Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004
1
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Chapters 6 and 7 problems.
001
(part 1 oF 1) 0 points
A cheerleader liFts his 70
.
9 kg partner straight
up o± the ground a distance oF 0
.
594 m beFore
releasing her.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
IF he does this 27 times, how much work has
he done?
Correct answer: 11143
.
5 J.
Explanation:
The work done in liFting the cheerleader
once is
W
1
=
mg h
= (70
.
9 kg)(9
.
8 m
/
s
2
)(0
.
594 m)
= 412
.
723 J
.
The work required to liFt her
n
= 27 times is
W
=
nW
1
= (27)(412
.
723 J) = 11143
.
5 J
.
002
(part 1 oF 5) 0 points
In the
parallel
spring system, the springs
are positioned so that the 44 N weight
stretches each spring equally. The spring con
stant For the leFthand spring is 3
.
5 N
/
cm and
the spring constant For the righhand spring
is 5
.
5 N
/
cm
.
3
.
5N
/
cm
5
44 N
How Far down will the 44 N weight stretch
the springs?
Correct answer: 4
.
88889 cm.
Explanation:
Let :
k
1
= 3
.
5 N
/
cm
,
k
2
= 5
.
5 N
/
cm
,
and
W
= 44 N
,
The springs stretch the same amount
x
be
cause oF the way they were positioned.
Then
F
1
=
k
1
x
and
F
2
=
k
2
x
, so the Force
equation For the suspended mass is
X
F
up
=
X
F
down
k
1
x
+
k
2
x
=
W
x
=
W
k
1
+
k
2
=
44 N
3
.
5 N
/
cm + 5
.
5 N
/
cm
= 4
.
88889 cm
.
Dimensional analysis
For
x
:
N
N
/
cm
= N
·
cm
N
= cm
003
(part 2 oF 5) 0 points
In this same
parallel
spring system, what is
the e±ective combined spring constant
k
parallel
oF the two springs?
Correct answer: 9 N
/
cm.
Explanation:
IF the springs were one spring, that spring
would react with a Force
F
=

k x
where
F
=

W
due the law oF action and reaction,
we have
k
parallel
=

W

x
=
W
x
=
44 N
4
.
88889 cm
= 9 N
/
cm
,
which happens to be the sum oF the individual
constants
k
parallel
=
k
1
+
k
2
= 3
.
5 N
/
cm + 5
.
5 N
/
cm
= 9 N
/
cm
.
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View Full DocumentAnswer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004
2
004
(part 3 of 5) 0 points
Now consider the same two springs in series.
3
.
5N
/
cm
5
44 N
What distance will the spring of constant
3
.
5 N
/
cm stretch?
Correct answer: 12
.
5714 cm.
Explanation:
In the
series
system, the springs stretch
a diFerent amount, but each carries the full
weight
W
= 44 N.
W
=
k
1
x
1
and
x
1
=
W
k
1
=
44 N
3
.
5 N
/
cm
= 12
.
5714 cm
.
005
(part 4 of 5) 0 points
In this same
series
spring system, what dis
tance will the spring of constant 5
.
5 N
/
cm
stretch?
Correct answer: 8 cm.
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 Spring '08
 RIJSSENBEEK
 Force, Mass, Work, Correct Answer, David McIntyre

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