honea (cth632) – H5Forces2 – Avram – (1955)
1
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18
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before answering.
001
10.0 points
Two blocks
m
1
(11 kg) and
m
2
(17 kg) are
connected by strings, as shown in the dia
gram. String 2 breaks when
T
2
= 4
.
9179 N.
Find
T
1
when string 2 breaks.
m
1
m
2
T
2
T
1
Correct answer: 12
.
5183 N.
Explanation:
Find
T
1
when
T
2
breaks at 4
.
9179 N.
summationdisplay
F
x
on m
1
=
T
2
=
m
1
a
⇒
a
=
T
2
m
1
=
4
.
9179 N
11 kg
= 0
.
447082 m
/
s
2
summationdisplay
F
x
on the whole system
=
T
1
= (
m
1
+
m
2
)
a
= (
m
1
+
m
2
)
T
2
m
1
= (
m
1
+
m
2
m
1
)
T
2
= (1 +
m
2
m
1
)
T
2
= (1 + 1
.
54545) 4
.
9179 N
= (2
.
54545) (4
.
9179 N) = 12
.
5183 N
002
(part 1 of 2) 10.0 points
A light, inextensible cord passes over a light,
frictionless pulley with a radius of 14 cm. It
has a(n) 23 kg mass on the left and a(n)
3
.
1 kg mass on the right, both hanging freely.
Initially their center of masses are a vertical
distance 1
.
9 m apart.
The acceleration of gravity is 9
.
8 m
/
s
2
.
1
.
9 m
14 cm
ω
23 kg
3
.
1 kg
At what rate are the two masses accelerat
ing when they pass each other?
Correct answer: 7
.
47203 m
/
s
2
.
Explanation:
Let :
R
= 14 cm
,
m
1
= 3
.
1 kg
,
m
2
= 23 kg
,
h
= 1
.
9 m
,
and
v
=
ω R .
Consider the free body diagrams
23 kg
3
.
1 kg
T
T
m
2
g
m
1
g
a
a
Since the larger mass will move down and
the smaller mass up, we can take motion
downward as positive for
m
2
and motion up
ward as positive for
m
1
.
Apply Newton’s
second law to
m
1
and
m
2
respectively and
then combine the results:
For mass 1:
summationdisplay
F
1
:
T

m
1
g
=
m
1
a
(1)
For mass 2:
summationdisplay
F
2
:
m
2
g

T
=
m
2
a
(2)
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honea (cth632) – H5Forces2 – Avram – (1955)
2
We can add Eqs. (1) and (2) above and obtain:
m
2
g

m
1
g
=
m
1
a
+
m
2
a
a
=
m
2

m
1
m
1
+
m
2
g
=
23 kg

3
.
1 kg
23 kg + 3
.
1 kg
(9
.
8 m
/
s
2
)
= 7
.
47203 m
/
s
2
.
003
(part 2 of 2) 10.0 points
What is the tension in the cord when they
pass each other?
Correct answer: 53
.
5433 N.
Explanation:
T
=
m
1
(
g
+
a
)
= (3
.
1 kg) (9
.
8 m
/
s
2
+ 7
.
47203 m
/
s
2
)
= 53
.
5433 N
.
004
10.0 points
A block is pushed up a frictionless incline by
an applied horizontal force as shown.
The acceleration of gravity is 9
.
8 m
/
s
2
.
5 kg
F
= 38 N
38
◦
What is the magnitude of the resulting ac
celeration of the block?
Correct answer: 0
.
0446044 m
/
s
2
.
Explanation:
Basic Concepts:
summationdisplay
vector
F
=
mvectora
Solution:
Consider the forces acting on the
block and their components along the incline.
The horizontal pushing force
vector
F
has upthe
incline component +
F
cos
θ
while the weight
force
vector
W
=
mvectorg
has uptheincline component

mg
sin
θ
. Consequently,
Ma
=
F
net
=
F
cos
θ

M g
sin
θ,
a
=
F
cos
θ
M

g
sin
θ
=
(38 N) cos 38
◦
5 kg

(9
.
8 m
/
s
2
) sin 38
◦
= 0
.
0446044 m
/
s
2
.
005
10.0 points
A 4
.
3 kg object hangs at one end of a rope that
is attached to a support on a railroad boxcar.
When the car accelerates to the right, the
rope makes an angle of 36
◦
with the vertical
The acceleration of gravity is 9
.
8 m
/
s
2
.
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 Spring '10
 Avram
 Physics, Force, Friction, Correct Answer, kg, Avram –

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