Page 136 Practice Problem 5.13: An average-sized cat of 4.5kg falls with a terminal velocity of approximately 27 m/s. What is the value of the proportionality constant D for the cat? Answer: 0.062 kg/m.
Page 136 Practice Problem 5.13: An average-sized cat of 4.5kg falls with a terminal velocity of approximately 27 m/s. What is the value of the proportionality constant D for the cat? Answer: 0.062 kg/m.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![Page 136 Practice Problem 5.13:
An average-sized cat of 4.5kg falls with a terminal velocity of approximately 27
m/s. What is the value of the proportionality constant D for the cat? Answer:
0.062 kg/m.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F28764828-4d1c-44b8-9544-7d79c8b69e6b%2Ffe27b147-47b0-41e5-aee5-1ed565ee71cd%2F718wluj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Page 136 Practice Problem 5.13:
An average-sized cat of 4.5kg falls with a terminal velocity of approximately 27
m/s. What is the value of the proportionality constant D for the cat? Answer:
0.062 kg/m.
![EXAMPLE 5.13 Fluid friction
In this example we are going to take a look at a different type of friction, one that is not described by Equa-
tions 5.1 and 5.2. When an object moves through a fluid (such as water or air), it exerts a force on the fluid
to push it out of the way. By Newton's third law, the fluid pushes back on the object, in a direction opposite
to the object's velocity relative to the fluid, always opposing the object's motion and usually increasing with
speed. In high-speed motion through air, the resisting force is approximately proportional to the square of
the object's speed u; it's called a drag force, or simply drag. We can represent its magnitude Fp by
FD = Du²
where D is a proportionality constant that depends on the shape and size of the object and the density of air.
When an object falls vertically through air, the drag force opposing the object's motion increases and
the downward acceleration decreases. Eventually, the object reaches a terminal velocity: Its acceleration
approaches zero and the velocity becomes nearly constant. Derive an expression for the terminal-velocity
magnitude UT in terms of D and the weight mg of the object.
ones as special cases. We employ this strategy throughout the t-starting with a simple
In this chapter, we started will
more and more general situations. Note that our most g
scenario and applying it to several situations.
SOLUTION
SET UP As shown in Figure 5.20, we take the positive y direction to
be downward, and we ignore any force associated with buoyancy in the
fluid. The net vertical component of force is mg - Dv².
SOLVE Newton's second law gives
mg - Du = may.
When the object has reached terminal velocity UT, the acceleration ay
is zero and
mg - Du 0
UT =
REFLECT For a skydiver in the spread-eagle position, the value of the
constant D is found experimentally to be about 0.25 kg/m. (Does this
number have the correct units?) If the skydiver's mass is 80 kg, the
terminal-velocity magnitude is
mg
or UT = VD
mg (80 kg) (9.8 m/s²)
0.25 kg/m
VD
This magnitude is about 125 mi/h. Does that seem reasonable?
QUANTITATIVE
= 56 m/s.
Before terminal
velocity: Object
accelerating; drag
force less than
weight
illi
Ø
回饋街口
0ZX%
Dv² < mg
de A
mg
ugh)
Video Tutor Solution
Dv² = mg
mg
At terminal velocity:
Object in equilibrium;
drag force equals
weight
A FIGURE 5.20 An object falling through air, before and after it
reaches its terminal velocity.
Practice Problem: An average-sized cat of 4.5 kg falls with a terminal
velocity of approximately 27 m/s. What is the value of the proportion-
ality constant D for the cat? Answer: 0.062 kg/m.
square
compar
5.4 E
In Sec
are ap
the us
magr
L
show
to ar
posi
com
stre
ob
far
ex
tu
of
W
is
t
C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F28764828-4d1c-44b8-9544-7d79c8b69e6b%2Ffe27b147-47b0-41e5-aee5-1ed565ee71cd%2F5c2dj4i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:EXAMPLE 5.13 Fluid friction
In this example we are going to take a look at a different type of friction, one that is not described by Equa-
tions 5.1 and 5.2. When an object moves through a fluid (such as water or air), it exerts a force on the fluid
to push it out of the way. By Newton's third law, the fluid pushes back on the object, in a direction opposite
to the object's velocity relative to the fluid, always opposing the object's motion and usually increasing with
speed. In high-speed motion through air, the resisting force is approximately proportional to the square of
the object's speed u; it's called a drag force, or simply drag. We can represent its magnitude Fp by
FD = Du²
where D is a proportionality constant that depends on the shape and size of the object and the density of air.
When an object falls vertically through air, the drag force opposing the object's motion increases and
the downward acceleration decreases. Eventually, the object reaches a terminal velocity: Its acceleration
approaches zero and the velocity becomes nearly constant. Derive an expression for the terminal-velocity
magnitude UT in terms of D and the weight mg of the object.
ones as special cases. We employ this strategy throughout the t-starting with a simple
In this chapter, we started will
more and more general situations. Note that our most g
scenario and applying it to several situations.
SOLUTION
SET UP As shown in Figure 5.20, we take the positive y direction to
be downward, and we ignore any force associated with buoyancy in the
fluid. The net vertical component of force is mg - Dv².
SOLVE Newton's second law gives
mg - Du = may.
When the object has reached terminal velocity UT, the acceleration ay
is zero and
mg - Du 0
UT =
REFLECT For a skydiver in the spread-eagle position, the value of the
constant D is found experimentally to be about 0.25 kg/m. (Does this
number have the correct units?) If the skydiver's mass is 80 kg, the
terminal-velocity magnitude is
mg
or UT = VD
mg (80 kg) (9.8 m/s²)
0.25 kg/m
VD
This magnitude is about 125 mi/h. Does that seem reasonable?
QUANTITATIVE
= 56 m/s.
Before terminal
velocity: Object
accelerating; drag
force less than
weight
illi
Ø
回饋街口
0ZX%
Dv² < mg
de A
mg
ugh)
Video Tutor Solution
Dv² = mg
mg
At terminal velocity:
Object in equilibrium;
drag force equals
weight
A FIGURE 5.20 An object falling through air, before and after it
reaches its terminal velocity.
Practice Problem: An average-sized cat of 4.5 kg falls with a terminal
velocity of approximately 27 m/s. What is the value of the proportion-
ality constant D for the cat? Answer: 0.062 kg/m.
square
compar
5.4 E
In Sec
are ap
the us
magr
L
show
to ar
posi
com
stre
ob
far
ex
tu
of
W
is
t
C
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