Muraj, Hamza – Homework 9 – Due: Feb 10 2006, 4:00 am – Inst: Florin
1
This printout should have 11 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
be±ore answering.
The due time is Central
time.
001
(part 1 o± 1) 10 points
In a certain system o± units, 37
.
8 thumbs = 1
meter, 1 pause = 25
.
6 s , and 1 glob = 20
.
5 kg.
The unit o± ±orce in this system is a perk (p).
How many kiloperks (kp) are in 49
.
5 N?
Correct answer: 59
.
8168 kp.
Explanation:
Basic concept:
The use o± conversion ±actors.
1 N = 1
kg
·
m
s
2
1 perk = 1
glob
·
thumb
pause
2
Solution:
N =
kgm
s
2
·
1glob
20
.
5 kg
·
37
.
8 thumbs
1m
·
µ
25
.
6 s
1pause
¶
2
≡
glob
·
thumb
pause
2
·
1kp
1000p
≡
kp
002
(part 1 o± 1) 10 points
A 4
.
2 g bullet leaves the muzzle o± a ri²e with
a speed o± 301 m
/
s.
What total constant ±orce is exerted on the
bullet while it is traveling down the 0
.
86 m
long barrel o± the ri²e?
Correct answer: 221
.
235 N.
Explanation:
Given :
m
= 4
.
2 g = 0
.
0042 kg
,
v
= 301 m
/
s
,
and
Δ
x
= 0
.
86 m
.
The acceleration o± the bullet is given by
v
2
=
v
2
i
+ 2
a
(Δ
x
) = 2
a
(Δ
x
)
since
v
i
= 0.
a
=
v
2
2(Δ
x
)
and the ±orce exerted on the bullet is
X
F
=
ma
=
m
v
2
2(Δ
x
)
=
(0
.
0042 kg)(301 m
/
s)
2
2(0
.
86 m)
=
221
.
235 N
.
003
(part 1 o± 1) 10 points
At the beginning o± a new school term, a
student moves a box o± books by attaching a
rope to the box and pulling with a ±orce o±
F
= 89
.
4 N at an angle o± 63
◦
, as shown in the
fgure.
The acceleration o± gravity is 9
.
8 m
/
s
2
.
The box o± books has a mass o± 14 kg and
the coe³cient o± kinetic ±riction between the
bottom o± the box and the ²oor is 0
.
34.
14 kg
F
63
◦
μ
= 0
.
34
What is the acceleration o± the box?
Correct answer: 1
.
50155 m
/
s
2
.
Explanation:
Given :
m
= 14 kg
,
μ
k
= 0
.
34
,
F
= 89
.
4 N
,
and
θ
= 63
◦
.
Basic Concepts:
Vertically there is no
acceleration, so
F
net
=
X
F
up

X
F
down
= 0
Horizontally,
F
net
=
ma
=
X
F
right

X
F
left
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View Full DocumentMuraj, Hamza – Homework 9 – Due: Feb 10 2006, 4:00 am – Inst: Florin
2
Solution
Consider the free body diagram
for the situation.
14 kg
F
63
◦
N
mg
μmg
The tension in the rope can be resolved
into a horizontal component of
F
cos
θ
and a
vertical component of
F
sin
θ
. The box is in
equilibrium vertically, so
F
net
=
N
+
F
sin
θ

mg
= 0
N
=
mg

F
sin
θ .
The frictional force is
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 Spring '09
 Turner
 Force, Friction, Correct Answer, Muraj, Sdown

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