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68
Chapter 6
6.29.
IDENTIFY
and
SET UP:
Use Eq.(6.8) to calculate
k
for the spring. Then Eq.(6.10), with
1
0,
x
=
can be used to
calculate the work done to stretch or compress the spring an amount
2
.
x
EXECUTE:
Use the information given to calculate the force constant of the spring.
x
Fk
x
=
gives
160 N
3200 N/m
0.050 m
x
F
k
x
==
=
(a)
(3200 N/m)(0.015 m)
48 N
x
x
=
(3200 N/m)(
0.020 m)
64 N
x
x
−
=
−
(magnitude 64 N)
(b)
22
11
(3200 N/m)(0.015 m)
0.36 J
Wk
x
=
(3200 N/m)(
0.020 m)
0.64 J
x
−
=
Note that in each case the work done is positive.
EVALUATE:
The force is not constant during the displacement so Eq.(6.2)
cannot
be used. A force in the
x
+
direction is required to stretch the spring and a force in the opposite direction to compress it. The force
x
F
is in the
same direction as the displacement, so positive work is done in both cases.
6.30.
IDENTIFY:
The magnitude of the work can be found by finding the area under the graph.
SET UP:
The area under each triangle is 1/2 base height
×
.
0
x
F
>
, so the work done is positive when
x
increases
during the displacement.
EXECUTE:
(a)
1/2 (8 m)(10 N)
40 J
=
.
(b)
1/2 (4 m)(10 N)
20 J
=
.
(c)
1/2 (12 m)(10 N)
60 J
=
.
EVALUATE:
The sum of the answers to parts (a) and (b) equals the answer to part (c).
6.31.
IDENTIFY:
Use the workenergy theorem and the results of Problem 6.30.
SET UP:
For
0
x
=
to
8.0 m
x
=
,
tot
40 J
W
=
. For
0
x
=
to
12.0 m
x
=
,
tot
60 J
W
=
.
EXECUTE:
(a)
(2)(40 J)
2.83 m/s
10 kg
v
(b)
(2)(60 J)
3.46 m/s
10 kg
v
.
EVALUATE:
F
"
is always in the
direction.
x
+
For this motion
F
"
does positive work and the speed continually
increases during the motion.
6.32.
IDENTIFY:
The force has only an
x
component and the motion is along the
x
direction, so
2
1
x
x
x
WF
d
x
=
∫
.
SET UP:
1
0
x
=
and
2
6.9 m
x
=
.
EXECUTE:
The work you do with your changing force is
2
1
2
( )
( 20.0 N)
(3.0 N/m)
( 20.0 N) 
(3.0 N/m)(
/2) 
xx
x
x
x
x
x
x
x
d
x
d
x
x
d
x
x
x
−
−
=
−
−
∫∫
∫
138 N m
71.4 N m
209 J
W
=−
⋅
−
⋅
.
EVALUATE:
The work is negative because the cow continues to move forward (in the
direction
x
+
) as you vainly
attempt to push her backward.
6.33.
IDENTIFY:
Apply Eq.(6.6) to the box.
SET UP:
Let point 1 be just before the box reaches the end of the spring and let point 2 be where the spring has
maximum compression and the box has momentarily come to rest.
EXECUTE:
tot
2
1
WK
K
2
1
10
2
,
K
mv
=
2
0
K
=
Work is done by the spring force.
2
1
tot
2
2
,
x
where
2
x
is the amount the spring is compressed.
20
kx
mv
−=
−
and
/
(3.0 m/s) (6.0 kg)/(7500 N/m)
8.5 cm
xvm
k
=
EVALUATE:
The compression of the spring increases when either
0
v
or
m
increases and decreases when
k
increases
(stiffer spring).
6.34.
IDENTIFY:
The force applied to the springs is
x
x
=
. The work done on a spring to move its end from
1
x
to
2
x
is
21
x
k
x
. Use the information that is given to calculate
k
.
SET UP:
When the springs are compressed 0.200 m from their uncompressed length,
1
0
x
=
and
2
0.200 m
x
.
When the platform is moved 0.200 m farther,
2
x
becomes
0.400 m
−
.
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View Full DocumentWork and Kinetic Energy
69
EVALUATE:
(a)
22
2
21
(
8
0
.
0
J
)
4000 N/m
(0.200 m)
0
W
k
xx
==
=
−−
.
(4000 N/m)( 0.200 m)
800 N
x
Fk
x
−
=
−
. The
magnitude of force that is required is 800 N.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Work

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