Version 052/AADBA – Test #2 – Antoniewicz – (56445)
1
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001
6.0 points
L
b
Mg
Two identical springs are in parallel. Each
spring has one oF its ends attached to the
ceiling and its other end attached to a weight.
The mass oF the springs is negligible. Consider
the situation in which the weight has a mass
M and the system is stretched by a length b.
Choose correct statements below
Ia. The sti±ness oF one spring is
M g
4
b
Ib. The sti±ness oF one spring is
M g
2
b
IIa. The amount oF stretch in one spring is
b
2
IIb. The amount oF stretch in one spring is
b
IIIa. The amount oF Force along one spring
is
M g
2
IIIb. The amount oF Force along one spring
is M g.
1.
Ia, IIb, IIIb
2.
Ib, IIa, IIIb
3.
Ia, IIa, IIIa
4.
Ia, IIb, IIIa
5.
Ib, IIb, IIIb
6.
Ia, IIa, IIIb
7.
Ib, IIb, IIIa
correct
8.
Ib, IIa, IIIa
Explanation:
The sti±ness oF the 2spring system For the
present setup is by defnition
k
=
M g
b
. With
the parallel setup, the Force is equally shared
between the two identical springs, i.e. each
spring experiences a Force oF
F
=
M g
2
. Since
the total stretch is given to be b, with the
parallel arrangement, each spring stretches
by the same amount.
The sti±ness oF one
spring is
M g
2
b
=
M g
2
b
.
002
6.0 points
A piece oF iron is Fastened on top oF a block oF
wood ²oating in water.
IF the block is inverted so the iron is below
the block oF wood, would the wood ²oat at
the same level, lower, or higher?
1.
Higher, because there is a buoyant Force
acting on the iron now.
correct
2.
It would ²oat lower, because the iron
displaces water and the overall water level
would rise.
3.
It depends on the ratio oF iron and wood
volumes.
4.
It would ²oat at the same level.
Explanation:
The ironwood unit displaces its combined
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2
weight and the same volume of water whether
the iron is on top or the bottom. When the
iron is on top, more wood is in the water.
When the iron is on the bottom, less wood is
in the water.
003
6.0 points
A ball of mass 350 g hangs from a spring whose
stiFness is 115 N
/
m. A string is attached to
the ball and you pull the string to the right,
so that the ball hangs motionless, as shown
in the ±gure.
In this situation, the spring
is stretched, and its length is 14 cm. What
would be the relaxed length of the spring, if
it were detached from the ball and laid on a
table? Use
g
= 9
.
8 m
/
s
2
.
14 cm
8 cm
1. 0.11465
2. 0.112717
3. 0.101251
4. 0.108795
5. 0.131846
6. 0.123672
7. 0.0951667
8. 0.12037
9. 0.0961802
10. 0.103656
Correct answer: 0
.
103656 m.
Explanation:
Let
θ
be the angle between the spring and
the vertical, and let
s
represent the amount
by which the spring has stretched under the
given circumstances. Then
sin
θ
=
8 cm
14 cm
and
cos
θ
=
11
.
4891 cm
14 cm
,
where 11
.
4891 cm is obtained from the
Pythagorean theorem. The vertical compo
nent of the spring tension must null out the
ball’s weight, so, using Hooke’s Law, we can
write
mg
=
k
s
s
cos
θ
⇒
s
=
k
s
14 cm
11
.
4891 cm
=
(350 g)(9
.
8 m
/
s
2
)(14 cm)
(115 N
/
m)(11
.
4891 cm)
= 0
.
0363444 m
.
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 Spring '08
 Turner
 Physics, Force, kg, Iiia, Version 052/AADBA

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