Transcript document

Work, Energy, and Power
H.W. Ch. 5
Read and Write for the whole chapter.
½ page per section.
5.1 1-18
5.3 34-46
5.4 48-57
5.5 59-78
5.6 80-91
All problems are due. Odds may be done
separately and copied from me in class.
Work
Underlined words are WOD
When I do a pushup, I move.
When Chuck Norris does a push up, what
happens?
Work
Underlined words are WOD
When Chuck Norris does a push up, what
happens? The earth moves.
What happens when my son Luke does a push
up?
Real answer: Use Newtons 3rd law.
Work
Underlined words are WOD
Luke-a-meter: The distance the earth moves
when my 100 lb son does a push up.
Work Demo
Have a student walk across the front of the room,
carrying the 10 kg brick, at a constant speed.
Did they do work?
Work

Work = Force ∙ displacement or W = F d cos Θ
(I have memorized “Work is Force times distance.”)



Guess, what are the units?
Note: Dot Product => mult by cos of angle between
Force and displacement. So W = F d cos Θ
Which means forces only do work in the direction of
motion.
Work

Work = Force ∙ displacement

Units = N m = Joules
Work Demo
Remember the moles we shot into space?
We applied a 50N force to them over a distance of
2 cm.
Work on Mole = F x d = 50N x .02m x cos 0
= 1 Joule
Have a student walk across the front of the room,
carrying the 10 kg brick, at a constant speed.
How much work did they do?
Guesses in Joules???
Work

Work = Force ∙ displacement

Units = N m = Joules

Force and displacement are both vectors.

How do you multiply vectors?
The Dot Product
One way of multiplying vectors is
“the dot product”.
If A & B are vectors,
A ∙ B = |A| * |B|* cos [Θ]
•
•
Θ = the angle between A and B
|A| is the length of A, |B| is the length of B
The result of the dot product is a SCALAR,
not a vector. Order of the vectors is important.
Θ Goes from 2nd to 1st vector.
The Dot Product
A ∙ B = |A| * |B|* cos [Θ]
70 Deg
If B is at 30 degrees and
30 Deg
A is at 70 degrees,
Then Θ is angle from 30 to 70 = 40 deg
Θ can be negative.
Order of the vectors is important.
Θ Goes from 2nd to 1st vector: from Displacement to
Force.
Dot Product examples
A = 3 @ 0, B = 4 @ 0
A ∙ B = __________
C = 3 @ 30, D = 4 @ 60
C ∙ D = __________
E = 3 @ 90, F = 4 @ 135
E ∙ F = __________
G = 3 @ 120, H = 4 @ 60
G ∙ H = __________
I = 3 @ 180, J = 4 @ 90
I ∙ J = __________
K = 3 @ 100, L = 4 @ 235 K ∙ L = __________
M = 3 @ 45, N = 4 @ 225 M ∙ N = __________
Dot Product examples
A = 3 @ 0, B = 4 @ 0
A ∙ B = ___12_____
C = 3 @ 30, D = 4 @ 60
C ∙ D =___10.4____
E = 3 @ 90, F = 4 @ 135
E ∙ F = ___8.5____
G = 3 @ 120, H = 4 @ 60
G ∙ H = ____6_____
I = 3 @ 180, J = 4 @ 90
I ∙ J = ____0_____
K = 3 @ 100, L = 4 @ 235 K ∙ L = __-8.5 ____
M = 3 @ 45, N = 4 @ 225 M ∙ N = _-12______
Work
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Work = Force ∙ displacement
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Units =

So, if Force & Displacement are
N m = Joules
–
In the same direction, work is +
–
In the opposite direction, work is -
–
Perpendicular, work is 0
Memorize this table:
Force:
Same
Opposite
90deg angle
Disp.:
Work:
Positive
Negative
Zero, Zed, Zilch
Nada, Nothing,
Zippo, Donut Hole
Example 1 Pulling a Suitcase-on-Wheels
Find the work done if the force is 45.0-N, the
angle is 50.0 degrees, and the displacement
is 75.0 m.
Example 1 Pulling a Suitcase-on-Wheels
Find the work done if the force is 45.0-N, the
angle is 50.0 degrees, and the displacement
is 75.0 m.
F
Remember:
Force is a vector
Fx = F cos θ
Motion
Example 1 Pulling a Suitcase-on-Wheels
Find the work done if the force is 45.0-N, the
angle is 50.0 degrees, and the displacement
is 75.0 m.


