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Physics 204A Class Notes
8-1
Section 8 – Work & Kinetic Energy
Section Outline
1.
The Work Done by a Constant Force
2.
Review of the Vector Dot Product
3.
The Work Done by a Varying Force
4.
The Work-Energy Theorem
5.
The Definition of Power
Why do objects do what they do?
Two ideas that help answer this question are force and linear
momentum.
However, both force and momentum have additional mathematical complications because
they are vectors.
Energy is another answer.
Energy has the advantage of being a scalar quantity.
As
with momentum, we will be able to build a conservation law that is a very powerful tool for
understanding why objects do what they do.
1. The Work Done by a Constant Force
Think about the last time you had to push a car.
Perhaps, the brake was on and you probably noticed
that the force you exerted resulted in no acceleration
of the car.
Maybe you remembered to release the
brake and the car slowly began to move.
The point is
that,
while
all
forces
in
principle
can
cause
acceleration, in some situations they just don’t result
in the object accelerating.
At the right is a block being pulled across a frictionless table by a
constant force, F, that is directed at an angle,
θ
, above the horizontal.
The component of the force that is along the motion, F
||
, causes the block
to accelerate, while the component perpendicular to the table, F
⊥
, does
not.
It is generally true that forces along the direction of motion cause
acceleration.
In analogy to developing linear momentum, where we defined a quantity called impulse which is the
force multiplied by the time, we define a quantity called “work” which is the force multiplied by the
distance.
However, we only count the component of the force along the motion of the object, so
work done by a constant force is defined as,
!
W
"
F
||
!
s
,
where
∆
s is the distance the object moves.
The units of work are the units of force times the units of distance so,
[
!
W]
=
[F][s]
=
N
"
m
.
This unit has its own name, 1N·m
≡
1 Joule = 1J.
!
F
F
F
"
||
!
s

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Physics 204A Class Notes
8-2
Example 8.1:
A 50.0N force is exerted at 30.0˚ above horizontal on a 20.0kg suitcase.
The
suitcase moves at a constant speed for 4.00m.
Find the work done by each force that acts on it.
Given: F
p
= 50.0N,
θ
= 50.0˚, m = 20.0kg, and
∆
s = 4.00m.
Find: W
p
= ?, W
kf
= ?, W
n
= ?, and W
g
= ?
To find the size of each force, apply the Second Law to each
direction separately,
!
F
x
=
ma
x
"
F
p
cos
#$
F
kf
=
0
"
F
kf
=
F
p
cos
#
,
!
F
y
=
ma
y
"
F
n
#
F
g
+
F
p
sin
$
=
0
"
F
n
=
F
g
#
F
p
sin
$
.
Using the mass/weight rule and plugging in the numbers,
F
g
=
mg
=
(20.0)(9.80)
!
F
g
=
196N
,
F
kf
=
50.0 cos30.0˚
!
F
kf
=
43.3N
,
F
n
=
F
g
!
F
p
sin
"
=
196
!
50.0sin30.0˚
#
F
n
=
171N
.
Using the definition of work,
!
W
"
F
||
!
s
#
W
p
=
(F
p
cos
$
)
!
s
=
(43.3)(4)
#
W
p
=
173J
.
Note that only the component along the motion counts.
!
W
"
F
||
!
s
#
W
kf
=
$
F
kf
!
s
=
(
$
43.3)(4)
#
W
kf
=
!
173J
.
Since the friction and motion are opposite, negative work is done.

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