baum (lmb2768) – HW05 – distler – (56295)
1
This printout should have 21 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
There are various ways to do part 2 oF the
”Wind and Pendulum” problem. The easiest
is to fnd the angle at which the three Forces
on the ball (tension, gravity and ±) add up to
zero.
In part 1 oF ”Pulling Two Masses”, you
are asked about the ”work done against Fric
tion”. This is very Funny phraseology. The
Friction does negative work, and what you are
being askedFor is the magnitude (a positive
number) oF the work done by Friction.
001
10.0 points
Ab
lockslidingonahorizontalsurFacehasan
initial speed oF 0.5 m/s. The block travels a
distance oF 1 m as it slows to a stop.
What distance would the block have trav
eled iF its initial speed had been 1 m/s?
1.
1m
2.
2m
3.
3m
4.
4m
correct
5.
more inFormation is needed to answer the
question
6.
0.5 m
Explanation:
W
nc
=

f
k
L
=0

1
2
mv
0
2

μmgL
=

1
2
0
2
L
=
v
0
2
2
μg
So the distance is proportional to the square
oF the initial speed.
002 (part 1 of 2) 10.0 points
Aba
l
lhav
ingma
s
s1
.
1kgi
sconn
e
c
t
edby
astr
ingo
Flength0
.
8mtoap
ivotpo
intand
held in place in a vertical position. A constant
wind Force oF magnitude 17
.
85 N blows From
leFt to right.
Pivot
Pivot
FF
(a)
(b)
H
m
m
L
L
IF the mass is released From the vertical po
sition, what maximum height above its initial
position will it attain? Assume that the string
does not break in the process. The accelera
tion oF gravity is 9
.
8m
/
s
2
.
Correct answer: 1
.
1724 m.
Explanation:
Let :
m
=1
.
1kg
,
L
.
,
F
=17
.
85 N
,
and
g
=9
.
/
s
2
.
Take the original point to be the one where
the ball is released and the fnal point where
its upward swing stops at height
H
.Thenthe
horizontal displacement is
x
=
±
L
2

(
L

H
)
2
Since the Force is purely horizontal, it does
work
W
=
²
F
·
d
s
=
F
³
2
LH

H
2
.
Applying the workenergy theorem,
(
K
+
U
)
i
+
W
wind
=(
K
+
U
)
f
0+0+
F
³
2
LH

H
2
=0+
mgH.
H
=
2
L
1+
´
mg
F
µ
2
·
F
2
F
2
=
2
LF
2
F
2
+(
mg
)
2
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View Full Documentbaum (lmb2768) – HW05 – distler – (56295)
2
=
2(0
.
8m)(17
.
85 N)
2
(17
.
85 N)
2
+(1
.
1kg)
2
(9
.
8m
/
s
2
)
2
=
1
.
1724 m
.
003 (part 2 of 2) 10.0 points
What will be the equilibrium height of the
mass?
Correct answer: 0
.
38643 m.
Explanation:
Considering components of the force equi
librium
±
F
=0,

T
sin
θ
+
F
=0 and
T
cos
θ

mg
=0
.
Dividing these,
tan
θ
=
F
θ
=arctan
²
F
³
´
17
.
85 N
(1
.
1kg)(9
.
/
s
2
)
µ
=58
.
8713
◦
,
so the equilibrium height is
h
=
L
(1

cos
θ
)
=(0
.
8m)(1

cos 58
.
8713
◦
)
=
0
.
38643 m
.
004
10.0 points
ANavyjetofmass10kglandsonana
ircraft
carrier and snags a cable to slow it down.
If the cable acts like a spring that has a
spring constant of 160 N
/
mandtheaircraftis
stopped by this spring in 25 m after hooking
onto the cable, what is the speed of the jet
when it Frst hooks on to the cable (assume
the jet does no breaking on its own)?
1.
70 m/s
2.
100 m/s
correct
3.
50 m/s
4.
150 m/s
5.
20 m/s
Explanation:
Let :
m
=10kg
,
k
=160N
/
m
,
and
x
=25m
Energy is conserved:
K
i
+
U
i
=
K
f
+
U
f
The initial kinetic energy is
K
i
=
1
2
mv
2
i
,
where
v
i
is the speed of the jet when it Frst
hooks on to the cable. The initial potential
energy stored in the cable is
U
i
since it is initially unstretched.
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 Spring '09
 IRENEPOLYCARPOU
 Physics, Energy, Mass, Potential Energy, Correct Answer, kg

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