Overton, Mays – Homework 10 – Due: Feb 14 2005, 4:00 am – Inst: Turner
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
A(n) 864 N crate is being pushed across a
level floor by a force of 396 N at an angle
of 25
◦
above the horizontal.
The coefficient
of kinetic friction between the crate and the
floor is 0.20.
The acceleration of gravity is 9
.
81 m
/
s
2
.
864 N
μ
k
= 0
.
20
396 N
25
◦
What is the acceleration of the box?
Correct answer: 1
.
73295 m
/
s
2
.
Explanation:
Let :
W
=
m g
= 864 N
,
F
applied
= 396 N
,
α
= 25
◦
,
μ
k
= 0
.
20
,
and
g
= 9
.
81 m
/
s
2
m
F
applied
α
m g
N
μ
k
N
Basic Concepts:
F
applied,x
=
F
applied
cos
α
F
applied,y
=
F
applied
sin
α
F
y,net
=
F
n
+
F
applied,y
 W
= 0
F
k
=
μ
k
F
n
F
x,net
=
m a
x
=
F
applied,x

F
k
Solution:
F
applied,x
= (396 N) cos 25
◦
= 358
.
898 N
F
applied,y
= (396 N) sin 25
◦
= 167
.
357 N
The normal force is
F
n
=
W
+
F
applied,y
= 864 N + 167
.
357 N
= 1031
.
36 N
.
The mass of the crate is
m
=
W
g
=
864 N
9
.
81 m
/
s
2
= 88
.
0734 kg
.
From the horizontal motion,
m a
x
=
F
applied,x

μ
k
F
n
a
x
=
F
applied,x

μ
k
F
n
m
=
358
.
898 N

(0
.
2) (1031
.
36 N)
88
.
0734 kg
=
1
.
73295 m
/
s
2
across the floor.
002
(part 1 of 1) 10 points
A 3.70 kg block is pushed along the ceiling
with a constant applied force of 90.0 N that
acts at an angle of 62.0
◦
with the horizontal.
The block accelerates to the right at 6.80
m/s
2
.
The acceleration of gravity is 9
.
81 m
/
s
2
.
3
.
7 kg
90 N
62
◦
μ
6
.
8 m
/
s
2
What is the coefficient of kinetic friction
between the block and the ceiling?
Correct answer: 0
.
395949 .
Explanation:
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Overton, Mays – Homework 10 – Due: Feb 14 2005, 4:00 am – Inst: Turner
2
Basic Concepts:
F
applied,x
=
F
applied
cos
θ
F
applied,y
=
F
applied
sin
θ
F
y,net
=
F
applied,y

mg

F
n
= 0
F
x,net
=
ma
x
=
F
applied,x

F
k
F
k
=
μ
k
F
n
m
F
θ
μ
a
Given:
m
= 3
.
70 kg
θ
= 62
.
0
◦
F
applied
= 90 N
a
x
= 6
.
80 m
g
= 9
.
81 m
/
s
2
Solution:
F
applied,x
= (90 N) cos 62
◦
= 42
.
2524 N
F
applied,y
= (90 N) sin 62
◦
= 79
.
4653 N
The normal force is
F
n
=
F
applied,y

mg
= 79
.
4653 N

(3
.
7 kg)(9
.
81 m
/
s
2
)
= 43
.
1683 N
From the horizontal motion,
ma
x
=
F
applied,x

μ
k
F
n
μ
k
=
F
applied,x

ma
x
F
n
=
42
.
2524 N

(3
.
7 kg)(6
.
8 m
/
s
2
)
43
.
1683 N
= 0
.
395949
003
(part 1 of 3) 10 points
The suspended 2
.
6 kg mass on the right is
moving up, the 1
.
4 kg mass slides down the
ramp, and the suspended 7
.
2 kg mass on the
left is moving down. The coefficient of friction
between the block and the ramp is 0
.
15
.
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 Spring '08
 Turner
 Physics, Friction, Work, kg, Overton, Mays

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