schneiter (tcs639) – HW08 – Ross – (89152)
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001
10.0 points
A block accelerates 3
.
5 m
/
s
2
down a plane
inclined at angle 26.0
◦
.
The acceleration of gravity is 9
.
81 m
/
s
2
.
m
μ
k
3
.
5 m
/
s
2
26
◦
Find
μ
k
between the block and the inclined
plane.
Correct answer: 0
.
0907798.
Explanation:
Given :
a
= 3
.
5 m
/
s
2
,
θ
= 26
◦
,
and
g
= 9
.
81 m
/
s
2
.
Consider the free body diagram for the
block
m g
sin
θ
N
=
m g
cos
θ
μ
N
a
m g
Basic Concepts:
vector
F
net
=
mvectora
Parallel to the ramp:
F
x,net
=
ma
x
=
F
g,x

F
k
F
g,x
=
mg
sin
θ
F
k
=
μ
k
F
n
Perpendicular to the ramp:
F
y,net
=
F
n

F
g,y
= 0
F
g,y
=
mg
cos
θ
Solution:
Consider the forces parallel to the ramp:
F
k
=
F
g,x

ma
x
=
mg
sin
θ

ma
x
Consider
the
forces
perpendicular
to
the
ramp:
F
n
=
F
g,y
=
mg
cos
θ
Thus the coefficient of friction is
μ
k
=
F
k
F
n
=
m g
sin
θ

m a
m g
cos
θ
=
g
sin
θ

a
g
cos
θ
=
9
.
81 m
/
s
2
sin 26
◦

3
.
5 m
/
s
2
9
.
81 m
/
s
2
cos 26
◦
=
0
.
0907798
.
002
10.0 points
The magnitude of each force is 283 N, the
force on the right is applied at an angle 35
◦
and the mass of the block is 23 kg.
The
coefficient of friction is 0
.
148.
The acceleration of gravity is 9
.
8 m
/
s
2
.
23 kg
μ
= 0
.
148
283 N
35
◦
283 N
What is the magnitude of the resulting ac
celeration?
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 Spring '08
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 mechanics, Force, Friction, Sin, Fnet

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