quiz 02 – BROWNING, ADAM – Due: Oct 17 2007, 10:00 pm
1
Question 1, chap 6, sect 1.
part 1 of 1
10 points
A box weighing 710 N is pushed along a
horizontal floor at constant velocity with a
force of 260 N parallel to the floor.
What is the coefficient of kinetic friction
between the box and the floor?
1.
0
.
34375
2.
0
.
354839
3.
0
.
366197
correct
4.
0
.
378378
5.
0
.
390625
6.
0
.
402778
7.
0
.
415385
8.
0
.
428571
9.
0
.
442623
10.
0
.
457627
Explanation:
Let :
W
= 710 N
and
F
app
= 260 N
.
vector
F
app
vector
f
k
vector
N
m
vector
W
Because the box is moving with constant
velocity, its acceleration is zero and the net
foce acting on it is zero.
Applying
summationdisplay
F
y
= 0 to the box,
N  W
= 0
N
=
W
= 710 N
.
Applying
summationdisplay
F
x
= 0 to the box,
F
app

f
k
= 0
f
k
=
F
app
μ
k
N
=
F
app
μ
k
=
F
app
N
=
260 N
710 N
=
0
.
366197
.
Question 2, chap 7, sect 1.
part 1 of 1
10 points
Starting from rest at a height equal to the
radius of the circular track, a block of mass
23 kg slides down a quarter circular track
under the influence of gravity with friction
present (of coefficient
μ
).
The radius of the
track is 25 m.
The acceleration of gravity is 9
.
8 m
/
s
2
.
25 m
23 kg
θ
If the kinetic energy of the block at the
bottom of the track is 2100 J, what is the
work done against friction?
1.
2641
.
6 J
2.
2738 J
3.
2827
.
6 J
4.
2915
.
2 J
5.
3006
.
4 J
6.
3105 J
7.
3214 J
8.
3320
.
8 J
9.
3423
.
8 J
10.
3535 J
correct
Explanation:
W
=
W
f
+
K
W
f
=
m g R

K
= (23 kg) (9
.
8 m
/
s
2
) (25 m)

(2100 J)
=
3535 J
.
Question 3, chap 8, sect 5.
part 1 of 1
10 points
A rain cloud contains 7.74
×
10
7
kg of water
vapor.
The acceleration of gravity is 9
.
81 m
/
s
2
.
How long would it take for a 3.59 kW pump
to raise the same amount of water to the
cloud’s altitude of 2.45 km?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
quiz 02 – BROWNING, ADAM – Due: Oct 17 2007, 10:00 pm
2
1.
4
.
85746
×
10
8
s
2.
5
.
01948
×
10
8
s
3.
5
.
18181
×
10
8
s
correct
4.
5
.
35718
×
10
8
s
5.
5
.
53175
×
10
8
s
6.
5
.
76078
×
10
8
s
7.
5
.
95765
×
10
8
s
8.
6
.
17526
×
10
8
s
9.
6
.
37904
×
10
8
s
10.
6
.
60288
×
10
8
s
Explanation:
Basic Concepts:
W
=
P
Δ
t
W
=
Fd
cos
θ
=
Fd
=
mgd
since
θ
= 0
◦
⇒
cos
θ
= 1.
Given:
m
= 7
.
74
×
10
7
kg
P
= 3
.
59 kW
d
= 2
.
45 km
g
= 9
.
81 m
/
s
2
Solution:
Δ
t
=
W
P
=
mgd
P
=
(7
.
74
×
10
7
kg)(9
.
81 m
/
s
2
)(2450 m)
3590 W
= 5
.
18181
×
10
8
s
Question 4, chap 5, sect 6.
part 1 of 1
10 points
Consider
the
following
system
of
two
masses and two pulleysl The pulleys that are
massless and frictionless, and the mass on the
left accelerates upward while the mass on the
right accelerates downward.
The acceleration of gravity is 9
.
8 m
/
s
2
.
6 kg
34 kg
a
Find the acceleration of the mass on the
left.
1.
6
.
88649 m
/
s
2
2.
7
.
12727 m
/
s
2
3.
7
.
43448 m
/
s
2
correct
4.
7
.
66957 m
/
s
2
5.
8
.
00563 m
/
s
2
6.
8
.
29231 m
/
s
2
7.
8
.
575 m
/
s
2
8.
8
.
90909 m
/
s
2
9.
9
.
22353 m
/
s
2
10.
9
.
52 m
/
s
2
Explanation:
Let :
m
1
= 6 kg
m
2
= 34 kg
and
g
= 9
.
8 m
/
s
2
.
m
1
m
2
1
2
T
T
T
a
Let
a
denote the upward acceleration of the
m
1
mass on the left and let
T
be the tension
of the long string attached to the
m
1
.
To
understand the motion of the
right
mass
m
2
,
consider the freebody diagram for the right
pulley:
The pulley is massless, so regardless of its
acceleration all forces acting on the pulley
must balance; the tension of the short string
connecting the right pulley to the
m
2
mass
must be 2
T
rather than
T
.
At the same
time, the unstretchability of the long string
requires the pulley — and hence the
m
2
mass
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Turner
 Force, Friction, Mass, kg, Adam –

Click to edit the document details