 2 + heat. Conservation of energy
demands that E0 = Ef .
Therefore, mgh = ½ mv2 + heat.
m
before
m
after
v
heat

Conservation of Energy vs. Kinematics
Many problems that we’ve been solving with kinematics can be solved using energy
methods. For many problems energy methods are easier, and for some it is the only
possible way to solve them. Let’s do one both ways:
A 185 kg orangutan drops from a 7 m high branch in a rainforest in Indonesia.
How fast is he moving when he hits the ground?
Note: the mass didn’t matter in either method. Also, we ignored air
resistance in each, meaning a is a constant in the kinematics
method and no heat is generated in the energy method.
kinematics:
vf
2 - v0
2 = 2a x
vf
2 = 2(-9.8)(-7)
vf = 11.71 m/s
Conservation of energy:
E0 = Ef
mgh = ½ mv2
2gh = v2
v = [2(9.8)(7)] ½ = 11.71 m/s

Solved Problems

Example W=Fxcos
A 70 kg base-runner begins to slide into second base when moving at a speed of 4.0
m/s. The coefficient of kinetic friction between his clothes and the earth is 0.70. He
slides so that his speed is zero just as he reaches the base (a) How much energy is
lost due to friction acting on the runner? (b) How far does he slide?
)
8
.
9
)(
70
)(
70
.
0
(


 mg
F
F n
f 

= 480.2 N







f
o
f
f
W
mv
W
K
W
a
2
2
)
4
)(
70
(
2
1
2
1
0
)
-560 J




x
x
x
F
W f
f
)
180
(cos
)
2
.
480
(
560
cos
1.17 m

Example
A 5.00 g bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm. (a) Use
the work-energy theorem, to determine the average frictional force that stops the
bullet.(b) Assuming that the frictional force is constant, determine how much time
elapses between the moment the bullet enters the tree and the moment it stops moving
2
1
0 (0.005)(600)
2
friction
W K
W
W
 
 
 -900 J
 
cos
900 0.04
f f
f
f
W F x
F
F



 22,500 N
6
22,500 (0.005)
0 600 ( 4.5 10 )
f NET
o
F F ma a
a
v v at x t
t
  

    

4.5x106 m/s/s
1.33x10-4 s

Lifting mass at a constant speed
Suppose you lift a mass upward at a constant
speed, v = 0 & K=0. What does the work
equal now?
Since you are lifting at a constant
speed, your APPLIED FORCE
equals the WEIGHT of the object
you are lifting.
Since you are lifting you are raising
the object a certain “y”
displacement or height above the
ground.
When you lift an object above the ground it is said to have POTENTIAL ENERGY

Suppose you throw a ball upward
What does work while it is
flying through the air?
Is the CHANGE in kinetic
energy POSITIVE or
NEGATIVE?
Is the CHANGE in potential
energy POSITIVE or
NEGATIVE?
W K U
   
GRAVITY
NEGATIVE
POSITIVE
( )
o o
o o
o o
BEFORE AFTER
K U
K K U U
K K U U
U K U K
Energy Energy
  
   
   
  

Note KE = K, PE = U; these symbols are
used interchangeably.

ENERGY IS CONSERVED
The law of conservation of mechanical energy states:
Energy cannot be created or destroyed, only
transformed!
Energy Before Energy After
Am I moving? If yes,
KEo
Am I above the
ground? If yes, PEo
Am I moving? If yes,
KE
Am I above the
ground? If yes, PE

Energy consistently changes forms

Position m v U K ME
1 60 kg 8 m/s
Am I above the ground?
Am I moving?
NO, h = 0, U = 0 J
0 J
Yes, v = 8 m/s, m = 60 kg
2 2
1 1 (60)(8)
2 2
1920
K mv
K J
 

1920 J
(= U+K)
1920 J

Position m v U K ME
1 60 kg 8 m/s 0 J 1920 J 1920 J
2 60 kg
Energy Before = Energy After
KO = U + K
1920= (60)(9.8)(1) + (.5)(60)v2
1920= 588 + 30v2
588 J
1332 = 30v2
44.4 = v2
v = 6.66 m/s
6.66 m/s 1920 J
1332 J

Position m v U K ME
1 60 kg 8 m/s 0 J 1920 J 1920 J
2 60 kg 6.66 m/s 588 J 1332 J 1920 J
3 60 kg 1920 J
Am I moving at the top? No, v = 0 m/s
0 m/s 0 J
1920 J
EB = EA
Using position 1
Ko = U
1920 = mgh
1920 =(60)(9.8)h
h = 3.27 m

Example
A 2.0 m pendulum is released from rest when the support string
is at an angle of 25 degrees with the vertical. What is the
speed of the bob at the bottom of the string?
L
 Lcos
h
h = L – Lcos
h = 2-2cos
h = 0.187 m
EB = EA
UO = K
mgho = 1/2mv2
gho = 1/2v2
2(1.83) = v2
1.94 m/s = v

Example
A load of 50 N attached to a spring hanging vertically stretches the
spring 5.0 cm. The spring is now placed horizontally on a table
and stretched 11.0 cm. What force is required to stretch the
spring this amount?



k
k
kx
Fs
)
05
.
0
(
50
1000 N/m



s
s
s
F
F
kx
F
)
11
.
0
)(
1000
(
110 N

Hooke’s Law from a Graphical Point of View
x(m) Force(N)
0 0
0.1 12
0.2 24
0.3 36
0.4 48
0.5 60
0.6 72
graph
x
vs.
F
a
of
Slope



k
x
F
k
kx
F
s
s
Suppose we had the following data:
Force vs. Displacement y = 120x + 1E-14
R2
= 1
0
10
20
30
40
50
60
70
80
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Displacement(Meters)
Force(Newtons)
k =120 N/m

We have seen F vs. x Before!!!!
Force vs. Displacement y = 120x + 1E-14
R2
= 1
0
10
20
30
40
50
60
70
80
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Displacement(Meters)
Force(Newtons)
Work or ENERGY = Fx
Since WORK or ENERGY
is the AREA, we must get
some type of energy when
we compress or elongate
the spring. This energy is
the AREA under the line!
Area = ELASTIC
POTENTIAL ENERGY
Since we STORE energy when the spring is compressed and
elongated it classifies itself as a “type” of POTENTIAL ENERGY, Us.
In this case, it is called ELASTIC POTENTIAL ENERGY (EPE).

Example
A slingshot consists of a light leather cup, containing a stone, that is pulled back
against 2 rubber bands. It takes a force of 30 N to stretch the bands 1.0 cm (a)
What is the potential energy stored in the bands when a 50.0 g stone is placed in
the cup and pulled back 0.20 m from the equilibrium position? (b) With what
speed does it leave the slingshot?











v
v
mv
U
K
U
E
E
c
k
kx
U
b
k
k
kx
F
a
s
s
A
B
s
s
2
2
2
)
050
.
0
(
2
1
2
1
)
)
20
)(.
(
5
.
0
2
1
)
)
01
.
0
(
30
) 3000 N/m
300 J
109.54 m/s

Work Power & Energy-SPP.pptx

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