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Friday, November
25 −
Lecture 32 :
Applications : Work and energy
(Refers to Section 7.4 in
your text)
Students who have mastered the content of this lecture know
:
what the “amount of work W performed by F(x) on
the mass m over the distance d ”
is.
Students who have practiced the techniques presented in this lecture will be able to
:
Compute the amount of work
done when a forces F(x) is exerted of an object over a distance d.
32.1
Work performed by a constant force on a mass over a distance d
.
−
Suppose a constant
force
F
is exerted on an object providing it with a
constant
acceleration
a
over a distance
d
.
Suppose its initial velocity is
v
i
while its final velocity is
v
f
and the force is exerted over the
period of time
t
f
–
t
i
=
∆
t
.
We will suppose that
F
is a “net force” in the sense that any force opposing the movement is
already taken into account inside
F
Since the acceleration
a
(
t
) is constant and that acceleration at time
t
is the rate of change of
the velocity at time
t
, then we can write
The distance
d
covered by the mass over the period of time
∆
t
is the “average velocity”
×
“the
time
∆
t
”. That is,
Recall Newton’s second law of motion states that
F
=
ma
.
The force will give to the mass
m
kinetic energy
K
= (1/2)
mv
2
. The energy provided to
m
by
this force is then
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View Full Documentwhere
K
f
is the final kinetic energy and
K
i
is the kinetic already possessed by the mass when
this net force starts acting on it. Then
So the net increase in energy
∆
K
provided to
m
by this net force
F
on a distance
d
is
∆
K
=
Fd
The units for
Fd
must be newtonsmeters. We define
1 joule = 1 newtonmeter.
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 Spring '11
 RobertAndre
 Calculus

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