Transcript Work - Northwest ISD Moodle

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

Did the object speed up or slow down?
 How far did the object travel?

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”

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

Physics Definition of “Work”

Like so many other things in physics, we have to
use an exact definition to really explain what
“work” is

PHYSICS DEFINITION


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

There are three key ingredients to work
Force
 Displacement
 Cause


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”

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


In each case described here there is a
force exerted upon an object to cause that
object to be displaced
Work

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).

Work



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.


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

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!
The units for work are Nm (newton ×
meter). As we did with newton (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

Defining Θ – “the angle”
This is a very specific angle
 Not just “any” angle - It is the angle
between the force and the displacement


Scenario A: A force acts rightward upon an
object as it is displaced rightward. The force
vector and the displacement vector are in the
same direction, therefore the angle between F
and d is 0 degrees
Because the force and the displacement
point in the same direction, there is
nothing between them – NO angle - 0°
“between” the vectors
F
Θ = 0 degrees
d
Defining Θ – “the angle”

Scenario B: A force acts upward upon an object
as it is displaced rightward. 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
Θ = 90 degrees
In order to pivot from F to d, you must
rotate 90°
d
Defining Θ – “the angle”

Scenario D: A pushing force acts leftward upon
an object as it is displaced up a ramp to the left.
The force vector and the displacement vector
need to be drawn starting from the same
location in order to find the angle between them.
d
d
F
F
In order to pivot from F to d, you must
rotate 28°
Θ = 28 degrees
Rules for Work
If the force and displacement are in the
same direction – the work done is
POSITIVE because it adds energy
 If the force and displacement are in
opposite directions – the work done is
NEGATIVE because it removes energy
 If the force and displacement are
perpendicular – the work done is equal to
ZERO

To Do Work,
Forces must CAUSE Displacement






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

On occasion, a force acts upon a moving object
to hinder a displacement





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
opposes the displacement
This results in negative work
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?

W = (620 N)(20 m) = 12,400 J
Practice Problem

Jessica Lee goes to the market and applies 200N of force at
an angle of 35° above horizontal on the handle of the 600N
shopping cart. She pushes the cart halfway down the freezer
aisle which is 12 m long and stops to grab a bag of peas. She
adds them to the cart which increases the amount of force
she must push with by 15 N. She continues to the end of the
freezer aisle.

How much work did Jessica Lee do before she got the peas?
 983.0

J
How much work did Jessica Lee do total (in the freezer aisle)?
 2039.7
J
Work Problems
Austin lifts a 200 N box 4 meters.
How much work did he do?
W = (200N)(4m)
W = (200N)(4m)
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 she do?
Work Done By “Lifting” Something



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:



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

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?
–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
Power!
We’ve talked about “work”, now
let’s talk about “power”
“Work” Refresher

Remember that the formula for finding the
amount of work done upon an object is:
W = (F)*(d)
W = the work done
 F = the force required to cause
displacement
 d = displacement of the object

Definition of Power
•Common definition usually includes
something about strength
•Physics definition - the timed rate at
which work is done
•depends directly upon the work and
inversely upon the time to do that work
If you increase work, then you increase power
BUT
If you increase time, then you decrease power
How Hard Are You Working?

The rate at which work is done is called
power:
(Power) = (Work Done) / (Time Spent Working)
P=W/t

Power is “how hard”
something is working.
Units for power:
1J
1s
=1W
one watt (W) equals one
joule (J) per second (s)
joule per second = watt
Power Example
Let’s say that it took us 40 s to move that 200 kg
stone block the 20 m.


Remember that we did 12,400 J of work on the stone
block
Given Information:



Basic Equation:


P = W/t
Working Equation:



W = 12,400 J
t = 40 s
P = 12,400 J / 40 s
P = 310 J/s
Since it took us 40 s to move the block, we were
doing 310 J of work per second OR generating 310 W
of power
Fast work isn’t more work
Go back to our 200 kg block example.
Remember that when it took us 40 s to
push the block the 20 m when we applied
a force of 620 N, that implied that we had
a power output of 310 W.
 If we exerted the same force (620 N) and
pushed the block the same distance (20
m), but took half as long to do so (20 s),
our power output would double to 620 W.

Fast work isn’t more work

*But* notice that the total work done
doesn’t change – we still exerted 620 N of
force over a distance of 20 m.

Copy this sentence in your notes:

So increasing power output doesn’t mean
you’re doing more work, it means you’re
doing the work faster.
What is energy??
Energy is the capacity of a physical system to
perform work.
 Or the ability to make things move

Energy comes in many forms:
mechanical, electrical , magnetic,
solar,
thermal, chemical, etc...

Energy is a scalar quantity…..remember what
scalar means??
How is energy divided?
All Energy
Potential Energy
Kinetic Energy
Gravitational
Potential
Energy
Elastic
Potential
Energy
Chemical
Potential
Energy
Kinetic Energy
Energy of motion
All moving objects that
have mass have kinetic energy.
K = 1/2
2
mv
m - mass of the object in kg
v - speed of the object in m/s
KE - the kinetic energy in J
Let’s practice a few problems….
Practice Problem 1: What is the kinetic energy
of a 150 kg object that is moving with a speed
of 15 m/s?
First!!! What are you trying to solve for?
KINETIC ENERGY
1.
2.
Remember that KE = 1/2mv2
Write down your known variables:
Mass is measured in kg so m = 150 kg
 Velocity is measured in m/s so v = 15 m/s

3.
Plug your numbers into the formula

4.
KE = ½ (150)(15)2
Solve your equation with units

KE = 16875 J – Don’t forget, Energy is measured in
Joules!
Practice Problem 2: What is the kinetic energy
of a 0.05 kg bullet traveling at 475 m/s?
1.
What are you trying to solve for?

