Download work - cloudfront.net

CHAPTER 5
5-1 work
Objectives
• Recognize the difference between the scientific and
ordinary definitions of work.
• Define work by relating it to force and displacement.
• Identify where work is being performed in a variety of
situations.
• Calculate the net work done when many forces are
applied to an object.
What is work ?
• When a force acts upon an object to cause a
displacement of the object, it is said that work was done
upon the object. There are three key ingredients to work force, displacement, and 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.
There are several good examples of work that can be
observed in everyday life - a horse pulling a plow through
the field, a father pushing a grocery cart down the aisle of
a grocery store, a freshman lifting a backpack full of
books upon her shoulder, a weightlifter lifting a barbell
above his head, an Olympian launching the shot-put, etc.
In each case described here there is a force exerted upon
an object to cause that object to be displaced.
Examples of work
• Read the following five statements and determine whether or
not they represent examples of work.
• 1. A teacher applies a force to a wall and
becomes: NO,
exhausted.
• Solution
no displacement
• 2 .A book falls off a table and free falls to the ground.
• Solution: yes, there is gravity acting on the object which causes
displacement
• 3. A waiter carries a tray full of meals above his head by one
arm straight across the room at constant speed. (Careful! This is
a very difficult question that will be discussed in more detail
later.)
• Solution: No, no displacement
• 4. rocket accelerates through space.
• solution: yes, displacement
Work
• Work is equal to the magnitude of the force f, times the
magnitude of the displacement d.
• W=f*d
Definition of work
Work
• where F is the force, d is the displacement, and the angle (theta) is
defined as the angle between the force and the displacement vector.
Perhaps the most difficult aspect of the above equation is the angle
"theta." The angle is not just any 'ole angle, but rather a very specific
angle. The angle measure is defined as the angle between the force and
the displacement. To gather an idea of it's meaning, consider the
following three scenarios.
• Scenario A: A force acts rightward upon an object as it is displaced
rightward. In such an instance, the force vector and the displacement
vector are in the same direction. Thus, the angle between F and d is 0
degrees.
• Scenario B: A force acts leftward upon an object that is displaced
rightward. In such an instance, the force vector and the displacement
vector are in the opposite direction. Thus, the angle between F and d is
180 degrees.
• Scenario C: A force acts upward on an object as it is displaced rightward.
In such an instance, the force vector and the displacement vector are at
right angles to each other. Thus, the angle between F and d is 90
degrees.
Work
Units of work
• Whenever a new quantity is introduced in physics, the standard
•
•
•
•
•
metric units associated with that quantity are discussed. In the
case of work (and also energy), the standard metric unit is the
Joule (abbreviated J). One Joule is equivalent to one Newton
of force causing a displacement of one meter. In other words,
The Joule is the unit of work.
1 Joule = 1 Newton * 1 meter
1J=1N*m
In fact, any unit of force times any unit of displacement is
equivalent to a unit of work. Some nonstandard units for work
are shown below. Notice that when analyzed, each set of units
is equivalent to a force unit times a displacement unit.
The meaning of negative work
• On occasion, a force acts upon a moving object to hinder a
displacement. Examples might include a car skidding to a stop
on a roadway surface or a baseball runner sliding to a stop on
the infield dirt. In such instances, the force acts in the direction
opposite the objects motion in order to slow it down. The force
doesn't cause the displacement but rather hinders it. These
situations involve what is commonly called negative work. The
negative of negative work refers to the numerical value that
results when values of F, d and theta are substituted into the
work equation. Since the force vector is directly opposite the
displacement vector, theta is 180 degrees. The cosine(180
degrees) is -1 and so a negative value results for the amount of
work done upon the object. Negative work will become
important (and more meaningful) in Lesson 2 as we begin to
discuss the relationship between work and energy.
•
Work
• In summary, work is done when a force acts upon an
object to cause a displacement. Three quantities must be
known in order to calculate the amount of work. Those
three quantities are force, displacement and the angle
between the force and the displacement.
• lets Investigate
Example#1
• How much work is done in a vacuum cleaner pulled 3m
by a force of 50 N at an angle 30 degree above the
horizontal?
Example#2
• Ben Travlun carries a 200-N suitcase up three flights of stairs
•
•
•
•
(a height of 10.0 m) and then pushes it with a horizontal force
of 50.0 N at a constant speed of 0.5 m/s for a horizontal
distance of 35.0 meters. How much work does Ben do on his
suitcase during this entire motion?
Solution
The motion has two parts: pulling vertically to displace the
suitcase vertically (angle = 0 degrees) and pushing horizontally
to displace the suitcase horizontally (angle = 0 degrees).
For the vertical part, W = (200 N) * (10 m) * cos (0 deg) = 2000
J. For the horizontal part, W = (50 N) * (35 m) * cos (0 deg) =
1750 J.
The total work done is 3750 J (the sum of the two parts).
Example#3
• A force of 50 N acts on the block at the angle shown in the
diagram. The block moves a horizontal distance of 3.0 m.
How much work is done by the applied force?
• Solution
• W = F * d * cos(Theta) W = (50 N) * (3 m) * cos (30
degrees) = 129.9 Joules
Example#4
• How much work is done by an applied force to lift a 15-
Newton block 3.0 meters vertically at a constant speed?
• solution
To lift a 15-Newton block at constant speed, 15-N of force
must be applied to it (Newton's laws). Thus,
W = (15 N) * (3 m) * cos (0 degrees
) = 45 Joules
Example#5
• A tired squirrel (mass of 1 kg) does push-ups by applying
a force to elevate its center-of-mass by 5 cm. Estimate the
number of push-ups that a tired squirrel must do in order
to do a approximately 5.0 Joules of work.
The squirrel applies a force of approximately 10 N (9.8 N to be exact) to raise its
body at constant speed. The displacement is 0.05 meters. The angle between the
upward force and the upward displacement is 0 degrees. The work for 1 push-up is
approximately
W = 10 N * 0.05 m * cos 0 degrees = 0.5 Joules
If the squirrel does a total of 5.0 Joules of work, then it must have done about 10
push-ups.
Homework
• Do problems 1-4 in your book page 156.
Closure
• Today we learned about work
• Next we are going to learned about energy
Have a great day