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Chapter 6 – Work and Energy
Midterm exams will be available next
Thursday.
Assignment 6
Textbook (Giancoli, 6th edition), Chapter 6:
Due on Thursday, November 5
1. On page 162 of Giancoli, problem 4.
2. On page 162 of Giancoli, problem 8.
3. On page 163 of Giancoli, problem 24.
4. A running man has half the kinetic energy that a boy of half
his mass has. The man speeds up by 1.00 m/s and then
has the same kinetic energy as the boy. What were
the original speeds of man and boy?
Chapter 6
• Work Done by a Constant Force
• Kinetic Energy, and the Work-Energy Principle
• Potential Energy
• Conservative and Nonconservative Forces
• Mechanical Energy and Its Conservation.
Recalling Last Lecture
Kinetic Energy, and the Work Energy Principle
In mechanics, we can use a less precise definition of energy as:
“The ability to do work”
The energy of motion is called kinetic energy (or KE for short).
Kinetic Energy, and the Work Energy Principle
Work-energy principle:
“ The net work done on a object is equal to the change in the object’s kinetic
energy ”
1) If Wnet > 0, then ∆KE > 0 implying that there was an increase in the object’s
velocity ( v2 > v1 ).
2) If Wnet < 0, then ∆KE < 0 implying that there was an decrease in the object’s
velocity ( v2 < v1 ).
Kinetic Energy, and the Work Energy Principle
Problem 6-25 (textbook): A 285-kg load is lifted 22.0 m vertically with an
acceleration a = 0.160g by a single cable. Determine
(a) the tension in the cable,
(b) the net work done on the load,
(c) the work done by the cable on the load,
(d) the work done by gravity on the load, and
(e) the final speed of the load assuming it started from rest.
Kinetic Energy, and the Work Energy Principle
Problem 6-25:
(a)
From the free-body diagram for the load being lifted, write Newton’s 2nd law for
the vertical direction, with up being positive.
Fnet =
∑F
= FT − m g = m a = 0.160 m g
(
→
)
FT = 1.16 m g = 1.16 ( 285 kg ) 9.80 m s 2 = 3.24 × 10 3 N
r
FT
r
mg
Kinetic Energy, and the Work Energy Principle
Problem 6-25:
(b)
The net work done on the load is found from the net force.
(
W net = Fnet d cos 0 o = ( 0.160 m g ) d = 0.160 ( 285 kg ) 9.80 m s 2
= 9.83 × 10 3 J
r
FT
r
mg
) ( 22.0 m )
Kinetic Energy, and the Work Energy Principle
Problem 6-25:
(c)
The work done by the cable on the load is
(
Wcable = FT d cos 0 o = (1.160 mg ) d = 1.16 ( 285 kg ) 9.80 m s 2
r
FT
r
mg
) ( 22.0 m ) =
7.13 × 10 4 J
Kinetic Energy, and the Work Energy Principle
Problem 6-25:
(d)
The work done by gravity on the load is
(
WG = mgd cos180 o = − mgd = − ( 285 kg ) 9.80 m s 2
r
FT
r
mg
) ( 22.0 m ) =
− 6.14 × 10 4 J
Work Done by a Constant Force
Problem 6-25:
(e)
Use the work-energy theory to find the final speed, with an initial speed of 0.
W net = K E 2 − K E 1 =
v2 =
2W n et
m
+ v
2
1
=
1
2
m v 22 −
1
2
m v 12
(
2 9 .8 3 × 1 0 3 J
285 kg
r
FT
r
mg
→
)+0
= 8 .3 1 m s
Potential Energy
Another form of energy “carried” by an object is called potential energy.
Potential energy is associated to forces that depend on the position or configuration
of an object.
There are several forces that depend on either the configuration or position of an
object
therefore there are several different kinds of potential energies, each
associated to a different force.
For example, we have seen that the gravitational force depends on the distance
between two objects
therefore, the energy of an object associated to the gravitational force
(gravitational potential energy) will consequently depend on the position of the
object.
Let’s analyze the gravitational potential energy …………..
Potential Energy
In raising a mass m to a height h (no net acceleration), the work done by the
external force is
(6-8)
Gravity also does work on the object of mass m:
Now, note that if at any instant of time you drop
the object, it can do work on something located
underneath it. Let’s say it strikes a nail, moving
it into the floor.
it will change the nail’s velocity, therefore it will change the nail’s
kinetic energy.
Potential Energy
How does it work?
As you drop the object, it will immediately
gain velocity due to the acceleration of
