Trevino, Denise – Midterm 1 – Due: Feb 16 2005, 10:00 pm – Inst: Turner
1
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The due time is Central
time.
001
(part 1 of 3) 10 points
Consider a force pulling 3 blocks along a rough
horizontal surface, where the masses are a
multiple of a given mass
m
, as shown in the
figure below.
The coefficient of the kinetic
friction is
μ
. The blocks are pulled by a force
of 48
μ m g
.
4
m
2
m
6
m
T
1
T
2
F
μ
Determine the acceleration.
1.
a
=
μ g
2.
a
= 4
μ g
3.
a
= 2
μ g
4.
a
= 6
μ g
5.
a
= 3
μ g
correct
6.
a
= 9
μ g
7.
a
= 5
μ g
8.
a
= 8
μ g
9.
a
= 7
μ g
Explanation:
m
1
m
2
m
3
T
1
T
2
F
μ
Given :
m
1
= 4
m ,
m
2
= 2
m ,
m
3
= 6
m ,
m
1
+
m
2
+
m
3
= 12
m ,
and
F
= 48
μ m g .
Basic Concept:
Newton’s 2nd law.
Solution:
Since all three blocks must have
the same acceleration (no block can “speed
ahead” or “lag behind”) Newton’s 2nd Law
applies to the whole system of three blocks.
(Take right to be positive.)
m
1
+
m
2
+
m
3
F
μ
(
m
1
+
m
2
+
m
3
)
g
F
f
F

F
f
= (
m
1
+
m
2
+
m
3
)
a
= 12
m a ,
where
F
f
=
μ
(
m
1
+
m
2
+
m
3
)
g
= 12
μ m g .
a
=
F

F
f
(
m
1
+
m
2
+
m
3
)
=
F

μ
(
m
1
+
m
2
+
m
3
)
g
(
m
1
+
m
2
+
m
3
)
=
48
μ m g

12
μ m g
12
m
=
36
μ g
12
=
3
μ g
.
002
(part 2 of 3) 10 points
The equation of motion for the central mass
2
m
is given by
1.
T
2
+
T
1

2
μ m g
= 2
m a
2.
T
2
+ 2
μ m g
= 2
m a
3.
T
2

T
1

2
μ m g
= 2
m a
correct
4.
T
2

T
1
+ 2
μ m g
= 2
m a
5.
T
2

2
μ m g
= 2
m a
6.
T
2

T
1

2
μ m g

4
μ m g
= 2
m a
7.
T
2
+
T
1

2
μ m g

4
μ m g
= 2
m a
8.
T
2
+
T
1
+ 2
μ m g

4
μ m g
= 2
m a
9.
T
2
+
T
1
+ 2
μ m g
= 2
m a
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Trevino, Denise – Midterm 1 – Due: Feb 16 2005, 10:00 pm – Inst: Turner
2
Explanation:
The acceleration is to the right.
m
2
T
1
T
2
μ m
2
g
The free body diagram for
m
2
shows
T
2

T
1

μ m
2
g
=
m
2
a
T
2

T
1

2
μ m g
=
2
m a
,
since
m
2
= 2
m .
003
(part 3 of 3) 10 points
Find the tension
T
1
in the rope between the
masses 4
m
and 2
m
in terms of
F
.
1.
T
1
=
F
2
2.
T
1
=
F
4
3.
T
1
=
F
6
4.
T
1
=
F
7
5.
T
1
=
F
9
6.
T
1
=
F
3
correct
7.
T
1
=
F
5
8.
T
1
=
F
8
9.
T
1
=
F
Explanation:
m
1
T
1
μ m
1
g
The free body diagram for
m
1
shows
T
1

μ m
1
g
=
m
1
a
= 4
m a
Since
F
f
1
=
μ m
1
g
= 4
μ m g
and from Part
1,
a
= 3
μ g ,
we have
T
1

4
μ m g
= 12
m μ g ,
so
T
1
= (12 + 4)
μ m g
T
1
= 16
μ m g
Since
F
= 48
μ m g
, we have
T
1
= 16
μ m g
F
48
μ m g
¶
=
16
48
F
=
F
3
.
004
(part 1 of 1) 10 points
At the beginning of a new school term, a
student moves a box of books by attaching a
rope to the box and pulling with a force of
F
= 83 N at an angle of 64
◦
, as shown in the
figure.
The acceleration of gravity is 9
.
8 m
/
s
2
.
The box of books has a mass of 15 kg and
the coefficient of kinetic friction between the
bottom of the box and the floor is 0
.
25.
15 kg
F
64
◦
μ
= 0
.
25
What is the acceleration of the box?
Correct answer: 1
.
21898 m
/
s
2
.
Explanation:
Given :
m
= 15 kg
,
μ
k
= 0
.
25
,
F
= 83 N
,
and
θ
= 64
◦
.
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 Spring '08
 Turner
 Physics, Force, Friction, Correct Answer, m/s, m/s2, Trevino

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