homework 05 – HILL, WILLIAM – Due: Oct 5 2007, 1:00 am
1
Question 1, chap 7, sect 1.
part 1 of 2
10 points
A 16
.
7 kg block is dragged over a rough,
horizontal surface by a constant force of 114 N
acting at an angle of 33
.
6
◦
above the horizon
tal.
The block is displaced 50
.
5 m, and the
coefficient of kinetic friction is 0
.
184.
The acceleration of gravity is 9
.
8 m
/
s
2
.
16
.
7 kg
μ
= 0
.
184
114 N
33
.
6
◦
Find the work done by the 114 N force.
Correct answer: 4795
.
13 J (tolerance
±
1 %).
Explanation:
Consider the force diagram
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
= (114 N) (50
.
5 m) cos 33
.
6
◦
=
4795
.
13 J
, .
Question 2, chap 7, sect 1.
part 2 of 2
10 points
Find the magnitude of the work done by
the force of friction.
Correct answer: 934
.
528 J (tolerance
±
1 %).
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
.
184 [(16
.
7 kg) (9
.
8 m
/
s
2
)

(114 N) sin 33
.
6
◦
] (50
.
5 m)
=

934
.
528 J
.
Question 3, chap 7, sect 1.
part 1 of 1
10 points
If 2.2 J of work is done in raising a 180 g
apple, how far is it lifted?
Correct answer:
1
.
24844
m (tolerance
±
1
%).
Explanation:
Basic Concepts:
W
applied
=
F
applied
d
cos
θ
=
mgd
since
θ
= 0
◦
⇒
cos
θ
= 1.
1 J = 1 N
·
m
F
=
mg
1 N = 1 kg
·
m
/
s
2
Given:
W
applied
= 2
.
2 J
m
= 180 g
g
= 9
.
81 m
/
s
2
Solution:
The force and displacement are
parallel, so the distance is given by
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homework 05 – HILL, WILLIAM – Due: Oct 5 2007, 1:00 am
2
d
=
W
applied
mg
=
2
.
2 J
(180 g)(9
.
79 m
/
s
2
)
·
1000 g
1 kg
= 1
.
24844 m
Question 4, chap 7, sect 1.
part 1 of 1
10 points
A tugboat exerts a constant force of 2900 N
on a ship moving at constant speed through a
harbor.
How much work does the tugboat do on the
ship if each moves a distance of 4
.
42 km?
Correct answer:
12
.
818
MJ (tolerance
±
1
%).
Explanation:
Given :
F
net
= 2900 N
and
d
= 4
.
42 km
.
The net force
F
net
= Σ
F
x
=
F .
The force and the displacement are parallel,
so the net work is
W
net
=
F
net
d
= (2900 N) (4
.
42 km)
parenleftbigg
1000 m
1 km
parenrightbigg
·
parenleftbigg
1 MJ
10
6
J
parenrightbigg
=
12
.
818 MJ
,
since 1 J = 1 N
·
m.
Question 5, chap 7, sect 1.
part 1 of 3
10 points
A 2
.
1 kg block is pushed 1
.
52 m up a vertical
wall with constant speed by a constant force
of magnitude
F
applied at an angle of 69
.
2
◦
with the horizontal.
The acceleration of gravity is 9
.
8 m
/
s
2
.
2
.
1 kg
F
69
.
2
◦
If the coefficient of kinetic friction between
the block and wall is 0
.
673, find the work done
by
F
.
Correct answer: 42
.
0253 J (tolerance
±
1 %).
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 Spring '08
 Turner
 Physics, Force, Mass, Potential Energy, Work, Correct Answer

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