homework 06 – FONTENOT, BRIAN – Due: Feb 25 2008, 4:00 am
1
Question 1, chap 7, sect 1.
part 1 of 3
10 points
Assume you are on a planet similar to
Earth where the acceleration of gravity
g
≈
10 m
/
s
2
.
A plane 50 m in length is inclined at an
angle of 16
.
3
◦
as shown.
A block of weight
500 N is placed at the top of the plane and
allowed to slide down.
500 N
μ
= 0
.
5
v
50 m
16
.
3
◦
48 m
14 m
The mass of the block is most nearly
1.
m
≈
25 kg
2.
m
≈
26 kg
3.
m
≈
39 kg
4.
m
≈
10 kg
5.
m
≈
52 kg
6.
m
≈
15 kg
7.
m
≈
20 kg
8.
m
≈
13 kg
9.
m
≈
50 kg
correct
Explanation:
Let :
W
= 500 N
and
g
≈
10 m
/
s
2
.
s
θ
x
y
The mass of the block is given by
m
=
W
g
=
500 N
10 m
/
s
2
=
50 kg
.
Question 2, chap 7, sect 1.
part 2 of 3
10 points
Calculate the
magnitude of the normal
force exerted on the block by the plane.
1.
bardbl
vector
N
bardbl ≈
240 N
2.
bardbl
vector
N
bardbl ≈
160 N
3.
bardbl
vector
N
bardbl ≈
360 N
4.
bardbl
vector
N
bardbl ≈
480 N
correct
5.
bardbl
vector
N
bardbl ≈
200 N
6.
bardbl
vector
N
bardbl ≈
400 N
7.
bardbl
vector
N
bardbl ≈
120 N
8.
bardbl
vector
N
bardbl ≈
80 N
Explanation:
Let :
x
= 48 m
and
y
= 14 m
.
s
=
radicalbig
x
2
+
y
2
=
radicalBig
(48 m)
2
+ (14 m)
2
= 50 m
.
cos
θ
=
x
s
=
48
50
.
The normal force is the force perpendicular
to the surface. It is given by
N
=
W
cos
θ
=
W
x
s
= (500 N)
parenleftbigg
48 m
50 m
parenrightbigg
≈
480 N
.
Question 3, chap 7, sect 1.
part 3 of 3
10 points
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homework 06 – FONTENOT, BRIAN – Due: Feb 25 2008, 4:00 am
2
Calculate the work done on the block by
the gravitational force during the 50 m slide
down the plane in J.
1.
W
≈
1750 J
2.
W
≈
3690 J
3.
W
≈
7000 J
correct
4.
W
≈
10400 J
5.
W
≈
1350 J
6.
W
≈
650 J
7.
W
≈
3750 J
8.
W
≈
5850 J
9.
W
≈
2400 J
Explanation:
Let :
d
= 50 m
.
sin
θ
=
y
s
=
14
50
.
Work is given by
W
=
vector
F
·
vector
d ,
where
vector
d
is the
displacement vector and
vector
F
is the force act
ing on the system.
The component of the
force pointing in the direction of the displace
ment is
F
bardbl
=
W
sin
θ
. The work done by the
gravitational force, then, is
W
=
F
bardbl
d
=
W
d
sin
θ
=
W
d
y
s
= (500 N) (50 m)
parenleftbigg
14 m
50 m
parenrightbigg
=
7000 J
.
Question 4, chap 7, sect 1.
part 1 of 3
10 points
Sally applies a horizontal force of 706 N
with a rope to drag a wooden crate across a
floor with a constant speed. The rope tied to
the crate is pulled at an angle of 42
◦
relative
to the floor.
The acceleration of gravity is 9
.
8 m
/
s
2
.
m
F
706 N
42
◦
How much force is exerted by the rope on
the crate?
Correct answer:
950
.
017
N (tolerance
±
1
%).
Explanation:
Given :
F
h
= 706 N
,
and
θ
= 42
◦
.
The horizontal component of the force exerted
by the rope is defined by
cos
θ
=
F
h
F
so that
F
=
F
h
cos
θ
=
706 N
cos 42
◦
=
950
.
017 N
.
Question 5, chap 7, sect 1.
part 2 of 3
10 points
What work is done by Sally if the crate is
moved 65
.
4 m?
Correct answer: 46172
.
4 J (tolerance
±
1 %).
Explanation:
The motion is in the direction of the hori
zontal component, so the work Sally does is
W
=
F
h
d
= (706 N) (65
.
4 m)
=
46172
.
4 J
.
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 Spring '08
 Turner
 Physics, Acceleration, Energy, Force, Gravity, Potential Energy, Work, Correct Answer, Fontenot, BRIAN

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