2.
What’s the equation???

3.
KE
KE = 1/2mv2
Write down your known variables:
Mass is measured in kg so m = 0.05 kg
 Velocity is measured in m/s so v = 475 m/s

4.
Plug your numbers into the formula

5.
KE = ½ (0.05)(475)2
Solve your equation with units

KE = 5640.63 J (round to the nearest hundredth
place)

1.
Practice Problem 3: A greyhound at a race track
can run at a speed of 17.0 m/s. What is the KE of
a 24 kg greyhound as it crosses the finish line?
What are you trying to solve for?

2.
What’s the equation???

3.
KE
KE = 1/2mv2
Write down your known variables:


4.
Plug your numbers into the formula

5.
Mass is measured in kg so m = 24 kg
Velocity is measured in m/s so v = 17 m/s
KE = ½ (24)(17)2
Solve your equation with units

KE = 3468 J (round to the nearest hundredth
place)
Think about this…

Two marbles, one twice as heavy as the other, are
dropped to the ground from the roof of a building.
Just before hitting the ground, the heavier marble
has
1. as much kinetic energy as the lighter one.
2. twice as much kinetic energy as the lighter one.
3. half as much kinetic energy as the lighter one.
4. four times as much kinetic energy as the lighter one.
5. impossible to determine
Think about the formula and write the question and the
answer you think is right in your journal.
Potential Energy

There are different ways that energy can
be stored:
Gravitational Potential Energy
 Elastic Potential Energy
 Chemical Potential Energy

Potential Energy
energy of position, location or condition
gravitational potential energy
Ug = mgh
m - mass of object in kg
g - acceleration of gravity in m/s2
h - height of object, in m
Ug – gravitational potential energy in
J
Let’s practice a few problems….
Practice Problem 1: Calculate the gravitational potential
energy of a 1,500 kg elevator that is 45 meters above the
ground floor. Choose the ground floor to be the reference
level.
First!!! What are you trying to solve for?
POTENTIAL ENERGY
1.
2.
Remember that PE = mgh
Write down your known variables:



3.
Plug your numbers into the formula

4.
Mass is measured in kg so m = 1500 kg
Gravity is ALWAYS 9.8 m/s2 (this is the acceleration due to
gravity on Earth)
Height is measured in meters so h = 45 m
PE = (1500)(9.8)(45)
Solve your equation with units

PE = 661500 J – Don’t forget, Energy is measured in Joules!
Practice Problem 2: The crystal ball which drops on
New Year’s Even in Times Square has a mass of 5,585
kg. If the ball descends 43 meters, what is its change in
potential energy?
1. What are you trying to solve for?
POTENTIAL ENERGY
2. What equation do you need to use to solve this?
PE = mgh
3. Write down your known variables:




Mass is measured in kg so m = 5585 kg
Gravity is ALWAYS 9.8 m/s2 (this is the acceleration due to
gravity on Earth)
Height is measured in meters so h = 43 m
4. Plug your numbers into the formula

PE = (5585)(9.8)(43)
5. Solve your equation with units

PE = 2353519 J – Don’t forget, Energy is measured in
Joules!
Practice Problem 3: Calculate the height above the Earth’s
surface of a 5.5 kg object whose gravitational potential
energy is equal to 5290 J.
1. What are you trying to solve for?
HEIGHT
2. What equation do you need to use to solve this?
PE = mgh
3. Write down your known variables:




Mass is measured in kg so m = 55 kg
Gravity is ALWAYS 9.8 m/s2 (this is the acceleration due to gravity
on Earth)
Gravitational PE is measured in Joules so PE = 5290 J
4. Notice you’ll have to re-arrange the formula to get height by
itself.
PE/mg = h
5. Plug your numbers into the formula

h = 5290/(5.5*9.8)
6. h = 98.14 m – Don’t forget, height is measured in
meters!
One more thing….





When you are calculating gravitational potential
energy, your reference point is important.
A zero height position has to be assigned
Typically, the ground is considered to be a
position of zero height, but not always
The tabletop is usually assigned as the zero
height position, since a lot of our labs are done on
the tabletop
This means if the potential energy is based upon
the height of the tabletop. If the object goes
beyond the tabletop, your height could be a
negative value, which also makes your GPE value
negative.
One more thing….
Make the lowest point of motion
your reference point
Potential Energy
energy of position, location or condition
elastic potential energy:
energy stored in elastic materials as the
result of their stretching or compressing
Us = ½
•
2
kx
k – elastic constant in N/m
x - elongation or compression in m
Us – elastic potential energy in J
Imagine the spring constant as how
“stiff” or hard to pull the spring is.
Think about this…

A spring-loaded toy dart gun is used to shoot a dart straight
up in the air, and the dart reaches a maximum height of 24
m. The same dart is shot straight up a second time from the
same gun, but this time the spring is compressed only half
as far before firing. How far up does the dart go this time,
neglecting friction and assuming an ideal spring?
1. 96 m
2. 48 m
3. 24 m
4. 12 m
5. 6 m
6. 3 m
7. impossible to determine
Work-Energy Theorem
the net work done on an object is
equal to its change in kinetic energy
W KE
W=½m
2
v
- ½ m v0
2