gravity.
Assuming its initial velocity to be zero, we have for
its velocity when it strikes the nail (travels a distance h):
Therefore, using the work-energy principle, we have:
0
2gh
This results tells us that an object located at a height h has the ability to do an
amount of work given by mgh
Potential Energy
The previous result can be generalized to any
height y such that we can define a
gravitational potential energy as
(6-9)
Given that h = y2 – y1 , it follows that:
which is exactly eq. 6.8 obtained two slides ago.
We can rewrite the above equation as:
(6-10)
Note 1: The choice of y=0 is arbitrary only changes in potential energy matter.
Note 2: Again, from two slides ago we can conclude that:
Potential Energy
Important:
1) Potential energy is a property of a system as a
whole.
For example, when you lift the object in the
figure from y1 to y2, the work done is given by
the difference between PE2 and PE1.
The values of PE1 and PE2 depend on the object
and on the Earth (responsible for the force of gravity),
therefore on the complete system: Earth + object.
This is NOT the case with kinetic energy where
ONLY the properties of the object matters
2) The work done by the gravitational force on an object depends only on the
height and not on the path taken by the object.
Potential Energy (Elastic force)
As mentioned before, there are other forms of potential energy. I will now discuss on
one of them associated with elastic materials.
A spring has potential energy when compressed or stretched.
For example, the figure below shows the potential energy stored in a spring when
it is compressed yielding kinetic energy when it is released.
Potential Energy (Elastic force)
The force needed to compress or stretch a string is written as:
Where,
k = spring stiffness constant
Newton’s 3rd law says that the spring will
exert a force FS in reaction to the force
FP such that:
Textbooks usually write:
(6-11)
Eq. 6.11 (spring equation) is known as Hooke’s law
FS is the force the spring exerts opposite to its displacement in an attempt to restore
its original length
Potential Energy (Elastic force)
Now that we know the force, we can define the potential energy of a spring
compressed or stretched by a length x from an initial reference position.
The work done to change the spring’s natural length, let’s say stretch it, will be given
by:
Where,
(We cannot use FP directly to calculate the work
done on the string since FP = kx varies with the
position x. Thus, we use the average force applied
on the spring).
Thus
We can define the elastic potential energy as:
(6-12)
Kinetic Energy, and the Work Energy Principle
Problem 6-30 (textbook): A 1.60-m tall person lifts a 2.10-kg book from the ground
so it is 2.20 m above the ground. What is the potential energy of the book relative to
(a) the ground
(b) and the top of the person’s head?
(c) How is the work done by the person related to the answers in parts (a) and (b)?
Kinetic Energy, and the Work Energy Principle
Problem 6-30:
(a)
Relative to the ground, the PE is given by
(
PE G = m g ( y book − y ground ) = ( 2.10 kg ) 9.80 m s 2
(b)
) ( 2.20 m ) =
Relative to the top of the person’s head, the PE is given by
(
PE G = m g ( y book − y head ) h = ( 2.10 kg ) 9.80 m s 2
(c)
45.3 J
) ( 0.60 m ) =
12 J
The work done by the person in lifting the book from the ground to the final
height is the same as the answer to part (a), 45.3 J.
In part (a), the PE is calculated relative to the starting location of the
application of the force on the book. The work done by the person is not
related to the answer to part (b).
Kinetic Energy, and the Work Energy Principle
Problem 6-29 (textbook): A 1200-kg car rolling on a horizontal surface has speed
v = 65 Km/h when it strikes a horizontal coiled spring and is brought to rest in a
distance of 2.2 m. What is the spring stiffness constant of the spring?
Kinetic Energy, and the Work Energy Principle
Problem 6-29:
Assume that all of the kinetic energy of the car becomes PE of the compressed
spring.
2
1
2
mv 2 = 12 kx 2

 1m s  
(1200 kg ) ( 65 km h ) 

2
3.6
km
h
mv

  = 8.1 × 10 4 N m

→ k= 2 =
2
x
( 2.2 m )