salazar (ggs332) – HW05 – TSOI – (58160)
1
This
printout
should
have
25
questions.
Multiplechoice questions may continue on
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before answering.
001
(part 1 of 2) 10.0 points
A 0
.
4693 kg raindrop falls vertically at con
stant speed under the influence of gravity and
air resistance.
After the drop has fallen 156 m, what is the
work done by gravity?
The acceleration of
gravity is 9
.
8 m
/
s
2
.
Correct answer: 717
.
466 J.
Explanation:
The work is
W
= Force
·
Distance
For the raindrop falling a distance
h
, the work
done by the force of gravity
m g
is
W
g
=
m g h
= (0
.
4693 kg) (9
.
8 m
/
s
2
) (156 m)
= 717
.
466 J
.
002
(part 2 of 2) 10.0 points
What is the work done by air resistance?
Correct answer:

717
.
466 J.
Explanation:
Since
the raindrop remains
at constant
speed, the acceleration of the raindrop is zero,
and Newton’s second law yields
summationdisplay
F
y
=
F
air

m g
= 0
.
It is evident that the force due to air resistance
is equal in magnitude to the force due to
gravity, but directed upward. The work done
by air resistance can then be found,
W
air
=
F
air
h
=

m g h
=

(0
.
4693 kg) (9
.
8 m
/
s
2
)
×
(156 m)
=

717
.
466 J
.
keywords:
003
10.0 points
A cheerleader lifts his 60
.
5 kg partner straight
up off the ground a distance of 0
.
845 m before
releasing her.
The acceleration of gravity is 9
.
8 m
/
s
2
.
If he does this 12 times, how much work has
he done?
Correct answer: 6012
.
01 J.
Explanation:
The work done in lifting the cheerleader
once is
W
1
=
m g h
= (60
.
5 kg)(9
.
8 m
/
s
2
)(0
.
845 m)
= 501
.
001 J
.
The work required to lift her
n
= 12 times is
W
=
n W
1
= (12)(501
.
001 J) = 6012
.
01 J
.
004
(part 1 of 3) 10.0 points
A block of mass
m
is pushed a distance
D
up
an inclined plane by a horizontal force
F
. The
plane is inclined at an angle
θ
with respect to
the horizontal. The block starts from rest and
the coefficient of kinetic friction is
μ
k
.
m
D
μ
k
F
θ
If
N
is the normal force, what is the work
done by friction?
1.
W
=

μ
k
(
N
+
m g
cos
θ
)
D
2.
W
=

μ
k
(
N 
m g
cos
θ
)
D
3.
W
= +
μ
k
N
D
4.
W
= +
μ
k
(
N 
m g
cos
θ
)
D
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salazar (ggs332) – HW05 – TSOI – (58160)
2
5.
W
= +
μ
k
(
N
+
m g
cos
θ
)
D
6.
W
= 0
7.
W
=

μ
k
N
D
correct
Explanation:
The
force
of
friction
has
a
magnitude
F
friction
=
μ
k
N
.
Since it is in the direc
tion opposite to the motion, we get
W
friction
=

F
friction
D
=

μ
k
N
D.
005
(part 2 of 3) 10.0 points
What is the work done by the normal force
N
?
1.
W
= (
m g
cos
θ
+
F
sin
θ
 N
)
D
2.
W
=
N
D
sin
θ
3.
W
= 0
correct
4.
W
=
N
D
5.
W
= (
N 
m g
cos
θ

F
sin
θ
)
D
6.
W
=
N
D
cos
θ
7.
W
= (
N
+
m g
cos
θ
+
F
sin
θ
)
D
8.
W
=
N
D
Explanation:
The normal force makes an angle of 90
◦
with the displacement, so the work done by it
is zero.
006
(part 3 of 3) 10.0 points
What is the final speed of the block?
1.
v
=
radicalbigg
2
m
(
F
cos
θ

m g
sin
θ

μ
k
N
)
D
correct
2.
v
=
radicalbigg
2
m
(
F
sin
θ

μ
k
N
)
D
3.
v
=
radicalbigg
2
m
(
F
cos
θ

μ
k
N
)
D
4.
v
=
radicalbigg
2
m
(
F
cos
θ
+
m g
sin
θ
)
D
5.
v
=
radicalbigg
2
m
(
F
sin
θ
+
μ
k
N
)
D
6.
v
=
radicalbigg
2
m
(
F
cos
θ

m g
sin
θ
+
μ
k
N
)
D
7.
v
=
radicalbigg
2
m
(
F
cos
θ
+
m g
sin
θ

μ
k
N
)
D
8.
v
=
radicalbigg
2
m
(
F
cos
θ

m g
sin
θ
)
D
Explanation:
The work done by gravity is
W
grav
=
m g D
cos(90
◦
+
θ
)
=

m g D
sin
θ .
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 Spring '10
 TSOI
 Physics, Force, Correct Answer, kg, Salazar

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