Transcript Work

Introduction to Work
Where we have been
Previously we used Newton’s Laws to
analyze motion of objects
 Force and mass information were used to
determine acceleration of an object
(F=ma)
 We could use the acceleration to
determine information about velocity or
displacement
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Did the object speed up or slow down?
 How far did the object travel?
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Where we are going
Now we will take a new approach to
looking at motion
 We will now look at work and power in
relation to motion
 Today we will focus on “work”
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Definition of “work”
The everyday definition of “work” and the
one that we use in physics are quite
different from each other
 When most people think about “work”,
they think of the job that they have
 Although it is possible that you are doing
the physics definition of work while at your
job, it is not always the case
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Physics Definition of “Work”
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Like so many other things in physics, we have to
use an exact definition to really explain what
“work” is
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PHYSICS DEFINITION
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Work happens when a force causes an
object to move through a displacement
When a force acts upon an object to cause a
displacement of the object, it is said that WORK
has been done upon the object
Work
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There are three key ingredients to work
Force
 Displacement
 Cause
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In order for a force to qualify as having
done “work” on an object, there must be a
displacement and the force must cause
the displacement
Everyday Examples of “Work”
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There are several good examples of work
which can be observed in everyday life
A horse pulling a plow through a field
 A person pushing a shopping cart
 A student lifting a backpack onto her shoulder
 A weightlifter lifting a barbell above his head
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In each case described here there is a
force exerted upon an object to cause that
object to be displaced
Work
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Work – Exerting force in a way that makes
a change in the world.
Throwing a rock is work: you’re exerting a
force, and the rock’s location changes (i.e.
“the world has been changed”)
 Pushing on a brick wall is not work: you’re
exerting a force, but “the world doesn’t
change” (the wall’s position doesn’t change).
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Work
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So exerting force alone isn’t enough. You have
to both exert a force, and make a change.
If you’re not exerting a force, you’re not doing
work.
Example: Throwing a ball.
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While you are “throwing the ball” (as opposed to just
holding it) you are exerting a force on the ball. And
the ball is moving. So you’re doing work.
After the ball leaves your hand, you are no longer
exerting force. The ball is still moving, but you’re no
longer doing work.
Work
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So, mathematically, we define work as “exerting a force
that causes a displacement”:
(Work) = (Force exerted) (Displacement of object) (cos Θ)
or
W = F*d*cosΘ
W = Work done (J)
F = Force exerted on object (N)
d = Displacement of object (m)
Θ = Angle between the force and the displacement
New Unit!
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The units for work are Nm (Newtons × meters).
As we did with Newtons (which are kg m/s2), we
will “define” the Newton-meter to be a new unit.
We’ll call this unit the Joule.
Abbreviation for Joule: J
So, 1 Nm = 1 J
 Example:
1 joule = work done to lift a
¼ lb hamburger (1 N) 1 meter
Defining Θ – “the angle”
This is a very specific angle
 Not just “any” angle - It is the angle
between the force and the displacement
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Scenario A: A force acts rightward (@ 0°) upon
an object as it is displaced rightward (@ 0°) .
The force vector and the displacement vector
are in the same direction, therefore the angle
between F and d is 0 degrees
0° - 0° = 0°
Subtract the smaller angle from the larger
angle to determine the angle “between”
the vectors
F
Θ = 0 degrees
d
Defining Θ – “the angle”
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Scenario B: A force acts leftward (@ 180°) upon
an object which is displaced rightward (@ 0°) .
The force vector and the displacement vector
are in the opposite direction, therefore the angle
between F and d is 180 degrees
F
180° - 0° = 180°
Subtract the smaller angle from the
larger angle to determine the angle
“between” the vectors
Θ = 180 degrees
d
Defining Θ – “the angle”
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Scenario C: A force acts upward (@ 90°) upon
an object as it is displaced rightward (@ 0°) .
The force vector and the displacement vector
are at a right angle to each other, therefore the
angle between F and d is 90 degrees
F
90° - 0° = 90°
Subtract the smaller angle from the
larger angle to determine the angle
“between” the vectors
Θ = 90 degrees
d
To Do Work,
Forces must CAUSE Displacement
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Consider scenario C from the previous slide
The situation is similar to a waiter who carried a
tray full of meals with one arm (F=20N) straight
across a room (d=10m) at constant speed
W = F*d*cosΘ
W = (20N)(10m)(cos 90°)
W = 0J
The waiter does not do work
upon the tray as he carries it
across the room
The Meaning of Negative Work
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On occasion, a force acts upon a moving object
to hinder a displacement
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A car skidding to a stop on a roadway surface
A baseball player sliding to a stop on the infield dirt
In such cases the force acts in the direction
opposite the objects motion in order to slow it
down
The force doesn’t cause the displacement, but it
hinders the displacement
This is commonly called negative work
The Meaning of Negative Work
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If you substitute the numerical values into the
work equation, you are left with a negative
answer
W = F*d*cosΘ
W = (40 N)(10 m)(cos 180°)
W = (40 N)(10 m)(-1)
W = -400 J
The 10 m displacement is hindered by a 40 N
force causing -400 J worth of work
This will be an important concept a little later
Example of Work
You are pushing a very heavy stone block
(200 kg) across the floor. You are exerting
620 N of force on the stone, and push it a
total distance of 20 m in 1 direction before
you get tired and stop.
 How much work did you just do?
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W = (620 N)(20 m) = 12,400 J
Work Problems
Austin lifts a 200 N box 4
meters. How much work did he
do?
W = (200N)(4m)(cos 0°)
W = (200N)(4m)(1)
W = 800 J
Work Problems
Caitlin pushes and pushes on a loaded
shopping cart for 2 hours with 100 N of
force. The shopping cart does not
move. How much work did Caitlin do?
Chase lifts a 100 kg (220 lbs) barbell 2
meters. How much work did he do?
Work Done By “Lifting” Something
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Notice that when we were pushing something
along the ground, the work done didn’t depend
on the mass.
Lifting up something does do work that depends
on mass.
Because of gravity:
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Gravity always pulls down with a force equal to m*ag,
where m is the mass, and ag = 9.8 m/s2.
So we must exert at least that much force to lift
something.
The more mass something has,
the more work required to lift it.
Work Done By “Lifting”
Something
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Example: A weightlifter lifts a barbell with a
mass of 280 kg a total of 2 meters off the floor.
What is the minimum amount of work the
weightlifter did?
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The barbell is “pulled” down by gravity with a force of
(280 kg)(9.8 m/s2) = 2,744 N
So the weightlifter must exert at least 2,744 N of
force to lift the barbell at all.
If that minimum force is used, the work done will be:
W = (2,744 N)(2 m) = 5,488 J
Questions???