quiz 02 – BAUTISTA, ALDO – Due: Mar 8 2006, 10:00 pm
1
Version number
encoded
for clicker
entry:
V1:1, V2:1, V3:1, V4:5, V5:5.
Question 1
Part 1 of 1.
10 points.
A(n) 81.2 kg Fsherman jumps from a dock
into a 126.8 kg rowboat at rest on the west
side of the dock.
If the velocity of the Fsherman is 5.41 m/s
to the west as he leaves the dock, what is the
Fnal velocity of the Fsherman and the boat?
1.

2
.
18063 m
/
s
2.

2
.
11198 m
/
s
correct
3.

2
.
0481 m
/
s
4.

1
.
98539 m
/
s
5.

1
.
92412 m
/
s
6.

1
.
86544 m
/
s
Explanation:
Let west be negative:
Let :
m
1
= 81
.
2 kg
,
m
2
= 126
.
8 kg
,
and
v
i,
1
=

5
.
41 m
/
s
.
The boat and Fsherman have the same Fnal
speed, and
v
i,
2
= 0 m/s, so
m
1
~v
i,
1
+
m
2
~v
i,
2
= (
m
1
+
m
2
)
~v
f
m
1
~v
i,
1
= (
m
1
+
m
2
)
~v
f
v
f
=
m
1
v
i
m
1
+
m
2
=
(81
.
2 kg) (

5
.
41 m
/
s)
81
.
2 kg + 126
.
8 kg
=

2
.
11198 m
/
s
,
which is 2
.
11198 m
/
s to the west.
Question 2
Part 1 of 1.
10 points.
A 2
.
18 kg block is pushed 1
.
58 m up a ver
tical wall with constant speed by a constant
force of magnitude
F
applied at an angle of
56
.
5
◦
with the horizontal.
The acceleration of gravity is 9
.
8 m
/
s
2
.
2
.
18 kg
F
56
.
5
◦
If the coe±cient of kinetic friction between
the block and wall is 0
.
511, Fnd the work done
by
F
.
1.
46
.
4583 J
2.
47
.
9087 J
3.
49
.
4048 J
4.
51
.
0068 J
correct
5.
52
.
6712 J
6.
54
.
3148 J
Explanation:
Given :
m
= 2
.
18 kg
,
μ
= 0
.
511
,
θ
= 56
.
5
◦
,
and
Δ
y
= 1
.
58 m
.
v
m g
f
k
n
The block is in equilibrium horizontally, so
X
F
x
=
F
cos
θ

n
= 0
,
so that
n
=
F
cos
θ
Since the block moves with constant velocity,
X
F
y
=
F
sin
θ

m g

f
k
= 0
F
sin
θ

m g

μ F
cos
θ
= 0
F
(sin
θ

μ
cos
θ
) =
m g ,
so