W  F cos  s  45.0 N  cos 50.0 75.0 m 
 2170 J
Work (example 1)
a 1.02 kg rock is in free fall. It falls for 5 meters.
How much work does gravity do on the rock?
Work (example 1)
a 1.02 kg rock is in free fall. It falls for 5 meters.
How much work does gravity do on the rock?

Why use 1.02kg instead of 1.00 kg???

(seems like a weird number?)
Work (example 1)
a 1.02 kg rock is in free fall. It falls for 5 meters.
How much work does gravity do on the rock?
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Why use 1.02kg?
(Answer: Makes for a round number when
multiplied by gravity.)
Work
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Example:
a 1.02 kg rock is in free fall. It falls for 5 meters.
How much work does gravity do on the rock?
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W = F*∆y = m*g*∆y = 1.02kg*9.80m/s2*5m
= 50 Nm = 50 Joules
Gravity is downward, so is ∆y, so work is positive.
Also, make sure the units are correct.
Work

Example:
a 1.02 kg rock is in free fall. It falls for 5 meters.
So how much work do you have to do on the rock
to lift it up the two flights of stairs to 5m height?

W = F*∆y = m*g*∆y = 1.02kg*9.80m/s2*5m
= 50 Nm = 50 Joules
Same. What comes out is what goes in!!!
Work


What if there are multiple forces acting on an
object?
We can find the work done by each force, and
we can find the total (net) work.

Step 1: Find the work each force does.

Step 2: Add those works together.
Work
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A 1.02 kg rock is accelerated upwards by a
string for a displacement of 30 meters. The
tension in the string is 20 N.

What is the work done by each force?

What is the was the net force during that 30 m?

How many forces? 2
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Tension in string

Gravity
Work
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


A 1.02 kg rock is accelerated upwards by a
string. The tension in the string is 20 N. It
travels 30 m up.
1.) What is the work done by tension?
2.) What is the work done by gravity during that
same 30 m?
3.) What is the net work?
Work


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A 1.02 kg rock is accelerated upwards by a
string. The tension in the string is 20 N. What
is the work done by the string's Tension to
move the rock up 30 meters?
WorkT = F d
=T*∆y cos [angle between them]
= 20 N * 30 m * cos(0)
=
600 J
Work


A 1.02 kg rock is accelerated upwards by a
string. The tension in the string is 20 N. What
is the work done by gravity during that same 30
m?
Workg = mg* ∆y * cos (180)
= 10N*30m * (-1)
= -300 Joules
Work
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A 1.02 kg rock is accelerated upwards by a
string. The tension in the string is 20 N. What
is the total amount of work done during that 30
m?
Work

A 1.02 kg rock is accelerated upwards by a
string. The tension in the string is 20 N. What
is the total amount of work done during that 30
m?
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Wnet = WT + Wg
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Wnet = 600 J – 300 J = 300 J
Work

A 1.02 kg rock is accelerated upwards by a
string. The tension in the string is 20 N.

Bonus question:

What is the was the net force during that 30 m?
Work

A 1.02 kg rock is accelerated upwards by a
string. The tension in the string is 20 N.

Bonus question:

What is the was the net force during that 30 m?
Wnet = Fnet * ∆y
Fnet = Wnet /∆y
= 300J / 30m = 10 N
Example 3 Accelerating a Crate
The truck is accelerating at
a rate of +1.50 m/s2. The mass
of the crate is 120-kg and it
does not slip. The magnitude of
the displacement is 65 m.
What is the total work done on
the crate by all of the forces
acting on it?
The angle between the
displacement and the normal
force is 90 degrees.
The angle between the
displacement and the weight
is 270 degrees.
W  F cos 90x  0
W  Weight cos 270x  0
Another way to think of this is:
Calculate work in the
direction of force.
W  F cos 0 y  0
because y  0
W  Weight cos180y  0
because y  0
In other words, because
there is no movement in the
y direction, there is no work
done by “y forces”.
The angle between the
displacement and the applied
force is 0 degrees.
ΣF = ma
So net force is just from friction.
Friction   Forces In X Direction


