fullwood (tf5349) – Homework 5 Forces – Catala – (49033)
1
This printout should have 20 questions.
Multiplechoice questions may continue on
the next column or page – ±nd all choices
before answering.
001
(part 1 of 2) 10.0 points
A man is pulling on a rope with a force of 51 N
directed at an angle of 65
◦
to the horizontal.
What is the
x
component of this force?
Correct answer: 21
.
5535 N.
Explanation:
Let :
F
= 51 N
and
θ
= 65
◦
.
51 N
Scale: 10 N
65
◦
F
x
=
F
cos
θ
= (51 N) cos65
◦
=
21
.
5535 N
.
002
(part 2 of 2) 10.0 points
What is the
y
component of this force?
Correct answer: 46
.
2217 N.
Explanation:
F
y
=
F
sin
θ
= (51 N) sin65
◦
=
46
.
2217 N
.
003
(part 1 of 4) 10.0 points
A 2.70 kg block starts from rest at the top of a
30.0
◦
incline and accelerates uniformly down
the incline, moving 2.36 m in 2.39 s.
The acceleration of gravity is 9
.
81 m
/
s
2
.
a) What is the magnitude of the accelera
tion of the block?
Correct answer: 0
.
826316 m
/
s
2
.
Explanation:
2
.
7 kg
F
k
a
n
g,y
g,x
F
g
30
◦
Note:
Figure is not drawn to scale.
Basic Concept:
Δ
x
=
1
2
a
(Δ
t
)
2
since
v
i
= 0 m/s.
Given:
Δ
x
= 2
.
36 m
Δ
t
= 2
.
39 s
Solution:
a
=
2Δ
x
(Δ
t
)
2
=
2(2
.
36 m)
(2
.
39 s)
2
= 0
.
826316 m
/
s
2
004
(part 2 of 4) 10.0 points
b) What is the coe²cient of kinetic friction
between the block and the incline?
Correct answer: 0
.
480088.
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View Full Documentfullwood (tf5349) – Homework 5 Forces – Catala – (49033)
2
Explanation:
Basic Concepts:
F
g,x
=
mg
sin
θ
F
g,y
=
mg
cos
θ
F
x,net
=
ma
x
=
F
g,x

F
k
F
y,net
=
F
n

F
g,y
= 0
F
k
=
μ
k
F
n
Given:
m
= 2
.
70 kg
θ
= 30
.
0
◦
g
= 9
.
81 m
/
s
2
Solution:
F
g,x
= (2
.
7 kg)
(
9
.
81 m
/
s
2
)
sin 30
◦
= 13
.
2435 N
F
g,y
= (2
.
7 kg)
(
9
.
81 m
/
s
2
)
cos 30
◦
= 22
.
9384 N
ma
x
=
F
g,x

μ
k
F
g,y
μ
k
=
F
g,x

ma
x
F
g,y
=
13
.
2435 N

(2
.
7 kg)(0
.
826316 m
/
s
2
)
22
.
9384 N
= 0
.
480088
005
(part 3 of 4) 10.0 points
c) What is the magnitude of the frictional
force acting on the block?
Correct answer: 11
.
0124 N.
Explanation:
Basic Concept:
F
k
=
μ
k
F
n
Solution:
F
k
= (0
.
480088)(22
.
9384 N)
= 11
.
0124 N
006
(part 4 of 4) 10.0 points
d) What is the speed of the block after it slides
the distance of 2.36?
Correct answer: 1
.
9749 m
/
s.
Explanation:
Basic Concept:
v
2
f
= 2
a
x
Δ
x
since
v
i
= 0 m/s.
Solution:
v
f
=
r
2
a
x
Δ
x
=
R
2 (0
.
826316 m
/
s
2
)(2
.
36 m)
= 1
.
9749 m
/
s
007
10.0 points
A 5.7 kg bucket of water is raised from a well
by a rope.
The acceleration of gravity is 9
.
81 m
/
s
2
.
If the upward acceleration of the bucket is
3.8 m/s
2
, ±nd the force exerted by the rope
on the bucket of water.
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 Spring '10
 PRO.KIROSHIMA
 Force, Friction, Work, Correct Answer, kg, FN

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