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alexander (jra2623) – homework 10 – Turner – (92510)
1
This printout should have 11 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 2) 10.0 points
A snowboarder oF mass
m
is at rest on the top
oF a sand dune. The coe±cients oF static and
kinetic Friction are
μ
s
and
μ
k
, respectively.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
m
μ
s
k
θ
What is the angle
θ
the incline must exceed
so that the snowboarder starts sliding?
1.
θ
=
1
arctan
μ
k
2.
θ
= arctan
μ
k
3.
θ
= arctan
μ
s
correct
4.
θ
= arcsin
μ
s
5.
θ
= arccos
μ
s
6.
θ
=
1
arctan
μ
s
7.
θ
= arcsin
μ
k
8.
θ
= arccos
μ
k
Explanation:
Consider the Free body diagram For the
block
m g
sin
θ
N
=
cos
m g
We know that the largest possible value the
static Friction Force can have is
F
f,max
=
μ
s
N
,
where the normal Force is
N
=
m g
cos
θ
.
Thus, since
F
f
=
m g
sin
θ
m g
sin
θ
=
μ
s
m g
cos
θ ,
and then
tan
θ
=
μ
s
θ
= arctan
μ
s
.
002
(part 2 oF 2) 10.0 points
What is the acceleration down the incline iF
θ
exceeds the value Found in the previous
question?
1.
a
=
g μ
s
sin
θ
2.
a
=
g
(
μ
k
sin
θ
−
cos
θ
)
3.
a
=
g
(
μ
k
sin
θ
−
μ
s
cosθ
)
4.
a
=
g μ
k
sin
θ
5.
a
=
g
(sin
θ
−
μ
k
cos
θ
)
correct
6.
a
=
g μ
s
cos
θ
7.
a
=
g
(
μ
s
sin
θ
−
cos
θ
)
8.
a
=
g
(
μ
s
sin
θ
−
μ
k
cosθ
)
9.
a
=
g μ
k
cos
θ
10.
a
=
g
(sin
θ
−
μ
s
cos
θ
)
Explanation:
When
θ
exceeds the value Found in Part 2,
the snowboarder starts moving and the Fric
tion Force is the kinetic Friction
F
k
=
μ
k
N
=
μ
k
m g
cos
θ
. Newton’s equation For the snow
boarder reads then
m a
=
m g
sin
θ
−
F
f
=
m g
sin
θ
−
μ
k
m g
cos
θ ,
and then
a
=
g
(sin
θ
−
μ
k
cos
θ
)
.
003
(part 1 oF 2) 10.0 points
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View Full Documentalexander (jra2623) – homework 10 – Turner – (92510)
2
A fastmoving VW Beetle traveling at 54 mph
hit a mosquito hovering at rest above the road.
Which bug experienced the largest force?
1.
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 Summer '08
 Kaplunovsky
 Force, Friction, Work

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