f s  ma  120 kg  1.5 m s 2  180 N
W  180 N cos 065 m  1.2 104 J
Work Demo Revisited.
Have a student walk across the front of the room,
carrying the 10 kg brick, at a constant speed.
How much work did they do?
Any new Guesses???
What is work, really?
Conceptually, what is work?
What is work, really?
Conceptually, what is work?
Work is the change of energy. From one form to
another or one object to another.
Add the following to your work def:
WORK: The amount of energy added to an
object.
Types of Energy
There are two types of energy that we're going to
be talking about in Mechanics. Energy can
easily be changed between these two forms.
MEMORIZE these equations:
Kinetic Energy: Energy of motion. KE = ½ mv2
Potential Energy(Gravitational): Energy of
position. PE = mgh
(Sometimes called GPE instead of PE)
Together, they are called
Mechanical Energy: ME = KE + PE
Talk about 1. other types of PE. 2. mechanical engineers and what they do.
Lets look at units.
Add Units to WOD.
KE = ½ mv2
kg * (m/s)2
(kg *m/s2 )*m
PE = mgh
kg *m/s2 * m
N *m
N*m
Both equal Joules
Add: “Units are Joules.”
to WOD.
Work Demo Revisited.
Have a student walk across the front of the room,
carrying the 10 kg brick, at a constant speed.
How much work did they do?
Any new Guesses???
Work Demo Revisited.
Have a student walk across the front of the room,
carrying the 10 kg brick, at a constant speed. How
much work did they do?
What was the ME of the brick before moving?
After moving?
What is the equation for the change in ME of the brick?
This is the amount of work done to it.
Work Demo Revisited.
Why do we talk about ME?
Add to WOD:
Work can also be defined as the change in ME of an
object. Work is in Joules. Energy is in Joules.
Work can be thought of as
Work is also “adding energy” to an object.
Most often, the equation used is:
Work = Change in KE and we ignore PE.
Add to WOD: W = ΔKE
Work Demo Revisited.
Have a student walk across the front of the room,
carrying the 10 kg brick, at a constant speed. How
much work did they do?
What is the equation for the change in KE of the brick?
Change in KE = KE2 – KE1 =
(½ mv22 – ½ mv12 ) but v2 = v1 so the change in KE = 0.
So no work is done because no kinetic energy is added
to the package. (But, don’t tell the UPS man he is
doing no work if you want to get your packages.)
Work Demo Revisited.
Have a student walk across the front of the room,
carrying the 10 kg brick, at a constant speed. How
much work did they do?
What is the equation for the change in PE of the brick?
Change in PE = PE2 – PE1 = (mgh2 – mgh1)
But h2 = h1, so no PE is added or change in PE = 0.
So no work is done because no potential energy is
added to the package. (Again, don’t tell the UPS
man he is doing no work if you want to get your
packages.)
Work Demo Revisited.
Have a student walk across the front of the room,
carrying the 10 kg brick, at a constant speed. How
much work did they do?
What about Work = Force . Displacement.
Again, the angle between them is 90 degrees. The
cos 90 is 0 and
Work = F x D Cos 90 = 0
So no work is done because the force and
displacement are 90 degrees to each other.
(STILL, don’t tell the UPS man he is doing no work if
you want to get your packages.)
Work Demo Revisited.
Question: I lift the 10 kg steel brick 1 foot. I walk
forward at a constant speed and quickly hand it to a
student. The students carries the brick around the
track outside for 5 miles at the same constant speed.
Who does the most work? Me or the student?
Answer: Who added energy to the brick?
Work Demo Revisited.
Revisit the red box motion. Think this time in terms of
Work as being a change in Kinetic energy.
Me Pushing the red box (on ice) adds Kinetic Energy,
when speeding it up.
Removes Kinetic Energy when slowing it down.
Now add friction and discuss.
2 Important Ideas
Law of conservation of energy 1.) Energy cannot be created or destroyed, but it can
change form.
2.) The total amount of energy in the universe is
constant. (Amount does not change.)
Mechanical Energy (KE + PE) is not always constant.
Things like friction can dissipate it, usually into heat.
Discuss E = mC^2
Work Energy Theorem
Work(net) = ∆K.E.
= K.E.f – K.E.i
= ½ m vf2 - ½ mvi2
Does not MEAN ∆ KE = ½ m (vf - vi )2
Because (vf - vi )2 Does not equal vf2 - vi2
Right is:
W=½m
2
(vf
-
2
vi )
Work Energy Theorem
Work “done against a non-friction force” = Potential
Energy.”
Work Energy Theorem
Work “done against a non-friction force” = Potential
Energy.”
Recall “Work = Force ∙ displacement”
Well, m*g = force
height = displacement (vertical)
So, work “done against gravity” = mgh