hiarker (srh959) – homework 14 – Turner – (58220)
1
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001
(part 1 of 2) 10.0 points
A crate is pulled by a force (parallel to the
incline) up a rough incline. The crate has an
initial speed shown in the figure below. The
crate is pulled a distance of 5
.
89 m on the
incline by a 150 N force.
The acceleration of gravity is 9
.
8 m
/
s
2
.
13 kg
μ
= 0
.
303
150 N
1
.
48 m
/
s
22
◦
a) What is the change in kinetic energy of
the crate?
Correct answer: 391
.
589 J.
Explanation:
Let :
F
= 150 N
,
d
= 5
.
89 m
,
θ
= 22
◦
,
m
= 13 kg
,
g
= 9
.
8 m
/
s
2
,
μ
= 0
.
303
,
and
v
= 1
.
48 m
/
s
.
F
μ N
N
m g
v
θ
The workenergy theorem with nonconser
vative forces reads
W
fric
+
W
appl
+
W
gravity
= Δ
K
To find the work done by friction we need the
normal force on the block from Newton’s law
summationdisplay
F
y
=
N −
m g
cos
θ
= 0
⇒ N
=
m g
cos
θ .
Thus
W
fric
=
−
μ m g d
cos
θ
=
−
(0
.
303) (13 kg) (9
.
8 m
/
s
2
)
×
(5
.
89 m) cos 22
◦
=
−
210
.
811 J
.
The work due to the applied force is
W
appl
=
F d
= (150 N) (5
.
89 m)
= 883
.
5 J
,
and the work due to gravity is
W
grav
=
−
m g d
sin
θ
=
−
(13 kg) (9
.
8 m
/
s
2
)
×
(5
.
89 m) sin 22
◦
=
−
281
.
1 J
,
so that
Δ
K
=
W
fric
+
W
appl
+
W
grav
= (
−
210
.
811 J) + (883
.
5 J)
+ (
−
281
.
1 J)
= 391
.
589 J
.
002
(part 2 of 2) 10.0 points
b) What is the speed of the crate after it is
pulled the 5
.
89 m?
Correct answer: 7
.
90158 m
/
s.
Explanation:
Since
1
2
m
(
v
2
f
−
v
2
i
) = Δ
K
v
2
f
−
v
2
i
=
2 Δ
K
m
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hiarker (srh959) – homework 14 – Turner – (58220)
2
v
f
=
radicalbigg
2 Δ
K
m
+
v
2
i
=
radicalBigg
2(391
.
589 J)
13 kg
+ (1
.
48 m
/
s)
2
= 7
.
90158 m
/
s
.
003
(part 1 of 3) 10.0 points
A block starts at rest and slides down a fric
tionless track. It leaves the track horizontally,
flies through the air, and subsequently strikes
the ground.
592 g
5
.
1 m
2
.
1 m
x
v
What is the speed of the ball when it leaves
the track?
The acceleration of gravity is
9
.
81 m
/
s
2
.
Correct answer: 7
.
67203 m
/
s.
Explanation:
Let :
g
=
−
9
.
81 m
/
s
2
,
m
= 592 g
,
and
h
1
= 3 m
.
m
h
h
1
h
2
5 m
v
Choose the point where the block leaves the
track as the origin of the coordinate system.
While on the ramp,
K
b
=
U
t
1
2
m v
2
x
=
−
m g h
1
v
2
x
=
−
2
g h
1
v
x
=
radicalbig
−
2
g h
1
=
radicalBig
−
2 (
−
9
.
81 m
/
s
2
) (3 m)
=
7
.
67203 m
/
s
.
004
(part 2 of 3) 10.0 points
What
horizontal
distance
does
the
block
travel in the air?
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 Spring '09
 NUI
 Physics, Energy, Force, Mass, Potential Energy, Work, eµ

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