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alexander (jra2623) – oldhomework 13 – Turner – (92510)
1
This printout should have 10 questions.
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beFore answering.
001
10.0 points
A block starts at rest and slides down a Fric
tionless track except For a small rough area on
a horizontal section oF the track (as shown in
the fgure below).
It leaves the track horizontally, ±ies through
the air, and subsequently strikes the ground.
The acceleration oF gravity is 9
.
81 m
/
s
2
.
μ
=0
.
2
1
.
1 m
b
b
b
b
b
b
b
b
b
b
b
b
484 g
h
2
.
2 m
5
.
12 m
9
81 m
/
s
v
At what height
h
above the ground is the
block released?
Correct answer: 5
.
39891 m.
Explanation:
Let :
x
= 5
.
12 m
,
g
= 9
.
81 m
/
s
2
,
m
= 484 g
,
μ
= 0
.
2
,
ℓ
= 1
.
1 m
,
h
2
=

2
.
2 m
,
h
=
h
1

h
2
,
and
v
x
=
v .
μ
ℓ
b
b
b
b
b
b
b
b
b
b
b
b
m
1
x
g
v
Basic Concepts:
Conservation oF Me
chanical Energy
U
i
=
U
f
+
K
f
+
W .
(1)
since
v
i
= 0 m/s.
K
=
1
2
mv
2
(2)
U
g
=
mg h
(3)
W
=
μmg ℓ .
(4)
Choosing the point where the block leaves the
track as the origin oF the coordinate system,
Δ
x
=
v
x
Δ
t
(5)
h
2
=

1
2
g
Δ
t
2
(6)
since
a
x
i
= 0 m/s
2
and
v
y
i
= 0 m/s.
Solution:
²rom energy conservation Eqs. 1,
2, 3, and 4, we have
1
2
mv
2
x
=
mg
(
h

h
2
)

μmg ℓ
v
2
x
= 2
g h
1

2
μg ℓ
h
1
=
v
2
x
2
g
+
μℓ
(7)
h
2
=

1
2
g t
2
(6)
x
=
v
x
t .
(5)
Using Eq. 6 and substituting
t
=
x
v
x
From
Eq. 5, we have
h
2
=

1
2
g
p
x
v
x
P
2
,
so
v
2
x
=

g x
2
2
h
2
.
(8)
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View Full Documentalexander (jra2623) – oldhomework 13 – Turner – (92510)
2
Using Eq. 6 and substituting
v
2
x
from Eq. 8,
we have
h
1
=

g x
2
2
h
2
2
g
+
μℓ
=

x
2
4
h
2
+
μℓ
(9)
=

(5
.
12 m)
2
4 (

2
.
2 m)
+ (0
.
2) (1
.
1 m)
= 3
.
19891 m
,
and
h
=
h
1

h
2
= (3
.
19891 m)

(

2
.
2 m)
=
5
.
39891 m
.
002
(part 1 of 2) 10.0 points
A block of mass
m
is pushed a distance
D
up
an inclined plane by a horizontal force
F
. The
plane is inclined at an angle
θ
with respect to
the horizontal. The block starts from rest and
the coeFcient of kinetic friction is
μ
k
.
m
D
μ
k
F
θ
If
N
is the normal force;
i.e.
, the sum of the
±component which is perpendicular to the
inclined plane and the
mg
component which
is perpendicular to the inclined plane, what is
the work done by friction?
1.
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 Summer '08
 Kaplunovsky
 Energy, Friction, Work

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