alameri (aoa434) – Hwk06 – Stokes – (19102)
1
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printout
should
have
22
questions.
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before answering.
This is ONLINE HOMEWORK No 6. It is
due before 7.00 am on Wednesday 3 March,
Abu Dhabi Time (9.00 pm on Tuesday in
Texas).
001
10.0 points
The
initial
kinetic
energy
imparted
to
a
0
.
51 kg bullet is 1859 J.
The acceleration of gravity is 9
.
81 m
/
s
2
.
Neglecting air resistance, find the range of
this projectile when it is fired at an angle such
that the range equals the maximum height
attained.
Correct answer: 0
.
348882 km.
Explanation:
Let :
m
= 0
.
51 kg
,
K
= 1859 J
,
and
g
= 9
.
81 m
/
s
2
.
Assume that the firing height is negligible and
that the bullet lands at the same elevation
from which it was fired. The range
R
is given
by
R
=
v
2
0
g
sin 2
θ .
From kinematics,
x
= (
v
0
cos
θ
)
t
and
y
= (
v
0
sin
θ
)
t
−
1
2
g t
2
.
R
=
h
max
=
(
v
0
sin
θ
)
2
2
g
, so
v
2
0
g
sin 2
θ
=
(
v
0
sin
θ
)
2
2
g
v
2
0
g
(2 sin
θ
cos
θ
) =
(
v
0
sin
θ
)
2
2
g
tan
θ
= 4
θ
= 76
◦
.
The kinetic energy is
K
=
1
2
m v
2
0
v
2
0
=
2
K
m
,
so
R
=
v
2
0
g
sin 2
θ
=
2
K
m g
sin 2
θ
=
2 (1859 J)
(0
.
51 kg) (9
.
81 m
/
s
2
)
sin 152
◦
·
km
1000 m
=
0
.
348882 km
.
002
(part 1 of 3) 10.0 points
A 18
.
7 kg block is dragged over a rough, hor
izontal surface by a constant force of 85
.
8 N
acting at an angle of 28
◦
above the horizon
tal.
The block is displaced 50
.
9 m, and the
coefficient of kinetic friction is 0
.
19.
The acceleration of gravity is 9
.
8 m
/
s
2
.
18
.
7 kg
μ
= 0
.
19
85
.
8 N
28
◦
Find the work done by the 85
.
8 N force.
Correct answer: 3856
.
03 J.
Explanation:
Let :
m
= 18
.
7 kg
,
F
= 85
.
8 N
,
s
x
= 50
.
9 m
,
μ
= 0
.
19
,
and
g
= 9
.
8 m
/
s
2
.
Consider the force diagram
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alameri (aoa434) – Hwk06 – Stokes – (19102)
2
F
θ
m g
n
f
k
Work is
W
=
vector
F
·
vectors
, where
vectors
is the distance
traveled. In this problem
vectors
= 5ˆ
ı
is only in the
x
direction.
⇒
W
F
=
F
x
s
x
=
F s
x
cos
θ
= (85
.
8 N) (50
.
9 m) cos 28
◦
=
3856
.
03 J
.
003
(part 2 of 3) 10.0 points
Find the work done by the force of friction.
Correct answer:
−
1382
.
75 J.
Explanation:
To find the frictional force,
F
friction
=
μ
N
,
we need to find
N
from vertical force balance.
Note that
N
is in the same direction as the
y
component of
F
and opposite the force of
gravity. Thus
F
sin
θ
+
N
=
m g
so that
N
=
m g
−
F
sin
θ .
Thus the friction force is
vector
F
friction
=
−
μ
N
ˆ
ı
=
−
μ
(
m g
−
F
sin
θ
)ˆ
ı .
The work done by friction is then
W
μ
=
vector
F
friction
·
vectors
=
−
f
μ
 
s

=
−
μ
(
m g
−
f
μ
sin
θ
)
s
x
=
−
(0
.
19) [(18
.
7 kg) (9
.
8 m
/
s
2
)
−
(85
.
8 N) sin 28
◦
] (50
.
9 m)
=
−
1382
.
75 J
.
004
(part 3 of 3) 10.0 points
If the block was originally at rest, determine
its final speed.
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 Spring '09
 NAJAFZADEH
 mechanics, Force, Friction, Mass, Work, Correct Answer, 85.8 N force

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