pandey (sp34566) – HW-17 – swinney – (57305)
1
This
print-out
should
have
29
questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
A skater pushes straight away from a wall.
She pushes on the wall with a force whose
magnitude is
F
, so the wall pushes on her
with a force
F
(in the direction of her motion).
As she moves away from the wall, her center
of mass moves a distance
d
.
Consider the
following statements regarding energy.
I Δ
K
trans
+ Δ
E
internal
=
Fd
II. Δ
K
trans
+ Δ
E
internal
=
−
Fd
III. Δ
K
trans
+ Δ
E
internal
= 0
IV. Δ
K
trans
=
Fd
V. Δ
K
trans
=
−
Fd
What is the correct form of the energy prin-
ciple for the skater as a real system and as a
point particle (PP) system?
F
is applied.
Thus, the work done is
Fd
.
Translational kinetic energy is the only type
of energy a point particle system can have. IV
is correct.
002
10.0points
Evaluate the cross product
(2ˆ
ı
+ 7ˆ
)
×
(
−
4ˆ
ı
+ 6ˆ
)
.
1.
−
16ˆ
ı
skater, i.e. by cooling down. The right hand
side of the energy principle must therefore
be zero for the real system.
Furthermore,
the left hand side must contain the skater’s
change in internal energy for a real system.
III is correct.
As a point particle system, the skater’s cen-
ter of mass moves a distance
d
while a force

8.
Real: III,
PP: V
9.
Real: II,
PP: V
10.
Real: IV,
PP: III
Explanation:
As a real system, the wall does no work
on the skater since the contact point of the
skater’s hand and the wall does not move.
Also, the wall does not transfer energy to the
Explanation:
Because both vectors are in the xy plane,
their cross product must be in the
±
z
di-
rection.
So we only need to look at the z-
component of the cross product.
If we label
the first vector
vector
A
and the second vector
vector
B
,
the
z
component of the cross product will be
A
x
B
y
−
A
y
B
x
= 40
.
So the resultant vector is
40
ˆ
k
.

pandey (sp34566) – HW-17 – swinney – (57305)
2
003(part1of2)10.0points
An airplane of mass 15375 kg flies level to the
ground at an altitude of 14 km with a constant
speed of 165 m
/
s relative to the Earth.
What is the magnitude of the airplane’s
angular momentum relative to a ground ob-
server directly below the airplane in kg
·
m
2
/
s?

Explanation:
Since the observer is directly below the air-
plane,
L
=
h m v
004(part2of2)10.0points