quiz 01 – BAUTISTA, ALDO – Due: Feb 15 2006, 10:00 pm
1
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Question 1
Part 1 of 1.
10 points.
A ball on the end of a string is whirled
around in a horizontal circle of radius 0
.
345 m.
The plane of the circle is 1
.
82 m above the
ground. The string breaks and the ball lands
2
.
26 m away from the point on the ground
directly beneath the ball’s location when the
string breaks.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Find the centripetal acceleration of the ball
during its circular motion.
1.
33
.
9499 m
/
s
2
2.
35
.
1241 m
/
s
2
3.
36
.
224 m
/
s
2
4.
37
.
4209 m
/
s
2
5.
38
.
6627 m
/
s
2
6.
39
.
8586 m
/
s
2
correct
Explanation:
In order to find the centripetal acceleration
of the ball, we need to find the initial velocity
of the ball.
Let
y
be the distance above the
ground. After the string breaks, the ball has
no initial velocity in the vertical direction, so
the time spent in the air may be deduced from
the kinematic equation,
y
=
1
2
g t
2
.
Solving for
t
,
⇒
t
=
r
2
y
g
.
Let
d
be the distance traveled by the ball.
Then
v
x
=
d
t
=
d
r
2
y
g
.
Hence, the centripetal acceleration of the ball
during its circular motion is
a
c
=
v
2
x
r
=
d
2
g
2
y r
=
39
.
8586 m
/
s
2
.
Question 2
Part 1 of 2.
10 points.
A book is at rest on an incline as shown
above.
A hand, in contact with the top of
the book, produces a constant force
F
hand
vertically downward.
F
hand
Book
The following figures show several attempts
at drawing freebody diagrams for the book.
Which figure has the correct directions for
each force?
Note:
The magnitude of the forces are not
necessarily drawn to scale.
1.
weight
force
normal
friction
2.
weight
friction
force
normal
3.
weight
friction
normal
force
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quiz 01 – BAUTISTA, ALDO – Due: Feb 15 2006, 10:00 pm
2
4.
weight
force
normal
friction
5.
weight
normal
friction
force
6.
weight
force
friction
normal
correct
Explanation:
The normal force points perpendicular to
the surface of the inclined plane. The weight
force points down.
The
F
hand
also points
down. The friction force keeps the book from
sliding and consequently points up the incline.
Question 3
Part 2 of 2.
10 points.
For the normal force exerted on the book
by the wedge in the diagram, which force(s)
complete(s) the force pair for Newton’s third
law (actionreaction)?
1.
the pull of the earth on the book
2.
the pull of the book on the earth
3.
the sum of the component of gravity per
pendicular to the surface of the incline and
the component of
F
hand
perpendicular to the
surface of the incline
4.
the normal force exerted on the wedge by
the book
correct
5.
the component of
F
hand
pointing perpen
dicular to the surface of the incline
6.
the component of gravity pointing parallel
to the surface of the incline
Explanation:
The force that completes the third law pair
with the normal force of the wedge on the
book is the normal force of the book on the
wedge.
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 Spring '08
 Turner
 Force, Friction, m/s, BAUTISTA

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