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Work
Physics 11
Comprehension Check
1. What is the impulse given to a golf ball of
mass 45.9g if it starts at rest and attains a
final velocity of 35m/s?
2. If the golf ball in problem 1 was in contact
with the golf club for 0.027s, what force
acted on the golf ball?
3. If there is no acceleration is there
momentum? Is there impulse?
4. Suppose that a 75.0kg goalkeeper catches
a 0.40 kg ball moving at 32m/s. With
what forward velocity must the goalkeeper
jump when she catches the ball so that she
and the ball have zero horizontal velocity?
Comprehension Check
1. 1.6Ns
2. 6.0x101N
3. Only momentum can have 0
acceleration. A force is needed for
impulse so there must be
acceleration.
4. V = + 0.17 m/s
Page 203, questions 33 to 35
Page 209, questions 37 to 46
Work
 In common language, work can mean
a variety of different things, however
when we describe work from a
scientific standpoint, work has a very
precise definition
 This means we must be careful not to
confuse work as used in the English
language and the work we will
determine in physics
Work – The physics definition
 Work is the transfer of mechanical energy (total
kinetic and potential energy). (We will learn more
about energy soon!)
 Work is done only if an object moves.
 When the work is done upon the object, that object
gains energy
 Work is only done on an object when the force
and displacement are in the same direction.
Work




The symbol is W
Units: Joule (J)
1 J = 1 Nm
Work is a
scalar***
 
W  F  d
W  Fd cos 
Zero Work Conditions
 How can we make work equal zero?
 1) Apply a force that does not cause
motion
 Example: Holding an object at the same
height for hours is not doing any work you may get tired but you are doing no
work on the object.
Zero Work Conditions
 2) Uniform motion in the absence of
a net force
 Example: If an object is already in
motion, it will continue in that same
motion (Newton’s First Law). If a
hockey puck is sliding across the ice
at a constant speed, no work is being
done.
Zero Work Conditions
 3) Applying a force that is perpendicular
to the motion
 Example: You are carrying a book down
the hallway. You are lifting the book (force
is upwards) but your motion is forwards
(perpendicular). Therefore there is no work
being done on the textbook by the person
once you are moving.
What about this situation… work or
no work?
 A) If you are pushing a grocery cart
through the store.
 B) If you are pushing a car and it is going
forward?
 C) “” going backward?
 D) “” not moving?
 E) If you are driving forward on cruise
control
 F) Swinging a ball on a rope in a circle
a) Yes – moving the cart, force is pushing the
cart in the same direction
• Yes – force is in same direction
• Yes since the car is moving in a direction
opposite this is called “negative work”
(discuss more later)
• No work done as car is not moving
• No work done as car is moving at constant
speed (so no net force)
• No work as the string’s tension force acts
in a direction perpendicular to the ball
Example 1: (page 220)
 A student is rearranging her room.
She decides to move her desk across
the room a distance of 3.00m. She
moves the desk at a constant velocity
by exerting a horizontal force of
200N. Calculate the amount of work
done on the desk by the student.
 W = Fd
 W = 200 x 3
 W = 600 J
Comprehension Check
1. How much work is done if you push on
a wall with 3500N but the wall does not
move?
2. How much work is done by you on the
book if you are carrying the book down
the hall at constant velocity?
3. How much work is done by you if you
push a box that has a mass of 50kg
down the hallway 45m with a force of
25N?
Comprehension Check
1. 0
2. 0
3. W = Fd = 25 x 45 =1125 N = 1200N
Practice Problems
 Page 221, questions 1, 2, 3
Example 2:
 An applied force of 20. N accelerates
a block across a level, frictionless
surface from rest to a velocity of
8.0 m/s in a time of 2.5 s.
 Calculate the work done by this force.
Practice Problems
 Page 225
 4-10
 Remember, work is done if a force is
exerted in the direction of motion.
 If you are pushing or pulling
something at an angle, only the
component that acts in the direction
of motion is doing work.
 W = F d cos Θ
Example 3:
 A person is doing work on the lawn mower
by pushing with 105N it at an angle of 40°
to the horizontal. If the person pushes the
mower for 5.00m, how much work is being
done on the mower? WHAT EQUATION DO
WE USE? WHY?
 W = F d cos Θ
 We use this equation because the handle
on the lawn mower is at an angle so we
only need the Fx value (cos).
 W = (105)(5)cos40 =
Practice Problems
1. If you pull a crate with a force of 550N
at an angle of 35° to the horizontal and
it moves 25m horizontally, how much
work was done?
(1.1x104J)
 Page 235, questions 16, 17
Using Graphs to Calculate Work
Estimating Work from a Graph
Positive and Negative Work
 When we consider work it is a scalar
so lacks direction
 How is it possible to have positive and
negative work?
Positive and Negative Work
 Positive work
occurs when the
angle between the
force and
displacement is 0°90°
 “Negative work”
occurs when the
angle between the
force and
displacement is
90°-180°
F
d
F
d
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.
Example:
 Imagine a weight lifter. She must lift
the bar-bell up and then lower it
down. Overall, she has done NO
WORK but she has done positive work
when she lifts the bar-bell and
negative work (the same amount) to
lower it.
Practice Problems
 Page 229
 Questions 11 (a, b, c only), 12, 13 in
Practice
 Page 235
 14, 15, 18
 Page 235
 Section Review q 3, 4, 5