hamdeh (ah37785) – HW 5 – opyrchal – (111213)
1
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printout
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have
10
questions.
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before answering.
001
10.0 points
A block is released from rest on an inclined
plane and moves 2
.
7 m during the next 3
.
8 s.
The acceleration of gravity is 9
.
8 m
/
s
2
.
13 kg
μ
k
28
◦
What is the coe
ffi
cient of kinetic friction
μ
k
for the incline?
Correct answer: 0
.
488492.
Explanation:
Given :
m
= 13 kg
,
= 2
.
7 m
,
θ
= 28
◦
,
and
t
= 3
.
8 s
.
Consider the free body diagram for the
block
m g
sin
θ
N
=
m g
cos
θ
μ
N
a
m g
The acceleration can be obtained through
kinematics. Since
v
0
= 0,
=
v
0
t
+
1
2
a t
2
=
1
2
a t
2
a
=
2
t
2
(1)
Applying Newton’s Second Law of Motion
F
i
=
m a
and Eq. 1, the sine component
of the weight acts down the plane and friction
acts up the plane. The block slides down the
plane, so
m a
=
m g
sin
θ

μ
k
m g
cos
θ
2
t
2
=
g
sin
θ

μ
k
cos
θ
2
=
g t
2
sin
θ

μ
k
cos
θ
μ
k
=
g t
2
sin
θ

2
g t
2
cos
θ
(2)
= tan
θ

2
g t
2
cos
θ
= tan 28
◦

2 (2
.
7 m)
(9
.
8 m
/
s
2
) (3
.
8 s)
2
cos 28
◦
=
0
.
488492
.
002
10.0 points
Two blocks are arranged at the ends of a mass
less string as shown in the figure. The system
starts from rest. When the 3
.
29 kg mass has
fallen through 0
.
405 m, its downward speed is
1
.
27 m
/
s.
The acceleration of gravity is 9
.
8 m
/
s
2
.
3
.
29 kg
4
.
88 kg
μ
a
What is the frictional force between the
4
.
88 kg mass and the table?
Correct answer: 15
.
9736 N.
Explanation:
Given :
m
1
= 3
.
29 kg
,
m
2
= 4
.
88 kg
,
v
0
= 0 m
/
s
,
and
v
= 1
.
27 m
/
s
.
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hamdeh (ah37785) – HW 5 – opyrchal – (111213)
2
Basic Concept:
Newton’s Second Law
F
=
M a
Solution:
The acceleration of
m
1
is obtained
from the equation
v
2

v
2
0
= 2
a
(
s

s
0
)
a
=
v
2

v
2
0
2
h
=
(1
.
27 m
/
s)
2

(0 m
/
s)
2
2 (0
.
405 m)
= 1
.
99123 m
/
s
2
.
Consider free
body diagrams for the
two
masses
T
m
1
g
a
T
N
μ
N
a
m
2
g
Because
m
1
and
m
2
are tied together with
string, they have same the speed and the same
acceleration, so the net force exerted on
m
2
is
F
2
=
m
2
a
The net force on
m
1
is
m
1
a
=
m
1
g

T ,
so
that
T
=
m
1
g

m
1
a
.
Thus
F
2
=
T

f
k
,
f
k
=
T

F
2
=
m
1
g

(
m
1
+
m
2
)
a
= (3
.
29 kg) (9
.
8 m
/
s
2
)

(3
.
29 kg + 4
.
88 kg)
×
(1
.
99123 m
/
s
2
)
=
15
.
9736 N
.
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 Spring '12
 James
 Physics, Force, Friction, kg, Fnet, µs

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