C13R.2
This is a conceptually tricky problem, and the book’s answer is incorrect. I will
instruct the grader to grade your problem on effort only.
To set the stage, let us first consider the situation where the wheels of the jetliner
are locked, so there is no rotation. Then, if we treat the jetliner plus wheels as our
system, there is a constant frictional force
vector
f
acting over a distance
d
. If the jet moves
in the +
x
direction, this gives rise to a constant acceleration in the
−
x
direction:
a
x
=
−
f
M
.
(1)
where
M
is the mass of the jet (including wheels).
This causes the velocity of the
jetliner to decrease:
v
2
f,x
=
v
2
i,x
+ 2
a
x
d
=
v
2
i,x
−
2
fd
M
.
(2)
The translational kinetic energy of the jet therefore decreases:
Δ
K
trans
=
1
2
Mv
2
f
−
1
2
Mv
2
i
=
−
fd.
(3)
(Here we are using a constant acceleration kinematic equation that we have not yet
covered in this course.) Since microscopically energy is conserved, this energy must
be converted into heat. However, this heat may be in the tires or in the runway. We
do not know how much goes into the wheels and how much goes into the tires. (The
book’s answer supposes that all of it goes into the tires, but that is not the most
important mistake in that solution!) We will avoid this issue by computing the total
amount of heat generated. In the case where the wheels are locked, this is
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 Fall '10
 MarkusLuty
 Physics, Energy, Kinetic Energy, Potential Energy

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