Dynamics
457
Problem 4.40
Compressed gas is used in many circumstances to propel things within tubes.
For example, you can still find pneumatic tubes in use in many banks to send
and receive items to drivethrough tellersand it is compressed gas that propels
a bullet out of a gun barrel. Let the crosssectional area of the tube be given
by
A
, the position of the cylinder by
s
, and assume that the compressed gas is
an ideal gas at constant temperature so that the pressure
P
times the volume
˝
is a constant, i.e.,
P ˝
D
constant
. Show that the potential energy of this
compressed gas is given by
V
D
P
0
s
0
A
ln
.s=s
0
/
, where
P
0
is the initial
pressure and
s
0
is the initial value of
s
. Model the cylinder as a particle and
assume that the forces resisting the motion of the cylinder are negligible.
Solution
We first realize that with
P ˝
D
C
. The force exerted on the cylinder can be written as
F
D
A
C
˝
:
(1)
Since
˝
D
As
, we can simplify the above equation to
F
D
C
s
:
(2)
The work done by the gas can now be calculated using
U
0s
D
Z
s
s
0
F ds
D
Z
s
s
0
C
s
ds
)
U
0s
D
C
ln
s
ˇ
ˇ
ˇ
s
s
0
:
(3)
Knowing that we can express
C
as the product of the initial pressure
P
0
and the initial volume
˝
D
s
0
A
, we
can simplify the expression for the work done by the gas to
U
0s
D
P
0
s
0
A
ln
s
s
0
:
(4)
Therefore we can write the potential energy as
U
12
D
Z
L
12
E
F
d
E
r
D
.V
2
V
1
/
)
V
D
P
0
s
0
A
ln
s
s
0
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Dynamics
465
Problems 4.46 through 4.48
A crate, initially traveling horizontally with a speed of
18
ft
=
s
, is made to slide
down a
14
ft
chute inclined at
35
ı
. The surface of the chute has a coefficient
of kinetic friction
k
and, at its lower end, it smoothly lets the crate onto
a horizontal trajectory. The horizontal surface at the end of the chute has a
coefficient of kinetic friction
k2
. Model the crate as a particle and assume that
gravity and the contact forces between the crate and the sliding surface are the
only relevant forces.
Problem 4.46
If
k
D
0:35
what is the speed with which the crate reaches
the bottom of the chute (immediately before the crate’s trajectory becomes
horizontal)?
Problem 4.47
Find
k
such that the crate’s speed at the bottom of the chute
(immediately before the crate’s trajectory becomes horizontal) is
15
ft
=
s.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 York
 Force, Eqs., mg cos, Workenergy principle

Click to edit the document details